Gravitational Fields & Escape Velocity

 

Gravitational Fields & Escape Velocity: Comprehensive Study Guide

Developed by: Oyewole Olatunbosun

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1. Definitions

Gravitational Field

A gravitational field is a region of space surrounding a body possessing mass, within which another mass experiences a gravitational force of attraction. It is a vector field, meaning it has both magnitude and direction (always directed towards the center of the mass creating the field).

Escape Velocity

Escape velocity is the minimum speed required for a body (such as a rocket or projectile) to escape from the gravitational influence of a massive body (like the Earth) without further propulsion.

2. Relevant Laws

Newton's Law of Universal Gravitation

This law states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

3. Relevant Formulas

Newton's Gravitational Force

F = (G * M * m) / r²

Gravitational Field Intensity (Strength)

g = (G * M) / r²

Escape Velocity

vₑ = √(2 * G * M / R)   or   vₑ = √(2 * g * R)

4. Explanation of Symbols & Formulas

• F = Gravitational force of attraction between two masses (measured in Newtons, N).
• g = Gravitational field intensity or acceleration due to gravity (measured in m/s² or N/kg).
• G = Universal gravitational constant ($\approx 6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$).
• M = Mass of the larger body creating the field, e.g., the Earth (measured in kg).
• m = Mass of the smaller object inside the field (measured in kg).
• r = Distance between the centers of the two masses (measured in meters, m).
• R = Radius of the planet/Earth (measured in meters, m).
• vₑ = Escape velocity (measured in meters per second, m/s).

Understanding the Mechanics:

The first formula shows that as distance ($r$) increases, the force drops off rapidly due to the inverse-square law. The escape velocity formulas show that a projectile's escape speed depends entirely on the mass ($M$) and radius ($R$) of the planet it is leaving—not the mass of the projectile itself. A tiny pebble and a massive rocket require the exact same escape velocity to leave Earth!

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5. WAEC Theory Questions & Solutions

Click the "View Solution" buttons below to reveal the step-by-step mathematical breakdowns.

Question 1: Calculate the gravitational force of attraction between the Earth of mass $5.97 \times 10^{24} \text{ kg}$ and a satellite of mass $1200 \text{ kg}$ orbiting at a distance of $7.0 \times 10^6 \text{ m}$ from the center of the Earth. $[G = 6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2]$

Data Given:

• Mass of Earth ($M$) = $5.97 \times 10^{24} \text{ kg}$
• Mass of satellite ($m$) = $1200 \text{ kg}$
• Distance ($r$) = $7.0 \times 10^6 \text{ m}$
• $G = 6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$

Formula:

F = (G * M * m) / r²

Substitution:

F = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 1200) / (7.0 × 10⁶)²

F = (4.7813 × 10¹⁷) / (4.9 × 10¹³)

F = 9,757.76 N

Final Answer: The gravitational force of attraction is approximately 9,757.8 N.

Question 2: (a) Explain why the acceleration due to gravity ($g$) varies on the surface of the Earth. (b) Calculate the gravitational field intensity at a height equal to the Earth's radius above the Earth's surface. [Take $g$ on Earth's surface = $10 \text{ m/s}^2$]

Part (a) Explanation:
The Earth is not a perfect sphere; it is an oblate spheroid, flattened at the poles and bulging at the equator. Since the polar radius is shorter than the equatorial radius, and $g \propto 1/R^2$, acceleration due to gravity is greater at the poles than at the equator.

Part (b) Solution:

• At the surface: $r = R \implies g_1 = \frac{GM}{R^2} = 10 \text{ m/s}^2$
• At height $h = R$: New distance from center $r = R + h = R + R = 2R$
• New field intensity $g_2 = \frac{GM}{(2R)^2} = \frac{GM}{4R^2}$

Comparing the two equations:

g₂ = g₁ / 4

g₂ = 10 / 4 = 2.5 m/s²

Final Answer: The gravitational field intensity at that height is 2.5 m/s².

Question 3: Determine the escape velocity of a rocket fired from the surface of Mars. Given that the mass of Mars is $6.42 \times 10^{23} \text{ kg}$ and its radius is $3.39 \times 10^6 \text{ m}$. $[G = 6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2]$

Data Given:

• Mass of Mars ($M$) = $6.42 \times 10^{23} \text{ kg}$
• Radius of Mars ($R$) = $3.39 \times 10^6 \text{ m}$
• $G = 6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$

Formula:

vₑ = √(2GM / R)

Substitution:

vₑ = √[ (2 × 6.67 × 10⁻¹¹ × 6.42 × 10²³) / (3.39 × 10⁶) ]

vₑ = √[ (8.566 × 10¹³) / (3.39 × 10⁶) ]

vₑ = √[ 25,268,436.58 ]

vₑ ≈ 5,026.77 m/s

Final Answer: The escape velocity from Mars is approximately 5,027 m/s (or 5.03 km/s).

Question 4: An astronaut on a distant planet notices that the acceleration due to gravity is $4.5 \text{ m/s}^2$. If the radius of the planet is $4.2 \times 10^6 \text{ m}$, calculate the escape velocity from the planet's surface.

Data Given:

• Acceleration due to gravity ($g$) = $4.5 \text{ m/s}^2$
• Radius of planet ($R$) = $4.2 \times 10^6 \text{ m}$

Formula:

When given $g$ and $R$ directly, use:

vₑ = √(2gR)

Substitution:

vₑ = √(2 × 4.5 × 4.2 × 10⁶)

vₑ = √(37,800,000)

vₑ ≈ 6,148.17 m/s

Final Answer: The escape velocity from the planet is 6,148.17 m/s (or 6.15 km/s).