Module 5: Advanced Mechanics | Year 12 Physics HSC | StudyCave

Pillar 1 of 3

Projectile Motion

Master the art of analysing objects in flight. From horizontal throws to angled launches, understand every aspect of parabolic motion.

Definition of a Projectile

A projectile is any object that moves through the air under the influence of gravity alone, with no thrust, propulsion, or lift forces acting on it after launch. Once released, the only force acting on a projectile (neglecting air resistance) is its weight ( $W = mg$ ), which acts vertically downward.

Key Characteristics:

• No engine or propulsion after launch
• Gravity is the only force (ideal case)
• Horizontal and vertical motions are independent
• Examples: Thrown ball, fired cannonball, kicked football

Practice Question

Q: A rocket fires its engines continuously while moving through the air. Is it a projectile? Explain.

Show Answer

No. A projectile only experiences gravity after launch. Since the rocket has continuous thrust from its engines, it is not a true projectile. The engines provide an additional force beyond gravity, violating the definition.

Vector Independence

The principle of vector independence states that horizontal ( $x$ ) and vertical ( $y$ ) motions of a projectile occur simultaneously but independently. What happens in one dimension does NOT affect the other.

Horizontal Motion (x-axis)

• Acceleration: $a_x = 0 \text{ m/s}^2$
• Velocity: $v_x = \text{constant}$
• Reason: No horizontal forces act

Vertical Motion (y-axis)

• Acceleration: $a_y = -9.8 \text{ m/s}^2$
• Velocity: $v_y = \text{changes}$
• Reason: Gravity acts downward

Mathematical Foundation

Initial velocity $\vec{u}$ at angle $\theta$ resolves into:

Horizontal Component

u_x = u \cos(\theta)

Vertical Component

u_y = u \sin(\theta)

Worked Example

A ball is thrown at $20 \text{ m/s}$ at an angle of $30°$ above the horizontal. Find the initial horizontal and vertical velocity components.

Given:

$u = 20 \text{ m/s}, \theta = 30°$

Find:

$u_x$ and $u_y$

Solution:

$u_x = u \cos \theta$

$u_x = 20 \times \cos 30°$

$u_x = 20 \times 0.866$

$u_x = \mathbf{17.3 \text{ m/s}}$

$u_y = u \sin \theta$

$u_y = 20 \times \sin 30°$

$u_y = 20 \times 0.5$

$u_y = \mathbf{10.0 \text{ m/s}}$

Equations of Motion (SUVAT)

The SUVAT equations apply independently to both horizontal and vertical motion. These five equations relate displacement ( $s$ ), initial velocity ( $u$ ), final velocity ( $v$ ), acceleration ( $a$ ) and time ( $t$ ).

The Three Primary SUVAT Equations

v = u + at

s = ut + \frac{1}{2}at^2

v^2 = u^2 + 2as

Applying SUVAT to Projectile Motion

Horizontal (x-direction)

• $s_x = u_x t$ (since $a_x = 0$ )
• $v_x = u_x$ (constant)
• $u_x = u \cos(\theta)$

Vertical (y-direction)

• $s_y = u_y t - \frac{1}{2}gt^2$
• $v_y = u_y - gt$
• $u_y = u \sin(\theta)$

Worked Example

A ball is thrown at $25 \text{ m/s}$ at $40°$ above horizontal. Calculate: (a) time to reach maximum height, (b) maximum height reached, (c) total time of flight.

Given:

$u = 25 \text{ m/s}, \theta = 40°, g = 9.8 \text{ m/s}^2$

Step 1: Find components

$u_x = 25 \cos(40°) = 19.15 \text{ m/s}$

$u_y = 25 \sin(40°) = 16.07 \text{ m/s}$

(a) Time to max height:

At max height, $v_y = 0$

Using $v = u + at$ :

$0 = 16.07 - 9.8t$

$t = \frac{16.07}{9.8}$

$t = \mathbf{1.64 \text{ s}}$

(b) Maximum height:

Using $v^2 = u^2 + 2as$ :

$0^2 = 16.07^2 - 2(9.8)s_y$

$s_y = \frac{258.2}{19.6}$

$s_y = \mathbf{13.2 \text{ m}}$

(c) Total flight time:

By symmetry:

$t_{\text{total}} = 2 \times t_{\text{up}}$

$t_{\text{total}} = 2 \times 1.64$

$t_{\text{total}} = \mathbf{3.28 \text{ s}}$

Practice Question

Q: A projectile is launched at $30 \text{ m/s}$ at $53°$ above horizontal. What is its vertical velocity after 2 seconds?

Show Answer

Solution:

$u_y = 30 \sin(53°)$

$u_y = 30 \times 0.8$

$u_y = 24 \text{ m/s}$

Using $v = u + at$ :

$v_y = 24 - 9.8(2)$

$v_y = 24 - 19.6$

$v_y = \mathbf{4.4 \text{ m/s upward}}$

The positive value indicates the projectile is still moving upward (hasn't reached peak yet).

Complementary Launch Angles

Two launch angles are complementary if they add up to $90°$ . A remarkable property: complementary angles (like $30°$ and $60°$ ) produce the same range when launched with the same initial speed.

Key Insight:

Angles $\theta$ and $(90° - \theta)$ give identical range because $\sin(2\theta) = \sin(180° - 2\theta)$ .

Mathematical Proof

The range formula is:

R = \frac{u^2 \sin(2\theta)}{g}

For complementary angles:

R_1 = \frac{u^2 \sin(2\theta)}{g}

R_2 = \frac{u^2 \sin(2(90° - \theta))}{g} = \frac{u^2 \sin(180° - 2\theta)}{g}

Since $\sin(180° - x) = \sin(x)$ :

R_2 = \frac{u^2 \sin(2\theta)}{g} = R_1 \,\checkmark

Worked Example

A projectile is launched at $20 \text{ m/s}$ at $30°$ . What other angle gives the same range? Calculate the range for both. ( $g = 9.8 \text{ m/s}^2$ )

Complementary angle:

$90° - 30° = \mathbf{60°}$

Range at 30°:

$R = \frac{u^2 \sin(2\theta)}{g}$

$R = \frac{20^2 \times \sin(60°)}{9.8}$

$R = \frac{400 \times 0.866}{9.8}$

$R = \mathbf{35.3 \text{ m}}$

Range at 60°:

$R = \frac{20^2 \times \sin(120°)}{9.8}$

$R = \frac{400 \times 0.866}{9.8}$

$R = \mathbf{35.3 \text{ m}} \,\checkmark$

Both angles produce identical range, as expected.

Practice Question

Q: Two projectiles are launched with the same speed. One at 25° travels 80 m. At what angle should the second be launched to also travel 80 m? Will their flight times be the same?

Show Answer

Angle: $90° - 25° = \mathbf{65°}$

Flight times: No, they will be different. The 65° launch has a larger vertical component ( $\sin(65°) > \sin(25°)$ ), so it reaches a greater maximum height and stays airborne longer. Although they travel the same horizontal distance, the 65° projectile has more "hang time."

Maximum Range (45° Optimal Angle)

For a projectile launched from level ground, the maximum range occurs at a launch angle of 45°.

Mathematical Proof

Given: $R = \frac{u^2 \sin(2\theta)}{g}$

To maximize: Find $\theta$ where $\frac{dR}{d\theta} = 0$

R = \frac{u^2}{g} \times \sin(2\theta)

Taking derivative:

\frac{dR}{d\theta} = \frac{u^2}{g} \times 2\cos(2\theta)

Set equal to zero:

\cos(2\theta) = 0 \implies 2\theta = 90° \implies \mathbf{\theta = 45°}

Alternative explanation: $\sin(2\theta)$ reaches its maximum value of 1 when $2\theta = 90°$ , giving $\theta = 45°$ .

Physical Interpretation:

45° represents the optimal balance between horizontal and vertical velocity components. At 45°: $u_x = u_y = \frac{u}{\sqrt{2}}$

Worked Example

A ball can be thrown at 15 m/s. Calculate (a) maximum possible range, (b) range at 30°, (c) percentage of maximum range achieved at 30°. ( $g = 9.8 \text{ m/s}^2$ )

(a) Maximum range (at 45°):

$R_{\max} = \frac{u^2 \sin(90°)}{g}$

$R_{\max} = \frac{15^2 \times 1}{9.8}$

$R_{\max} = \frac{225}{9.8}$

$R_{\max} = \mathbf{23.0 \text{ m}}$

(b) Range at 30°:

$R_{30} = \frac{15^2 \times \sin(60°)}{9.8}$

$R_{30} = \frac{225 \times 0.866}{9.8}$

$R_{30} = \mathbf{19.9 \text{ m}}$

(c) Percentage:

$\frac{19.9}{23.0} \times 100\%$

$= \mathbf{86.5\%}$

Practice Question

Q: A projectile launched at 40 m/s achieves a range of 120 m. Is this the maximum possible range for this launch speed? If not, what is the maximum? ( $g = 9.8 \text{ m/s}^2$ )

Show Answer

No, this is not maximum.

Maximum range:

$R_{\max} = \frac{u^2}{g}$

$R_{\max} = \frac{40^2}{9.8}$

$R_{\max} = \frac{1600}{9.8}$

$R_{\max} = \mathbf{163 \text{ m}}$

The actual range of 120 m is only 74% of the maximum possible (120/163 = 0.74). This means the launch angle was not 45°.

Trajectory Equation

The trajectory equation describes the path as a function $y(x)$ , producing a parabolic equation.

Derivation

Step 1: Write SUVAT equations

x = (u \cos \theta)t

y = (u \sin \theta)t - \frac{1}{2}gt^2

Step 2: Eliminate time

t = \frac{x}{u \cos \theta}

Step 3: Substitute into y equation

y = (u \sin \theta) \cdot \frac{x}{u \cos \theta} - \frac{1}{2}g\left(\frac{x}{u \cos \theta}\right)^2

Step 4: Simplify to final equation

y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}

This is a parabola in the form $y = ax + bx^2$

Worked Example

A ball is thrown from ground level at 20 m/s at 30° above horizontal. Find the height when the ball has traveled 10 m horizontally. ( $g = 9.8 \text{ m/s}^2, \tan 30° = 0.577, \cos 30° = 0.866$ )

Given:

$u = 20 \text{ m/s}, \theta = 30°, x = 10 \text{ m}$

Using trajectory equation:

$y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2\theta}$

$y = 10(0.577) - \frac{9.8 \times 10^2}{2 \times 20^2 \times 0.866^2}$

$y = 5.77 - \frac{980}{2 \times 400 \times 0.75}$

$y = 5.77 - 1.63$

$y = \mathbf{4.1 \text{ m}}$

Practice Question

Q: Show that the trajectory equation $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$ is indeed a parabola by identifying it in the standard form $y = ax + bx^2$ .

Show Answer

Comparison with $y = ax + bx^2$ :

Coefficient of x: $a = \tan \theta$ (constant)

Coefficient of $x^2$ : $b = -\frac{g}{2u^2 \cos^2\theta}$ (constant, negative)

Since both coefficients are constants (for given launch conditions), this is a quadratic function in x. The negative coefficient of $x^2$ confirms the parabola opens downward, matching the physical trajectory.

Air Resistance (Qualitative Effects)

In reality, projectiles experience air resistance (drag force), which opposes motion. While HSC Physics doesn't require quantitative drag calculations, you must understand the qualitative effects.

Without Air Resistance

• Symmetric parabola
• Maximum range
• Maximum peak height
• Launch angle = landing angle

With Air Resistance

• Asymmetric (steeper descent)
• Reduced range
• Reduced peak height
• Landing angle > launch angle

Why These Effects Occur:

Drag force opposes velocity, removing kinetic energy. Horizontal velocity decreases (unlike ideal projectiles). On ascent, both gravity and drag act downward. On descent, gravity pulls down but drag opposes downward velocity.

Practice Question

Q: A projectile is launched at 45° with initial speed 30 m/s. Sketch two trajectories on the same axes: one neglecting air resistance, one including air resistance. Label key differences.

Show Answer

Sketch should show:

• Ideal trajectory: symmetric parabola
• Real trajectory with drag: lower peak height, shorter range, asymmetric (steeper on descent)
• Both trajectories start at same point and angle
• Real trajectory lands at steeper angle than launch
• Ideal trajectory has launch angle = landing angle (45°)

Non-Level Ground (Cliff Launch)

When a projectile is launched from an elevated position, the trajectory is no longer symmetric. These problems require careful application of quadratic displacement equations.

Key Strategy:

• Define coordinate system (origin at launch, +y upward)
• Final $s_y$ is negative if landing below launch
• Use $s_y = u_y t - \frac{1}{2}gt^2$ and solve for time
• Calculate range using $x = u_x t$

Worked Example

A ball is thrown horizontally at $15 \text{ m/s}$ from a cliff $80 \text{ m}$ high. Calculate: (a) time to hit ground, (b) horizontal distance.

Given:

$u = 15 \text{ m/s}$ horizontal, $h = 80 \text{ m}, \theta = 0°$

$u_x = 15 \text{ m/s}, u_y = 0, s_y = -80 \text{ m}$

(a) Time of flight:

$s_y = u_y t - \frac{1}{2}gt^2$

$-80 = 0 - \frac{1}{2}(9.8)t^2$

$t^2 = \frac{160}{9.8} = 16.3$

$t = \mathbf{4.04 \text{ s}}$

(b) Horizontal range:

$R = u_x t$

$R = 15 \times 4.04$

$R = \mathbf{60.6 \text{ m}}$

Pillar 2 of 3

Circular Motion

Understand the physics of rotation. From banked curves to vertical loops, master centripetal forces and torque.

Uniform Circular Motion

Uniform Circular Motion (UCM) occurs when an object moves in a circular path at constant speed. Although the speed is constant, the velocity is changing because direction is continuously changing. This change in velocity means the object is accelerating.

Key Characteristics:

• Constant speed: $|v| = \text{constant}$
• Changing velocity (direction changes continuously)
• Centripetal acceleration always directed toward center
• Period ( $T$ ): Time for one complete revolution
• Frequency ( $f$ ): Number of revolutions per second, $f = 1/T$

Fundamental Relationships

Speed

v = \frac{2\pi r}{T}

Angular velocity

\omega = \frac{2\pi}{T} = 2\pi f

Linear and Angular Speed Relationship

v = r\omega

Worked Example

A satellite orbits Earth at a radius of $7.0 \times 10^6 \text{ m}$ with a period of 90 minutes. Calculate: (a) its orbital speed, (b) its angular velocity.

Given:

$r = 7.0 \times 10^6 \text{ m}, \quad T = 90 \text{ min} = 5400 \text{ s}$

(a) Orbital speed:

$v = \frac{2\pi r}{T}$

$v = \frac{2\pi \times 7.0 \times 10^6}{5400}$

$v = \mathbf{8140 \text{ m/s}}$

(b) Angular velocity:

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{5400}$

$\omega = \mathbf{1.16 \times 10^{-3} \text{ rad/s}}$

Practice Question

Q: A car tire has a radius of 0.30 m and rotates at 10 revolutions per second. Calculate the speed of a point on the tire's edge.

Show Answer

Solution:

$f = 10 \text{ Hz}, \quad T = \frac{1}{f} = 0.1 \text{ s}$

$v = \frac{2\pi r}{T} = \frac{2\pi \times 0.30}{0.1}$

$v = \mathbf{18.8 \text{ m/s}}$

Alternatively: $\omega = 2\pi f = 20\pi \text{ rad/s}$ , then $v = r\omega = 18.8 \text{ m/s}$

Centripetal Calculations

An object in uniform circular motion experiences centripetal acceleration directed toward the center of the circle. This acceleration requires a net inward force.

Key Equations

Centripetal Acceleration:

a_c = \frac{v^2}{r} = \omega^2 r = \frac{4\pi^2 r}{T^2}

Centripetal Force:

F_c = ma_c = \frac{mv^2}{r} = m\omega^2 r

Critical Understanding: $F_c$ is a net force provided by physical forces (tension, friction, gravity, normal force), NOT a force on its own.

Worked Example

A 1500 kg car travels around a circular track of radius 80 m at 25 m/s. Calculate: (a) the centripetal acceleration, (b) the centripetal force required.

Given:

$m = 1500 \text{ kg}, \quad r = 80 \text{ m}, \quad v = 25 \text{ m/s}$

(a) Centripetal acceleration:

$a_c = \frac{v^2}{r}$

$a_c = \frac{25^2}{80}$

$a_c = \mathbf{7.81 \text{ m/s}^2}$

(b) Centripetal force:

$F_c = ma_c = \frac{mv^2}{r}$

$F_c = \frac{1500 \times 25^2}{80}$

$F_c = \mathbf{11,720 \text{ N}}$

This force is provided by friction between the tires and the road.

Practice Question

Q: A 0.50 kg ball is attached to a 1.2 m string and whirled in a horizontal circle at 4.0 m/s. Calculate the tension in the string.

Show Answer

Solution:

Tension provides the centripetal force:

$T = F_c = \frac{mv^2}{r}$

$T = \frac{0.50 \times 4.0^2}{1.2}$

$T = \mathbf{6.67 \text{ N}}$

Horizontal Circles (Friction on Flat Tracks)

For objects moving in horizontal circles on flat surfaces, friction provides the centripetal force.

Car on Flat Turn

Friction provides centripetal force:

f = \frac{mv^2}{r}

Maximum friction available:

f_{\max} = \mu N = \mu mg

Maximum Speed Before Skidding

Set friction equal to required centripetal force:

\mu mg = \frac{mv^2_{\max}}{r}

Solve for maximum speed:

v_{\max} = \sqrt{\mu gr}

Worked Example

A car travels around a flat circular track of radius 50 m. The coefficient of friction between the tires and road is 0.60. Calculate the maximum speed before the car skids.

Given:

$r = 50 \text{ m}, \quad \mu = 0.60, \quad g = 9.8 \text{ m/s}^2$

Solution:

At maximum speed, friction equals required centripetal force:

$\mu mg = \frac{mv^2_{\max}}{r}$

$v_{\max} = \sqrt{\mu gr}$

$v_{\max} = \sqrt{0.60 \times 9.8 \times 50}$

$v_{\max} = \mathbf{17.1 \text{ m/s}}$

Practice Question

Q: A 0.20 kg object sits on a turntable 0.15 m from the center. The coefficient of static friction is 0.25. What is the maximum angular velocity before the object slides off?

Show Answer

Solution:

Maximum friction: $f_{\max} = \mu mg = 0.25 \times 0.20 \times 9.8 = 0.49 \text{ N}$

This must equal centripetal force: $f_{\max} = m\omega^2 r$

$0.49 = 0.20 \times \omega^2 \times 0.15$

$\omega^2 = \frac{0.49}{0.03} = 16.3$

$\omega = \mathbf{4.04 \text{ rad/s}}$

Alternatively: $\omega_{\max} = \sqrt{\frac{\mu g}{r}} = \sqrt{\frac{0.25 \times 9.8}{0.15}} = 4.04 \text{ rad/s}$

Conical Pendulum

A conical pendulum consists of a mass attached to a string, moving in a horizontal circle while the string makes an angle $\theta$ with the vertical.

Free Body Diagram Analysis

Forces acting:

• Tension $T$ at angle $\theta$ to vertical
• Weight $mg$ acting downward

Resolve tension components:

Vertical component:

T_y = T\cos\theta

Horizontal component:

T_x = T\sin\theta

Apply Newton's Laws:

Vertical equilibrium (no vertical acceleration):

T\cos\theta = mg

Horizontal (provides centripetal force):

T\sin\theta = \frac{mv^2}{r}

Derive speed equation:

Divide the horizontal by vertical equation:

\frac{T\sin\theta}{T\cos\theta} = \frac{mv^2/r}{mg}

\tan\theta = \frac{v^2}{rg}

v = \sqrt{rg\tan\theta}

Worked Example

A 0.80 kg mass is attached to a 1.5 m string and swings in a horizontal circle with the string making a 30° angle with the vertical. Calculate: (a) the tension in the string, (b) the speed of the mass, (c) the radius of the circle.

Given:

$m = 0.80 \text{ kg}, \quad L = 1.5 \text{ m}, \quad \theta = 30°$

(a) Tension:

From vertical equilibrium:

$T\cos\theta = mg$

$T = \frac{mg}{\cos\theta} = \frac{0.80 \times 9.8}{\cos 30°}$

$T = \frac{7.84}{0.866} = \mathbf{9.05 \text{ N}}$

$r = L\sin\theta = 1.5 \times \sin 30°$

$r = \mathbf{0.75 \text{ m}}$

(b) Speed:

$v = \sqrt{rg\tan\theta}$

$v = \sqrt{0.75 \times 9.8 \times \tan 30°}$

$v = \sqrt{0.75 \times 9.8 \times 0.577} = \mathbf{2.06 \text{ m/s}}$

Practice Question

Q: A conical pendulum with a 2.0 m string makes a 45° angle with the vertical. Calculate the speed of the mass.

Show Answer

Solution:

Radius: $r = L\sin\theta = 2.0 \times \sin 45° = 1.414 \text{ m}$

Speed: $v = \sqrt{rg\tan\theta}$

$v = \sqrt{1.414 \times 9.8 \times \tan 45°}$

$v = \sqrt{1.414 \times 9.8 \times 1} = \mathbf{3.72 \text{ m/s}}$

Banked Curves (Ideal - Frictionless)

A banked curve is a road or track that is tilted at an angle $\theta$ to the horizontal. In the ideal (frictionless) case, the normal force alone provides the necessary centripetal force for circular motion at a specific design speed.

Key Insight:

At the design speed $v_{\text{design}}$ , a vehicle can safely navigate the curve without relying on friction. This design speed depends only on the banking angle, radius, and gravity - not on the mass of the vehicle.

Derivation of Design Speed

Step 1: Draw the Free Body Diagram

• Normal force $N$ acts perpendicular to the banked surface (at angle $\theta$ to vertical)
• Weight $mg$ acts vertically downward
• No friction force (ideal case)

Step 2: Resolve Normal Force Components

Vertical component:

N_y = N\cos\theta

Horizontal component (toward center):

N_x = N\sin\theta

Step 3: Apply Newton's Second Law

Vertical direction (no vertical acceleration):

N\cos\theta = mg

Horizontal direction (centripetal acceleration):

N\sin\theta = \frac{mv^2}{r}

Step 4: Eliminate N and Solve for v

Divide the horizontal equation by the vertical equation:

\frac{N\sin\theta}{N\cos\theta} = \frac{mv^2/r}{mg}

\tan\theta = \frac{v^2}{rg}

Rearranging for the design speed:

v_{\text{design}} = \sqrt{rg\tan\theta}

Worked Example

A highway curve has a radius of 120 m and is banked at 15° to the horizontal. Calculate: (a) the design speed for this curve, (b) verify that a 1500 kg car and a 3000 kg truck both have the same design speed.

Given:

$r = 120 \text{ m}, \quad \theta = 15°, \quad g = 9.8 \text{ m/s}^2$

(a) Design speed:

$v_{\text{design}} = \sqrt{rg\tan\theta}$

$v_{\text{design}} = \sqrt{120 \times 9.8 \times \tan 15°}$

$v_{\text{design}} = \sqrt{120 \times 9.8 \times 0.268}$

$v_{\text{design}} = \sqrt{315}$

$v_{\text{design}} = \mathbf{17.7 \text{ m/s}}$ (approximately $64 \text{ km/h}$ )

(b) Mass independence verification:

For 1500 kg car: $v = \sqrt{rg\tan\theta} = 17.7 \text{ m/s}$ (mass doesn't appear in formula)

For 3000 kg truck: $v = \sqrt{rg\tan\theta} = 17.7 \text{ m/s}$ (same result)

Both vehicles have the same design speed because mass cancels in the derivation.

Practice Question

Q: A racetrack curve has a radius of 200 m and is designed for a speed of 25 m/s. At what angle should the curve be banked? ( $g = 9.8 \text{ m/s}^2$ )

Show Answer

Solution:

Using $\tan\theta = \frac{v^2}{rg}$ :

$\tan\theta = \frac{25^2}{200 \times 9.8}$

$\tan\theta = \frac{625}{1960} = 0.319$

$\theta = \arctan(0.319) = \mathbf{17.7°}$

The curve should be banked at approximately 18° to the horizontal.

Banked Curves (with Friction)

In reality, banked curves have friction between the tires and road. When a vehicle travels at a speed different from the design speed, friction must act to prevent sliding. The direction of friction depends on whether the vehicle is traveling faster or slower than the design speed.

Key Concept - Friction Direction:

Speed Greater Than Design Speed

$v > v_{\text{design}}$

• Vehicle needs more centripetal force
• Friction acts down the slope (toward center)
• Friction helps Normal force provide centripetal force
• Without friction, vehicle would slide outward

Speed Less Than Design Speed

$v < v_{\text{design}}$

• Vehicle needs less centripetal force
• Friction acts up the slope (away from center)
• Friction opposes Normal force component
• Without friction, vehicle would slide inward/downward

Force Analysis with Friction

Forces on banked curve with friction:

• Normal force $N$ perpendicular to surface
• Weight $mg$ vertically downward
• Friction force $f$ parallel to surface (direction varies)

Case 1: Speed Greater Than Design Speed ( $v > v_{\text{design}}$ )

Friction acts down the slope. Horizontal centripetal force equation:

N\sin\theta + f\cos\theta = \frac{mv^2}{r}

Vertical equilibrium:

N\cos\theta = mg + f\sin\theta

Case 2: Speed Less Than Design Speed ( $v < v_{\text{design}}$ )

Friction acts up the slope. Horizontal centripetal force equation:

N\sin\theta - f\cos\theta = \frac{mv^2}{r}

Vertical equilibrium:

N\cos\theta + f\sin\theta = mg

Determining Friction Direction - Simple Rule:

Imagine the curve with no friction. If $v > v_{\text{design}}$ , the vehicle would slide outward and up the slope. Therefore, friction must act down the slope to prevent this. If $v < v_{\text{design}}$ , the vehicle would slide inward and down the slope, so friction acts up the slope.

Worked Example

A banked curve has a radius of 100 m and is banked at 20° to the horizontal. The design speed is 18 m/s. A car travels around the curve at 25 m/s. (a) In which direction does friction act? (b) Explain why friction is needed.

Given:

$r = 100 \text{ m}, \quad \theta = 20°, \quad v_{\text{design}} = 18 \text{ m/s}, \quad v = 25 \text{ m/s}$

(a) Friction direction:

Compare speeds: $v = 25 \text{ m/s} > v_{\text{design}} = 18 \text{ m/s}$

Therefore: Friction acts DOWN the slope (toward the center of the curve).

(b) Explanation:

At the design speed (18 m/s), the horizontal component of the Normal force alone provides exactly the needed centripetal force:

$F_{c,\text{design}} = \frac{m(18)^2}{100} = 3.24m \text{ N}$

At the actual speed (25 m/s), the required centripetal force is:

$F_{c,\text{actual}} = \frac{m(25)^2}{100} = 6.25m \text{ N}$

The Normal force component alone is insufficient ( $3.24m < 6.25m$ ). Friction must act down the slope to provide the additional inward force needed to keep the car in circular motion. Without friction, the car would slide outward.

Practice Question

Q: A banked curve is designed for 30 m/s. A nervous driver navigates the curve at only 15 m/s. (a) In which direction does friction act on the car's tires? (b) What would happen if the road were icy (frictionless)?

Show Answer

(a) Friction direction:

$v = 15 \text{ m/s} < v_{\text{design}} = 30 \text{ m/s}$

Friction acts UP the slope (away from the center).

(b) What happens on icy road:

At 15 m/s, the required centripetal force is less than what the Normal force component provides at the design banking angle.

Without friction, the car would slide down and inward toward the center of the curve, because the Normal force's horizontal component is too large for the slow speed.

The car cannot maintain the circular path at the banked radius - it would spiral inward and down the slope.

Vertical Loops

A vertical loop (or vertical circle) occurs when an object travels in a circular path in a vertical plane. Common examples include roller coasters and stunt planes. The key challenge is that gravity's effect changes at different points around the loop.

Key Insight:

The normal force varies around the loop. At the bottom, you feel heaviest (maximum Normal force). At the top, you feel lightest (minimum Normal force). The minimum speed at the top occurs when the Normal force just reaches zero.

Force Analysis at Key Points

At the TOP of the loop:

• Both Normal force $N$ and Weight $mg$ point toward center (downward)
• Both forces contribute to centripetal force

Centripetal force equation at top:

F_c = N_{\text{top}} + mg = \frac{mv_{\text{top}}^2}{r}

Minimum speed condition: When $N_{\text{top}} = 0$ (barely maintains contact):

mg = \frac{mv_{\text{min}}^2}{r} \implies v_{\text{min}} = \sqrt{gr}

At the BOTTOM of the loop:

• Normal force $N$ points toward center (upward)
• Weight $mg$ points away from center (downward)
• Normal force must overcome weight AND provide centripetal force

Centripetal force equation at bottom:

F_c = N_{\text{bottom}} - mg = \frac{mv_{\text{bottom}}^2}{r}

Rearranging for Normal force:

N_{\text{bottom}} = mg + \frac{mv_{\text{bottom}}^2}{r}

This shows $N_{\text{bottom}} > N_{\text{top}}$ - you feel heaviest at the bottom!

Using Conservation of Energy:

To find the speed at different points in the loop, use conservation of mechanical energy. Taking the bottom as the reference level ( $h = 0$ ):

Energy at bottom = Energy at top:

\frac{1}{2}mv_{\text{bottom}}^2 = \frac{1}{2}mv_{\text{top}}^2 + mg(2r)

Simplifying:

v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gr

Worked Example

A roller coaster car travels around a vertical loop of radius 8.0 m. Calculate: (a) the minimum speed at the top for the car to maintain contact with the track, (b) the speed at the bottom if the car has the minimum speed at the top, (c) the Normal force on a 60 kg passenger at the bottom.

Given:

$r = 8.0 \text{ m}, \quad m = 60 \text{ kg}, \quad g = 9.8 \text{ m/s}^2$

(a) Minimum speed at top:

At minimum speed, $N = 0$ , so:

$v_{\text{min}} = \sqrt{gr} = \sqrt{9.8 \times 8.0} = \sqrt{78.4}$

$v_{\text{min}} = \mathbf{8.85 \text{ m/s}}$

(b) Speed at bottom:

Using energy conservation:

$v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gr = (8.85)^2 + 4(9.8)(8.0)$

$v_{\text{bottom}}^2 = 78.3 + 313.6 = 391.9$

$v_{\text{bottom}} = \sqrt{391.9} = \mathbf{19.8 \text{ m/s}}$

(c) Normal force at bottom:

$N_{\text{bottom}} = mg + \frac{mv_{\text{bottom}}^2}{r}$

$N_{\text{bottom}} = 60(9.8) + \frac{60(19.8)^2}{8.0}$

$N_{\text{bottom}} = 588 + \frac{60(391.9)}{8.0}$

$N_{\text{bottom}} = 588 + 2939 = \mathbf{3527 \text{ N}}$

The passenger experiences a force 6 times their weight!

Practice Question

Q: A 0.50 kg ball on a 1.2 m string is swung in a vertical circle. At the top of the circle, the tension in the string is 2.0 N. Calculate: (a) the speed of the ball at the top, (b) the speed at the bottom using energy conservation.

Show Answer

(a) Speed at top:

At the top: $T + mg = \frac{mv_{\text{top}}^2}{r}$

$2.0 + 0.50(9.8) = \frac{0.50 \times v_{\text{top}}^2}{1.2}$

$2.0 + 4.9 = \frac{0.50 \times v_{\text{top}}^2}{1.2}$

$6.9 = \frac{0.50 \times v_{\text{top}}^2}{1.2}$

$v_{\text{top}}^2 = \frac{6.9 \times 1.2}{0.50} = 16.56$

$v_{\text{top}} = \mathbf{4.07 \text{ m/s}}$

(b) Speed at bottom:

$v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gr$

$v_{\text{bottom}}^2 = 16.56 + 4(9.8)(1.2)$

$v_{\text{bottom}}^2 = 16.56 + 47.04 = 63.6$

$v_{\text{bottom}} = \sqrt{63.6} = \mathbf{7.98 \text{ m/s}}$

Torque

Torque (symbol $\tau$ , Greek letter tau) is the rotational equivalent of force. It measures the turning effect of a force about a pivot point or axis of rotation.

Definition and Formula

Torque is defined as:

\tau = rF\sin\theta

Where:

• $\tau$ = torque (N·m)
• $r$ = distance from pivot to point where force is applied (m)
• $F$ = magnitude of applied force (N)
• $\theta$ = angle between position vector $\vec{r}$ and force vector $\vec{F}$

Alternative form using lever arm:

\tau = F \times d_{\perp}

Where $d_{\perp} = r\sin\theta$ is the perpendicular distance (lever arm) from the pivot to the line of action of the force.

Key Concepts:

• Direction: Torque is positive for counterclockwise rotation, negative for clockwise
• Maximum torque: Occurs when $\theta = 90°$ (force perpendicular to position vector)
• Zero torque: Occurs when $\theta = 0°$ or $180°$ (force parallel to position vector)
• Units: Newton-meters (N·m), NOT Joules (though dimensionally equivalent)

Identifying the Lever Arm

The lever arm is the shortest distance from the pivot to the line of action of the force.

Strategy:

1. Identify the pivot point
2. Draw the line of action of the force
3. Draw perpendicular from pivot to this line
4. This perpendicular distance is $d_{\perp}$

Sign Convention

Positive torque: Causes counterclockwise rotation (viewed from above)

Negative torque: Causes clockwise rotation

Always establish your convention clearly in problems!

Worked Example

A force of 50 N is applied to a door handle 0.80 m from the hinge. Calculate the torque when: (a) the force is applied perpendicular to the door, (b) the force is applied at 60° to the door's surface.

Given:

$F = 50 \text{ N}, \quad r = 0.80 \text{ m}$

(a) Force perpendicular to door ( $\theta = 90°$ ):

$\tau = rF\sin\theta$

$\tau = 0.80 \times 50 \times \sin(90°)$

$\tau = 0.80 \times 50 \times 1$

$\tau = \mathbf{40 \text{ Nm}}$

This is the maximum possible torque for this force and distance.

(b) Force at 60° to door surface:

$\tau = rF\sin\theta$

$\tau = 0.80 \times 50 \times \sin(60°)$

$\tau = 0.80 \times 50 \times 0.866$

$\tau = \mathbf{34.6 \text{ Nm}}$

The torque is less because the force is not fully perpendicular.

Practice Question

Q: A wrench handle is 0.25 m long. A mechanic applies a 100 N force at the end of the wrench. (a) What is the maximum torque? (b) At what angle must the force be applied to produce a torque of 20 N·m?

Show Answer

(a) Maximum torque:

Maximum occurs when $\theta = 90°$ :

$\tau_{\max} = rF = 0.25 \times 100 = \mathbf{25 \text{ Nm}}$

(b) Angle for 20 N·m:

$\tau = rF\sin\theta$

$20 = 0.25 \times 100 \times \sin\theta$

$20 = 25\sin\theta$

$\sin\theta = 0.8$

$\theta = \arcsin(0.8) = \mathbf{53.1°}$

Static Equilibrium

An object is in static equilibrium when it is at rest and remains at rest. This requires two conditions to be satisfied simultaneously: no net force AND no net torque.

Conditions for Static Equilibrium

Condition 1: Translational Equilibrium

The sum of all forces in each direction must be zero:

\sum F_x = 0

\sum F_y = 0

This prevents linear acceleration - the object won't start moving.

Condition 2: Rotational Equilibrium

The sum of all torques about any pivot point must be zero:

\sum \tau = 0

This prevents angular acceleration - the object won't start rotating.

Problem-Solving Strategy:

1. Draw a clear free body diagram showing all forces
2. Choose a convenient pivot point (often where an unknown force acts to eliminate it from torque equation)
3. Establish sign conventions for forces and torques
4. Apply $\sum F_x = 0$ and $\sum F_y = 0$
5. Apply $\sum \tau = 0$ about your chosen pivot
6. Solve the system of equations for unknown forces/distances

Choosing the Pivot Point

The choice of pivot is arbitrary - torques sum to zero about ANY point for an object in equilibrium.

Strategic choice: Choose the pivot where an unknown force acts. This force produces zero torque ( $r = 0$ ), eliminating it from the torque equation and simplifying calculations.

Worked Example

A uniform beam of length 4.0 m and mass 20 kg is supported at its left end by a hinge and at a point 3.0 m from the left by a vertical support. A 50 kg mass hangs from the right end. Calculate the force exerted by the vertical support. ( $g = 9.8 \text{ m/s}^2$ )

Given:

$L = 4.0 \text{ m}, \quad m_{\text{beam}} = 20 \text{ kg}, \quad m_{\text{load}} = 50 \text{ kg}, \quad d_{\text{support}} = 3.0 \text{ m}$

Forces and their positions:

• Weight of beam: $W_{\text{beam}} = 20 \times 9.8 = 196 \text{ N}$ at center (2.0 m from pivot)

• Weight of load: $W_{\text{load}} = 50 \times 9.8 = 490 \text{ N}$ at 4.0 m from pivot

• Support force: $F_{\text{support}}$ (upward) at 3.0 m from pivot

Apply $\sum \tau = 0$ about left end (counterclockwise positive):

$+F_{\text{support}}(3.0) - W_{\text{beam}}(2.0) - W_{\text{load}}(4.0) = 0$

$F_{\text{support}}(3.0) - 196(2.0) - 490(4.0) = 0$

$F_{\text{support}}(3.0) - 392 - 1960 = 0$

$3.0 F_{\text{support}} = 2352$

$F_{\text{support}} = \frac{2352}{3.0}$

$F_{\text{support}} = \mathbf{784 \text{ N}}$

The support must provide 784 N upward to maintain equilibrium.

Practice Question

Q: A 3.0 m uniform ladder of mass 15 kg leans against a frictionless wall at an angle of 60° to the horizontal. The ladder's base rests on the ground 1.5 m from the wall. Calculate: (a) the normal force from the wall, (b) the normal force from the ground.

Show Answer

Setup:

Choose pivot at base of ladder.

Height of wall contact: $h = 1.5\tan(60°) = 2.598 \text{ m}$

Weight acts at center: 1.5 m horizontally from base, $0.75 \text{ m}$ horizontal distance to pivot

(a) Wall normal force $N_W$ :

Apply $\sum \tau = 0$ about base:

$N_W(2.598) - (15 \times 9.8)(0.75) = 0$ (Clockwise torque from weight vs CCW from Wall)

$2.598 N_W = 110.25$

$N_W = \frac{110.25}{2.598} = \mathbf{42.4 \text{ N}}$

(b) Ground normal force:

Apply $\sum F_y = 0$ :

$N_G - mg = 0$

$N_G = 15(9.8) = \mathbf{147 \text{ N}}$

Pillar 3 of 3

Gravitational Motion

Explore the universal force that governs planets, satellites, and orbital mechanics through Newton's laws and Kepler's principles.

Universal Gravitation

Newton's Law of Universal Gravitation states that every point mass attracts every other point mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Newton's Universal Law

F = \frac{GMm}{r^2}

Where:

$F$ = gravitational force between the masses (N)
$G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$ (universal gravitational constant)
$M$ and $m$ = masses of the two objects (kg)
$r$ = distance between their centers of mass (m)

Key Properties of Gravitational Force:

Always Attractive

Gravitational force is always attractive, never repulsive. It acts along the line joining the centers of the two masses, pulling them toward each other.

Action-Reaction Pair

By Newton's Third Law, the force on mass A due to mass B equals the force on mass B due to mass A: $F_{AB} = -F_{BA}$ . They are equal in magnitude but opposite in direction.

Inverse Square Law

Force decreases with the square of distance. If you double the distance $(r \to 2r)$ , the force becomes $1/4$ of the original. If you triple the distance $(r \to 3r)$ , force becomes $1/9$ .

Universal Application

This law applies to all masses in the universe, from falling apples to orbiting planets to distant galaxies. The same equation governs all gravitational interactions.

Worked Example

Two masses of 50 kg and 100 kg are placed 2.0 m apart. Calculate the gravitational force between them. Then determine what happens to this force if the distance is doubled to 4.0 m.

Given:

$m_1 = 50 \text{ kg}, \quad m_2 = 100 \text{ kg}, \quad r = 2.0 \text{ m}$

Step 1: Apply Newton's Universal Law

$F = \frac{GMm}{r^2} = \frac{(6.67 \times 10^{-11})(50)(100)}{(2.0)^2}$

$F = \frac{3.335 \times 10^{-7}}{4.0} = 8.34 \times 10^{-8} \text{ N}$

Step 2: Apply inverse square law when distance doubles

When $r \to 2r$ , the new force becomes:

$F_{\text{new}} = \frac{GMm}{(2r)^2} = \frac{GMm}{4r^2} = \frac{F}{4}$

$F_{\text{new}} = \frac{8.34 \times 10^{-8}}{4} = \mathbf{2.09 \times 10^{-8} \text{ N}}$

Original force: $8.34 \times 10^{-8} \text{ N}$

Force at double distance: $2.09 \times 10^{-8} \text{ N}$ (exactly 1/4 of original)

Practice Question

The Earth has a mass of $6.0 \times 10^{24} \text{ kg}$ and the Moon has a mass of $7.35 \times 10^{22} \text{ kg}$ . They are separated by an average distance of $3.84 \times 10^8 \text{ m}$ . Calculate the gravitational force between Earth and Moon. If Earth were three times further from the Moon, what would the new force be?

Show Answer

Solution:

$F = \frac{GMm}{r^2} = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(7.35 \times 10^{22})}{(3.84 \times 10^8)^2}$

$F = \frac{2.94 \times 10^{37}}{1.475 \times 10^{17}} = \mathbf{1.99 \times 10^{20} \text{ N}}$

Force at triple distance:

When $r \to 3r$ : $F_{\text{new}} = \frac{GMm}{(3r)^2} = \frac{F}{9}$

$F_{\text{new}} = \frac{1.99 \times 10^{20}}{9} = \mathbf{2.21 \times 10^{19} \text{ N}}$

The force is exactly 1/9 of the original, demonstrating the inverse square relationship.

Gravitational Field Strength

Gravitational field strength at a point in space is defined as the gravitational force per unit mass experienced by a small test mass placed at that point. It is a vector quantity pointing toward the source of the gravitational field.

Derivation of Field Strength Formula

Step 1: Start with Newton's Universal Law

The gravitational force on a test mass $m$ due to a source mass $M$ :

F = \frac{GMm}{r^2}

Step 2: Define field strength

Field strength $g$ is force per unit mass:

g = \frac{F}{m}

Step 3: Substitute and simplify

g = \frac{F}{m} = \frac{GMm/r^2}{m} = \frac{GM}{r^2}

Final result:

g = \frac{GM}{r^2}

Key observation: Field strength depends only on the source mass $M$ and distance $r$ , not on the test mass $m$ .

Understanding Gravitational Field Strength:

At Earth's Surface

At Earth's surface, $r = R_E = 6.37 \times 10^6 \text{ m}$ and $M = M_E = 6.0 \times 10^{24} \text{ kg}$ :

g = \frac{GM_E}{R_E^2} \approx 9.86 \text{ m/s}^2

Above Earth's Surface

At altitude $h$ above Earth's surface, $r = R_E + h$ :

g = \frac{GM_E}{(R_E + h)^2}

Field strength decreases with altitude.

Inverse Square Relationship

Like gravitational force, field strength follows an inverse square law. Doubling the distance from Earth's center reduces $g$ to 1/4 of its original value.

Vector Nature

Field strength is a vector pointing toward the source mass. Near Earth, it points toward Earth's center (radially inward).

Worked Example

The International Space Station (ISS) orbits at an altitude of 400 km above Earth's surface. Calculate the gravitational field strength at the ISS's orbital altitude. Compare this to the standard $g = 9.8 \text{ m/s}^2$ at Earth's surface. ( $R_E = 6.37 \times 10^6 \text{ m}, \quad M_E = 6.0 \times 10^{24} \text{ kg}$ )

Given:

$h = 400 \text{ km} = 4.0 \times 10^5 \text{ m}, \quad R_E = 6.37 \times 10^6 \text{ m}$

Step 1: Calculate orbital radius

$r = R_E + h = 6.37 \times 10^6 + 4.0 \times 10^5$

$r = 6.77 \times 10^6 \text{ m}$

Step 2: Apply field strength formula

$g = \frac{GM_E}{r^2} = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{(6.77 \times 10^6)^2}$

$g = \frac{4.00 \times 10^{14}}{4.58 \times 10^{13}} = \mathbf{8.73 \text{ m/s}^2}$

Step 3: Compare to surface gravity

$\frac{g_{\text{ISS}}}{g_{\text{surface}}} = \frac{8.73}{9.8} = 0.89 = \mathbf{89\%}$

This is 89% of surface gravity. Astronauts experience microgravity not because gravity is absent, but because they are in free fall!

Practice Question

The Moon has a mass of $7.35 \times 10^{22} \text{ kg}$ and a radius of $1.74 \times 10^6 \text{ m}$ . Calculate the gravitational field strength at the Moon's surface. How does this compare to Earth's surface gravity?

Show Answer

Solution:

$g_{\text{Moon}} = \frac{GM_{\text{Moon}}}{R_{\text{Moon}}^2}$

$g_{\text{Moon}} = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{(1.74 \times 10^6)^2}$

$g_{\text{Moon}} = \frac{4.90 \times 10^{12}}{3.03 \times 10^{12}} = \mathbf{1.62 \text{ m/s}^2}$

Comparison to Earth:

$\frac{g_{\text{Moon}}}{g_{\text{Earth}}} = \frac{1.62}{9.8} = 0.165 \approx \frac{1}{6}$

The Moon's surface gravity is approximately 1/6 of Earth's. This is why astronauts can jump much higher on the Moon!

Kepler's Third Law

Kepler's Third Law states that the square of the orbital period of a planet is proportional to the cube of its average distance from the Sun. This law applies to any object orbiting under gravitational attraction.

Derivation of Kepler's Third Law

Step 1: Centripetal force equals gravitational force

For a circular orbit, gravitational force provides the centripetal force:

\frac{GMm}{r^2} = \frac{mv^2}{r}

Step 2: Express velocity in terms of period

For circular motion: $v = \frac{2\pi r}{T}$

Substitute into the force equation:

\frac{GMm}{r^2} = \frac{m(2\pi r / T)^2}{r}

Step 3: Simplify

\frac{GM}{r^2} = \frac{4\pi^2 r}{T^2}

Step 4: Rearrange to get Kepler's Third Law

T^2 = \frac{4\pi^2}{GM} r^3

Final form:

T^2 \propto r^3

or equivalently

\frac{T^2}{r^3} = \frac{4\pi^2}{GM} = \text{constant}

Key insight: The ratio $T^2/r^3$ is constant for all objects orbiting the same central mass.

Applications:

• Compare orbital periods of different satellites around Earth
• Determine the mass of a planet from its moons' orbits
• Predict orbital characteristics of newly discovered exoplanets
• The law works for elliptical orbits using the semi-major axis

Worked Example

A satellite orbits Earth at a radius of $8.0 \times 10^6 \text{ m}$ with a period of 120 minutes. A second satellite orbits at $1.6 \times 10^7 \text{ m}$ . Calculate the orbital period of the second satellite.

Given:

$r_1 = 8.0 \times 10^6 \text{ m}, \quad T_1 = 120 \text{ min}, \quad r_2 = 1.6 \times 10^7 \text{ m}$

Step 1: Apply Kepler's Third Law

Since both satellites orbit Earth: $\frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3}$

Step 2: Solve for $T_2$

$T_2^2 = T_1^2 \times \frac{r_2^3}{r_1^3}$

$T_2 = T_1 \times \left(\frac{r_2}{r_1}\right)^{3/2}$

Step 3: Calculate the radius ratio

$\frac{r_2}{r_1} = \frac{1.6 \times 10^7}{8.0 \times 10^6} = 2$

Step 4: Calculate $T_2$

$T_2 = 120 \times (2)^{3/2}$

$T_2 = 120 \times 2.828$

$T_2 = \mathbf{339 \text{ min}}$

The second satellite has a period of 339 minutes (approximately 5.7 hours).

Notice: Doubling the radius increases the period by a factor of $2^{3/2} \approx 2.83$ .

Practice Question

Mars orbits the Sun at an average distance of $2.28 \times 10^{11} \text{ m}$ , while Earth orbits at $1.50 \times 10^{11} \text{ m}$ with a period of 1 year. Calculate the orbital period of Mars in Earth years.

Show Answer

Solution:

$T_{\text{Mars}} = T_{\text{Earth}} \times \left(\frac{r_{\text{Mars}}}{r_{\text{Earth}}}\right)^{3/2}$

$T_{\text{Mars}} = 1 \times \left(\frac{2.28 \times 10^{11}}{1.50 \times 10^{11}}\right)^{3/2}$

$T_{\text{Mars}} = (1.52)^{3/2}$

$T_{\text{Mars}} = \mathbf{1.87 \text{ years}}$

Mars takes approximately 1.87 Earth years to complete one orbit around the Sun, which matches astronomical observations!

Orbital Velocity

For a satellite to maintain a stable circular orbit, it must travel at a specific velocity that depends only on the orbital radius and the mass of the central body. The orbital velocity is independent of the satellite's own mass.

Derivation of Orbital Velocity Formula

Step 1: Equate gravitational and centripetal forces

For a circular orbit, the gravitational force provides the centripetal force:

\frac{GMm}{r^2} = \frac{mv^2}{r}

Step 2: Simplify by canceling mass

Notice that the satellite mass $m$ appears on both sides and cancels:

\frac{GM}{r^2} = \frac{v^2}{r}

Step 3: Multiply both sides by $r$

\frac{GM}{r} = v^2

Step 4: Take the square root

v = \sqrt{\frac{GM}{r}}

Key Insight: The orbital velocity depends only on $G$ , $M$ (mass of central body), and $r$ (orbital radius). The satellite's mass does not affect its orbital velocity!

Key Properties of Orbital Velocity

Mass Independent

A small satellite and a large space station at the same altitude travel at the same orbital velocity. The satellite's mass cancels out in the derivation.

Inverse Relationship

Since $v \propto 1/\sqrt{r}$ , satellites in higher orbits travel slower. The ISS (400 km altitude) travels faster than geostationary satellites (36,000 km altitude).

Constant Speed

In a perfectly circular orbit, the speed remains constant throughout the orbit. Only the direction of velocity changes, creating centripetal acceleration.

Typical Values

Low Earth Orbit satellites (LEO) travel at approximately 7.8 km/s, while geostationary satellites travel at about 3.1 km/s. Both are much slower than Earth's escape velocity of 11.2 km/s.

Worked Example

Calculate the orbital velocity of the International Space Station (ISS), which orbits at an altitude of 400 km above Earth's surface.

Given: $M_{\text{Earth}} = 6.0 \times 10^{24} \text{ kg}$ , $R_{\text{Earth}} = 6.37 \times 10^6 \text{ m}$ , $G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$

Step 1: Calculate orbital radius

$r = R_{\text{Earth}} + h$

$r = 6.37 \times 10^6 + 4.00 \times 10^5$

$r = 6.77 \times 10^6 \text{ m}$

Step 2: Apply orbital velocity formula

$v = \sqrt{\frac{GM}{r}}$

$v = \sqrt{\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6.77 \times 10^6}}$

$v = \sqrt{5.91 \times 10^7}$

$v = 7.69 \times 10^3 \text{ m/s}$

Step 3: Convert to km/s

$v = \mathbf{7.69 \text{ km/s}}$

The ISS travels at approximately 7.69 km/s (or about 27,680 km/h). At this speed, it completes one orbit around Earth every 90 minutes!

Practice Question

A GPS satellite orbits Earth at an altitude of 20,200 km. Calculate its orbital velocity and compare it to the ISS velocity calculated above. Use the same values for Earth's mass and radius.

Show Answer

Step 1: Calculate orbital radius

$r = R_{\text{Earth}} + h$

$r = 6.37 \times 10^6 + 2.02 \times 10^7$

$r = 2.66 \times 10^7 \text{ m}$

Step 2: Apply orbital velocity formula

$v = \sqrt{\frac{GM}{r}}$

$v = \sqrt{\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{2.66 \times 10^7}}$

$v = \sqrt{1.50 \times 10^7}$

$v = 3.88 \times 10^3 \text{ m/s} = \mathbf{3.88 \text{ km/s}}$

Step 3: Compare to ISS

$\frac{v_{\text{GPS}}}{v_{\text{ISS}}} = \frac{3.88}{7.69} \approx 0.50$

The GPS satellite travels at 3.88 km/s, which is approximately half the velocity of the ISS. This demonstrates the inverse relationship between orbital radius and velocity - higher orbits require slower speeds.

Escape Velocity

Escape velocity is the minimum speed an object must have to break free from a planet's gravitational pull without any further propulsion. Unlike orbital velocity, escape velocity applies to objects traveling away from a planet, not orbiting it.

Derivation of Escape Velocity Formula

Step 1: Apply energy conservation

For an object to escape, its total energy at the surface must equal zero (kinetic energy at infinity is zero, potential energy at infinity is zero):

E_{\text{total}} = KE + PE = 0

Step 2: Write kinetic and potential energy

At the surface (or distance $r$ from center):

\frac{1}{2}mv^2 - \frac{GMm}{r} = 0

Step 3: Solve for velocity

Rearrange to isolate $v$ :

\frac{1}{2}mv^2 = \frac{GMm}{r}

v^2 = \frac{2GM}{r}

Step 4: Take the square root

v_{\text{escape}} = \sqrt{\frac{2GM}{r}}

Key Insight: Comparing escape velocity to orbital velocity: $v_{\text{escape}} = \sqrt{2} \times v_{\text{orbital}} \approx 1.41 \times v_{\text{orbital}}$ . Notice both are independent of the object's mass!

Understanding Escape Velocity

Direction Independent

Escape velocity is the same regardless of launch direction. However, launching in the direction of Earth's rotation (eastward) allows you to take advantage of Earth's rotational velocity, reducing fuel requirements.

Minimum Speed

Escape velocity is the minimum initial speed needed with no further propulsion. Rockets can escape at lower speeds by continuing to thrust as they climb, but this requires more fuel.

Black Holes

A black hole's event horizon is defined as the radius where escape velocity equals the speed of light. Inside this radius, not even light can escape, making the black hole truly "black."

Atmospheric Consideration

In reality, objects cannot reach escape velocity at Earth's surface due to atmospheric friction. Spacecraft achieve escape velocity gradually as they climb through the atmosphere.

Worked Example

Calculate Earth's escape velocity from its surface. Compare this to the orbital velocity of a satellite just above Earth's surface.

Given: $M_{\text{Earth}} = 6.0 \times 10^{24} \text{ kg}$ , $R_{\text{Earth}} = 6.37 \times 10^6 \text{ m}$ , $G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$

Step 1: Calculate escape velocity

$v_{\text{escape}} = \sqrt{\frac{2GM}{r}}$

$v_{\text{escape}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6.37 \times 10^6}}$

$v_{\text{escape}} = \sqrt{1.256 \times 10^8}$

$v_{\text{escape}} = 1.12 \times 10^4 \text{ m/s} = \mathbf{11.2 \text{ km/s}}$

Step 2: Calculate orbital velocity at surface

$v_{\text{orbital}} = \sqrt{\frac{GM}{r}}$

$v_{\text{orbital}} = \sqrt{\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6.37 \times 10^6}}$

$v_{\text{orbital}} = \sqrt{6.28 \times 10^7}$

$v_{\text{orbital}} = 7.93 \times 10^3 \text{ m/s} = \mathbf{7.93 \text{ km/s}}$

Step 3: Compare the two velocities

$\frac{v_{\text{escape}}}{v_{\text{orbital}}} = \frac{11.2}{7.93} = 1.41 \approx \sqrt{2}$

Earth's escape velocity is 11.2 km/s (about 40,300 km/h), which is exactly √2 ≈ 1.41 times the orbital velocity of 7.93 km/s. This confirms the theoretical relationship between the two velocities.

Practice Question

The Moon has a mass of $7.35 \times 10^{22} \text{ kg}$ and a radius of $1.74 \times 10^6 \text{ m}$ . Calculate the escape velocity from the Moon's surface. Why is this significantly lower than Earth's escape velocity?

Show Answer

Step 1: Apply escape velocity formula

$v_{\text{escape}} = \sqrt{\frac{2GM}{r}}$

$v_{\text{escape}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^6}}$

$v_{\text{escape}} = \sqrt{5.63 \times 10^6}$

$v_{\text{escape}} = 2.37 \times 10^3 \text{ m/s} = \mathbf{2.37 \text{ km/s}}$

Step 2: Compare to Earth

$\frac{v_{\text{Moon}}}{v_{\text{Earth}}} = \frac{2.37}{11.2} \approx 0.21$

The Moon's escape velocity is about 21% of Earth's escape velocity.

The Moon's escape velocity is only 2.37 km/s, much lower than Earth's 11.2 km/s. This is because the Moon has both a smaller mass (about 1/81 of Earth's mass) and a smaller radius. The weaker gravitational field makes it easier for objects to escape. This is why the Moon has no atmosphere - gas molecules can reach escape velocity through thermal motion alone!

Gravitational Potential Energy

Gravitational potential energy describes the energy stored in an object due to its position in a gravitational field. Unlike the simple $U = mgh$ formula used near Earth's surface, the universal formula accounts for the inverse square nature of gravity and extends to infinity.

Understanding the Potential Energy Formula

The Universal Formula

U = -\frac{GMm}{r}

Where:

$U$ = gravitational potential energy (J)
$G$ = gravitational constant
$M$ = mass of central body (kg)
$m$ = mass of object (kg)
$r$ = distance from center of mass (m)

Why is it negative?

The negative sign is crucial and has three important meanings:

1. Zero at infinity: We define potential energy to be zero when objects are infinitely far apart. Since gravitational force is attractive, bringing objects closer requires external work, making the energy more negative.

2. Bound systems: A negative total energy means the object is gravitationally bound (in orbit). A positive or zero total energy means the object can escape to infinity.

3. Energy must be added: To move an object from distance $r$ to infinity, you must add energy equal to $+GMm/r$ to overcome the attractive force.

Relationship to $U = mgh$

Near Earth's surface, where $h \ll R_{\text{Earth}}$ :

\Delta U = -\frac{GMm}{R + h} - \left(-\frac{GMm}{R}\right) \approx mgh

This shows that the familiar $mgh$ formula is an approximation that only works for small height changes.

Important Properties

Increases with Distance

As $r$ increases, $U$ becomes less negative (increases). At infinity, $U = 0$ . Moving away from a planet always increases potential energy.

Inverse Relationship

Potential energy is inversely proportional to distance: $U \propto -1/r$ . Doubling the distance makes the potential energy half as negative (less bound).

Work and Energy

The work done by gravity moving an object from $r_1$ to $r_2$ equals $W = U_1 - U_2$ . If moving closer ( $r_2 < r_1$ ), gravity does positive work.

Gravitational Wells

Massive objects create "gravitational wells" - regions of low (very negative) potential energy. Escaping requires climbing out of this well by adding kinetic energy.

Worked Example

Calculate the gravitational potential energy of a 1000 kg satellite at an altitude of 400 km above Earth's surface. Compare this to the energy needed to lift it from the surface to this altitude.

Given: $M_{\text{Earth}} = 6.0 \times 10^{24} \text{ kg}$ , $R_{\text{Earth}} = 6.37 \times 10^6 \text{ m}$ , $G = 6.67 \times 10^{-11} \ \text{N}\text{m}^2/\text{kg}^2$

Step 1: Calculate PE at 400 km altitude

$r = R_{\text{Earth}} + h = 6.37 \times 10^6 + 4.00 \times 10^5 = 6.77 \times 10^6 \text{ m}$

$U_{\text{orbit}} = -\frac{GMm}{r}$

$U_{\text{orbit}} = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1000}{6.77 \times 10^6}$

$U_{\text{orbit}} = -5.91 \times 10^{10} \text{ J}$

Step 2: Calculate PE at Earth's surface

$U_{\text{surface}} = -\frac{GMm}{R_{\text{Earth}}}$

$U_{\text{surface}} = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1000}{6.37 \times 10^6}$

$U_{\text{surface}} = -6.28 \times 10^{10} \text{ J}$

Step 3: Calculate energy needed to lift satellite

$\Delta U = U_{\text{orbit}} - U_{\text{surface}}$

$\Delta U = (-5.91 \times 10^{10}) - (-6.28 \times 10^{10})$

$\Delta U = \mathbf{3.7 \times 10^9 \text{ J}}$

The satellite has a potential energy of -59.1 GJ at orbital altitude. To lift it from the surface requires 3.7 GJ of energy. Note: If we incorrectly used $U = mgh$ , we'd get $3.92 \times 10^9 \text{ J}$ , which is 6% higher - demonstrating why the universal formula is needed for high altitudes!

Practice Question

A 500 kg spacecraft is initially at rest on Earth's surface. Calculate how much energy is required to move it to a position infinitely far from Earth (ignoring air resistance and Earth's rotation). This represents the minimum energy needed for the spacecraft to escape Earth's gravity.

Show Answer

Step 1: Calculate initial potential energy

$U_{\text{initial}} = -\frac{GMm}{R_{\text{Earth}}}$

$U_{\text{initial}} = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 500}{6.37 \times 10^6}$

$U_{\text{initial}} = -3.14 \times 10^{10} \text{ J}$

Step 2: Determine final potential energy

At infinity: $U_{\text{final}} = 0 \text{ J}$

Step 3: Calculate required energy

$\Delta U = U_{\text{final}} - U_{\text{initial}}$

$\Delta U = 0 - (-3.14 \times 10^{10})$

$\Delta U = \mathbf{3.14 \times 10^{10} \text{ J}}$

31.4 GJ of energy is required. This is exactly equal to $\frac{1}{2}mv_{\text{escape}}^2$ , confirming the relationship between potential energy and escape velocity!

Total Energy in Orbit

The total mechanical energy of a satellite in orbit is the sum of its kinetic and potential energies. This total energy determines whether the satellite is bound to orbit, will escape, or will fall back to Earth.

Derivation of Total Energy Formula

Step 1: Write total energy

E_{\text{total}} = KE + PE = \frac{1}{2}mv^2 - \frac{GMm}{r}

Step 2: Substitute orbital velocity

For a circular orbit, we know $v = \sqrt{GM/r}$ . Therefore $v^2 = GM/r$ :

E_{\text{total}} = \frac{1}{2}m \cdot \frac{GM}{r} - \frac{GMm}{r}

Step 3: Simplify

E_{\text{total}} = \frac{GMm}{2r} - \frac{GMm}{r}

Step 4: Combine fractions

E_{\text{total}} = \frac{GMm}{2r} - \frac{2GMm}{2r} = -\frac{GMm}{2r}

Key Relationships:

$E_{\text{total}} = -\frac{GMm}{2r}$ (always negative for bound orbits)
$E_{\text{total}} = \frac{1}{2}PE$ (total energy is half the potential energy)
$KE = -E_{\text{total}}$ (kinetic energy equals the magnitude of total energy)
$KE = -\frac{1}{2}PE$ (kinetic energy is half the magnitude of potential energy)

Understanding Total Energy

Negative Energy = Bound

A satellite in orbit always has negative total energy, meaning it's gravitationally bound. To escape, energy must be added to bring the total to zero or positive.

Higher Orbit = Less Negative

Since $E \propto -1/r$ , satellites in higher orbits have less negative (higher) total energy. They're less tightly bound and closer to escaping.

The Virial Theorem

The relationship $E_{\text{total}} = \frac{1}{2}PE$ is an example of the virial theorem, which applies to many bound systems in physics, from atoms to galaxies.

Energy to Escape

The energy needed to move a satellite from orbit to infinity equals $-E_{\text{total}} = GMm/2r$ . This is exactly half the energy needed to lift it from the surface to that orbit!

Energy Comparison for a Satellite in Orbit

Energy Type	Formula	Sign	Relative Magnitude
Kinetic Energy	$KE = \frac{GMm}{2r}$	Positive (+)	$1$ unit
Potential Energy	$PE = -\frac{GMm}{r}$	Negative (−)	$-2$ unit
Total Energy	$E = -\frac{GMm}{2r}$	Negative (−)	$1$ unit

Notice: $KE + PE = E_{\text{total}}$ → $1 + (-2) = -1$

Worked Example

A 1200 kg satellite orbits Earth at an altitude of 800 km. Calculate its kinetic energy, potential energy, and total mechanical energy. Verify the energy relationships.

Given: $M_{\text{Earth}} = 6.0 \times 10^{24} \text{ kg}$ , $R_{\text{Earth}} = 6.37 \times 10^6 \text{ m}$ , $G = 6.67 \times 10^{-11} \text{ N}\text{m}^2/\text{kg}^2$

Step 1: Calculate orbital radius

$r = R_{\text{Earth}} + h = 6.37 \times 10^6 + 8.00 \times 10^5$

$r = 7.17 \times 10^6 \text{ m}$

Step 2: Calculate kinetic energy

$KE = \frac{GMm}{2r}$

$KE = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1200}{2 \times 7.17 \times 10^6}$

$KE = 3.35 \times 10^{10} \text{ J}$

Step 3: Calculate potential energy

$PE = -\frac{GMm}{r}$

$PE = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1200}{7.17 \times 10^6}$

$PE = -6.70 \times 10^{10} \text{ J}$

Step 4: Calculate total energy

$E_{\text{total}} = KE + PE$

$E_{\text{total}} = 3.35 \times 10^{10} + (-6.70 \times 10^{10})$

$E_{\text{total}} = -3.35 \times 10^{10} \text{ J}$

Step 5: Verify relationships

$E_{\text{total}} = \frac{1}{2}PE$ : $-3.35 \times 10^{10} = \frac{1}{2} \times (-6.70 \times 10^{10})$ ✓

$KE = -E_{\text{total}}$ : $3.35 \times 10^{10} = -(-3.35 \times 10^{10})$ ✓

KE = 33.5 GJ, PE = −67.0 GJ, E_total = −33.5 GJ. All three energy relationships are verified!

Practice Question

A geostationary satellite orbits at an altitude of 35,800 km above Earth's equator. If the satellite has a mass of 2000 kg, calculate how much additional energy would be needed to move it from this orbit to infinity (to escape Earth's gravity completely).

Show Answer

Step 1: Calculate orbital radius

$r = R_{\text{Earth}} + h = 6.37 \times 10^6 + 3.58 \times 10^7$

$r = 4.23 \times 10^7 \text{ m}$

Step 2: Calculate current total energy

$E_{\text{orbit}} = -\frac{GMm}{2r}$

$E_{\text{orbit}} = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 2000}{2 \times 4.23 \times 10^7}$

$E_{\text{orbit}} = -9.46 \times 10^9 \text{ J}$

Step 3: Calculate energy needed

At infinity: $E_{\infty} = 0 \text{ J}$

$\Delta E = E_{\infty} - E_{\text{orbit}}$

$\Delta E = 0 - (-9.46 \times 10^9)$

$\Delta E = \mathbf{9.46 \times 10^9 \text{ J}}$

9.46 GJ of energy is needed. This is significantly less than the 31.4 GJ needed from Earth's surface.

Changing Orbits: The Burn

Satellites don't simply "move" to different orbits - they must fire thrusters in controlled "burns" to change their velocity and energy. Understanding how orbital velocity and energy change with radius reveals a counterintuitive result: to speed up, you must first slow down!

The Orbital Paradox

Key Observations:

Satellites in lower orbits travel faster: $v = \sqrt{GM/r}$
Satellites in lower orbits have less total energy: $E = -GMm/2r$

The Paradox:

To move to a higher orbit (where you'll travel slower), you must add energy by firing your thrusters forward, which temporarily increases your speed! Conversely, to drop to a lower orbit (where you'll travel faster), you must fire thrusters backward to slow down.

Resolution:

When you fire thrusters forward at point A in a circular orbit:

Your velocity increases at point A
This creates an elliptical transfer orbit
You coast to a higher altitude at point B
At point B, your velocity is actually lower than in your original orbit
A second burn circularizes you into the higher orbit at the slower speed

Hohmann Transfer Orbit

The most fuel-efficient way to change orbits is the Hohmann transfer, which uses two burns:

Step 1: First Burn (at point A)

Fire thrusters in the direction of motion to increase velocity

$\Delta v_1 = v_{\text{transfer,A}} - v_{\text{initial}}$

Step 2: Coast

Follow the elliptical transfer orbit to point B (no thrust)

Energy is conserved during coast phase

Step 3: Second Burn (at point B)

Fire thrusters again in direction of motion to circularize

$\Delta v_2 = v_{\text{final}} - v_{\text{transfer,B}}$

The total velocity change required is:

\Delta v_{\text{total}} = \Delta v_1 + \Delta v_2

Understanding Orbit Changes

Increasing Orbit Radius

Burn prograde (forward) → Add energy → Enter elliptical transfer orbit → Coast to higher altitude → Burn prograde again → Circularize at higher orbit with lower velocity.

Decreasing Orbit Radius

Burn retrograde (backward) → Reduce energy → Enter elliptical transfer orbit → Coast to lower altitude → Burn retrograde again → Circularize at lower orbit with higher velocity.

Energy Changes

The energy difference between orbits is $\Delta E = GMm/2 \times (1/r_1 - 1/r_2)$ . Moving to a higher orbit always requires adding energy, regardless of the velocity changes.

Timing Matters

Burns must occur at specific points in the orbit. For maximum efficiency, burn at periapsis (closest point) to raise apoapsis (farthest point), or vice versa.

Worked Example

A satellite in a circular orbit at 300 km altitude needs to move to a 600 km altitude circular orbit. Calculate: (a) the initial and final orbital velocities, (b) the change in total energy, and (c) whether the satellite travels faster or slower after the transfer.

Given: $M_{\text{Earth}} = 6.0 \times 10^{24} \text{ kg}$ , $R_{\text{Earth}} = 6.37 \times 10^6 \text{ m}$ , satellite mass = 1000 kg

Step 1: Calculate initial orbital velocity (300 km)

$r_1 = 6.37 \times 10^6 + 3.00 \times 10^5 = 6.67 \times 10^6 \text{ m}$

$v_1 = \sqrt{\frac{GM}{r_1}} = \sqrt{\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6.67 \times 10^6}}$

$v_1 = 7.75 \times 10^3 \text{ m/s}$

Step 2: Calculate final orbital velocity (600 km)

$r_2 = 6.37 \times 10^6 + 6.00 \times 10^5 = 6.97 \times 10^6 \text{ m}$

$v_2 = \sqrt{\frac{GM}{r_2}} = \sqrt{\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6.97 \times 10^6}}$

$v_2 = 7.58 \times 10^3 \text{ m/s}$

Step 3: Calculate energy change

$E_1 = -\frac{GMm}{2r_1} = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1000}{2 \times 6.67 \times 10^6} = -3.00 \times 10^{10} \text{ J}$

$E_2 = -\frac{GMm}{2r_2} = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1000}{2 \times 6.97 \times 10^6} = -2.87 \times 10^{10} \text{ J}$

$\Delta E = E_2 - E_1 = (-2.87 - (-3.00)) \times 10^{10} = \mathbf{1.3 \times 10^9 \text{ J}}$

(a) Initial velocity: 7.75 km/s, Final velocity: 7.58 km/s

(b) Energy added: 1.3 GJ

(c) The satellite travels slower in the higher orbit, even though we added energy! The energy went into increasing potential energy.

Practice Question

A space station in a circular orbit at 400 km needs to perform an emergency de-orbit to drop to 300 km altitude. Will the astronauts need to fire thrusters forward or backward? After the maneuver, will they be traveling faster or slower than before?

Show Answer

Answer:

Thruster direction: BACKWARD (retrograde burn). To drop to a lower orbit, you must reduce your orbital energy by slowing down.

Final speed: FASTER. This demonstrates the paradox - you fire thrusters backward to slow down initially, but end up traveling faster once you've settled into the lower orbit!

Low Earth Orbit vs Geostationary Orbit

Two important types of satellite orbits serve different purposes: Low Earth Orbit (LEO) for observation and communication, and Geostationary Orbit (GEO) for weather monitoring and telecommunications. Understanding their differences reveals the trade-offs in satellite design.

Orbit Comparison

Property	Low Earth Orbit (LEO)	Geostationary Orbit (GEO)
Altitude	200-2,000 km (typical: 400 km)	~35,800 km above equator
Orbital Velocity	~7.8 km/s (28,000 km/h)	~3.1 km/s (11,000 km/h)
Orbital Period	90-120 minutes	24 hours (synchronous with Earth)
Coverage Area	Limited (needs constellation)	40% of Earth's surface (1 satellite)
Signal Latency	~1-4 ms (very low)	~250 ms (noticeable delay)
Launch Energy	Lower (easier to reach)	Much higher (~50× more)
Atmospheric Drag	Significant (requires boost)	Negligible

Why Geostationary Orbit is Special

A geostationary satellite appears to hover motionless above a fixed point on Earth's equator because its orbital period exactly matches Earth's rotation period.

Required Conditions:

Period = 24 hours: Must orbit once per day
Equatorial orbit: Must orbit directly above the equator (0° inclination)
Circular orbit: Must maintain constant altitude
Prograde motion: Must orbit in the same direction as Earth's rotation

Calculating GEO Altitude:

Using Kepler's Third Law with $T = 24 \text{ hours} = 86,400 \text{ s}$ :

T^2 = \frac{4\pi^2}{GM}r^3

$r^3 = \frac{GMT^2}{4\pi^2} = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times (86400)^2}{4\pi^2}$

$r^3 \approx 7.57 \times 10^{22} \text{ m}^3$

$r \approx 4.23 \times 10^7 \text{ m} = 42,300 \text{ km from Earth's center}$

$h = r - R_{\text{Earth}} = 42,300 - 6,370 = 35,930 \text{ km altitude}$

Key Insight: There is only ONE altitude where geostationary orbit is possible - approximately 35,800 km above the equator. This makes GEO "orbital real estate" extremely valuable and internationally regulated!

Applications and Trade-offs

LEO Applications

Best for: Earth observation, imaging, low-latency communication, space stations

Examples: ISS, Starlink constellation, Hubble Space Telescope. The close proximity provides high resolution and low signal delay.

GEO Applications

Best for: Weather monitoring, telecommunications, broadcast TV

Examples: GOES weather satellites, broadcast satellites. The fixed position allows ground antennas to point in one direction permanently.

Worked Example

Compare the orbital velocity and energy requirements for satellites in LEO (400 km) and GEO (35,800 km).

Given: $M_{\text{Earth}} = 6.0 \times 10^{24} \text{ kg}$ , $R_{\text{Earth}} = 6.37 \times 10^6 \text{ m}$

LEO Calculations (h = 400 km):

$r_{\text{LEO}} = 6.37 \times 10^6 + 4.00 \times 10^5 = 6.77 \times 10^6 \text{ m}$

$v_{\text{LEO}} = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6.77 \times 10^6}}$

$v_{\text{LEO}} = 7,690 \text{ m/s}$

$E_{\text{LEO}}/m = -\frac{GM}{2r} = -\frac{4.00 \times 10^{14}}{2 \times 6.77 \times 10^6}$

$E_{\text{LEO}}/m = -2.96 \times 10^7 \text{ J/kg}$

GEO Calculations (h = 35,800 km):

$r_{\text{GEO}} = 6.37 \times 10^6 + 3.58 \times 10^7 \approx 4.22 \times 10^7 \text{ m}$

$v_{\text{GEO}} = \sqrt{\frac{GM}{r}} = \sqrt{\frac{4.00 \times 10^{14}}{4.22 \times 10^7}}$

$v_{\text{GEO}} = 3,080 \text{ m/s}$

$E_{\text{GEO}}/m = -\frac{GM}{2r} = -\frac{4.00 \times 10^{14}}{2 \times 4.22 \times 10^7}$

$E_{\text{GEO}}/m = -4.74 \times 10^6 \text{ J/kg}$

Energy Comparison:

Energy to reach LEO from surface:

$E_{\text{surface}}/m = -\frac{GM}{R} = -6.28 \times 10^7 \text{ J/kg}$

$\Delta E_{\text{LEO}} = (-2.96 \times 10^7) - (-6.28 \times 10^7) = 3.32 \times 10^7 \text{ J/kg}$

Energy to reach GEO from surface:

$\Delta E_{\text{GEO}} = (-4.74 \times 10^6) - (-6.28 \times 10^7) = 5.81 \times 10^7 \text{ J/kg}$

LEO velocity: 7.69 km/s, GEO velocity: 3.08 km/s (2.5× slower)

Energy needed: LEO = 33.2 MJ/kg, GEO = 58.1 MJ/kg

GEO requires 1.75× more energy per kilogram than LEO, making it significantly more expensive to launch satellites into geostationary orbit!

Practice Question

A telecommunications company wants to provide global coverage. Option A: Use 3 GEO satellites positioned 120° apart above the equator. Option B: Use a constellation of LEO satellites at 550 km altitude. For the LEO option, calculate how many satellites would be needed if each satellite can communicate with ground stations within a 2,000 km radius on Earth's surface. Which option makes more sense for global internet coverage?

Show Answer

GEO Analysis:

✓ 3 satellites provide nearly complete coverage (except poles)

✓ Fixed position - no tracking needed

✗ 250 ms latency (poor for real-time applications)

✗ High launch costs (~$50M+ per satellite)

✗ Cannot cover polar regions

LEO Analysis:

Coverage area per satellite: Limited to ~12.5 million km² visible area

Earth's surface area: ~510 million km²

Rough estimate: Need 40-60 satellites for continuous global coverage

✓ Very low latency (1-4 ms)

✓ Lower launch costs per satellite

✓ Can cover polar regions

✗ Requires complex satellite constellation

✗ Satellites must track and handoff connections

✗ Atmospheric drag requires periodic boosts

For modern internet: LEO constellation (like Starlink with ~4,000 satellites) is superior due to low latency critical for video calls, gaming, and interactive applications. The higher number of satellites is offset by mass production and reusable rockets.

For traditional TV broadcast: GEO remains ideal - only 3 satellites needed, latency doesn't matter for one-way transmission, and fixed antennas are simple and cheap.

The choice depends on the application! This is why both LEO and GEO satellites continue to serve important roles in our communication infrastructure.

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