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

v_e = √(2 × G × M / R)   or   v_e = √(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 (≈ 6.67 × 10^-11 N·m²/kg²).
• 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).
• v_e = 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 Objective Practice Questions (Interactive)

Click on an option to check if your answer is correct. Explanations will automatically reveal themselves.

1. Which of the following statements about a gravitational field is correct?

A. It is a scalar field directed away from the mass

B. It is a vector field directed towards the center of the mass

C. Its strength increases as distance from the mass increases

D. It can experience forces of repulsion

Explanation: A gravitational field is always attractive (directed towards the center of the mass creating it) and possesses both magnitude and direction, making it a vector field.

2. If the distance between two masses is doubled, the gravitational force of attraction between them becomes:

A. 2 times greater

B. 4 times greater

C. halved

D. one-quarter of its original value

Explanation: According to Newton's law, F ∝ 1/r². Doubling the distance (2r)² creates a factor of 4 in the denominator, reducing the force to 1/4th.

3. The escape velocity of a projectile from the Earth's surface depends on which of the following properties?

A. Mass of the projectile

B. Mass and radius of the Earth

C. Angle of projection

D. Duration of the launch window

Explanation: The formula v_e = √(2GM/R) demonstrates that escape velocity depends solely on the planet's mass (M) and radius (R), completely independent of the projectile's mass.

4. What is the appropriate SI unit for gravitational field intensity?

A. N·m

B. N/m

C. N/kg

D. kg/N

Explanation: Gravitational field intensity is force per unit mass (g = F/m), yielding Newtons per kilogram (N/kg), which is equivalent to m/s².

5. A planet has twice the mass of Earth and twice its radius. The acceleration due to gravity on this planet compared to Earth's acceleration (g) is:

A. 2g

B. 4g

C. 0.5g

D. g

Explanation: g = GM/R². For the new planet, g' = G(2M)/(2R)² = 2GM/4R² = 0.5 × (GM/R²) = 0.5g.

6. Calculate the escape velocity from a planet's surface where g = 10 m/s² and the planet's radius is 6.4 × 10⁶ m.

A. 1.13 × 10⁴ m/s

B. 8.00 × 10³ m/s

C. 1.28 × 10⁸ m/s

D. 5.65 × 10³ m/s

Explanation: v_e = √(2gR) = √(2 × 10 × 6.4 × 10⁶) = √(1.28 × 10⁸) ≈ 11,313.7 m/s = 1.13 × 10⁴ m/s.