3.4 Gravitational Field and Acceleration Due to Gravity on Different Planets

Overview

The gravitational field, denoted by $g$, represents the force per unit mass acting on an object in a specific region of space. This field is responsible for the gravitational force $F_g = mg$ that acts on objects, causing them to accelerate toward the center of a planet or star.

On Earth, $g$ has a value of 9.8 m/s², but it varies on other planets depending on their mass and radius. Let’s explore how gravitational fields work and how to calculate $g$ for any celestial body.

What Is a Gravitational Field?

A gravitational field is a region around a mass where another mass experiences a force due to gravity. Its strength is expressed as the gravitational force per unit mass:

$$g = \frac{F_g}{m}$$

In simpler terms, $g$ is the acceleration due to gravity an object experiences in a given gravitational field.

Key Properties

1. Direction: Always points toward the center of the mass generating the field.
2. Magnitude: Determined by the mass of the celestial body and the distance from its center.
3. Radial Nature: For spherically symmetric objects, the gravitational field decreases as the inverse square of the distance from the center.

Deriving Gravitational Field Strength

Using Newton’s Universal Law of Gravitation:

$$F_g = G \frac{m_1 m_2}{r^2}$$

Combine this with $F_g = m_1 g$:

$$g = G \frac{m}{r^2}$$

Where:

• $g$ is the gravitational field strength (m/s²),
• $G$ is the gravitational constant ($6.67 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2$),
• $m$ is the mass of the planet (kg),
• $r$ is the distance from the planet’s center (m).

This equation allows us to calculate $g$ for any celestial body.

Key Insights

1. Mass-Independent: $g$ does not depend on the mass of the object experiencing the gravitational field.
2. Distance Matters: $r$ includes the radius of the planet and any additional height above the surface.
3. Universal Application: The equation works for all spherically symmetric masses, from planets to stars.

Gravitational Acceleration on Different Planets

The value of $g$ varies based on the mass and radius of a planet. For example:

• Earth: $g = 9.8 \, \text{m/s}^2$
  
• Mars: $g \approx 3.7 \, \text{m/s}^2$
  
• Jupiter: $g \approx 24.8 \, \text{m/s}^2$
  

The larger and denser a planet, the stronger its gravitational field.

Practice Problems

Problem 1:

How does $g$ change if the mass of a planet is doubled?

• a) It remains the same.
• b) It is halved.
• c) It is doubled.
• d) It is quadrupled.

Answer: c) It is doubled.

Explanation: Since $g \propto m$, doubling the mass doubles $g$.

Problem 2:

A planet has a mass of $6 \times 10^{24} \, \text{kg}$ and a radius of $6 \times 10^6 \, \text{m}$. Calculate $g$.

• a) $7.2 \, \text{m/s}^2$
  
• b) $9.8 \, \text{m/s}^2$
  
• c) $10.4 \, \text{m/s}^2$
  
• d) $18.6 \, \text{m/s}^2$
  

Answer: b) $9.8 \, \text{m/s}^2$

Explanation:

$$g=G\frac{m}{r^2}=\frac{(6.67\times 10^{-11})(6\times 10^{24})}{(6\times 10^6{)}^2}=9.8{\text{m/s}}^2$$

Problem 3:

How does $g$ change if the distance between two masses is doubled?

• a) It remains the same.
• b) It is halved.
• c) It is quartered.
• d) It is doubled.

Answer: c) It is quartered.

Explanation: Since $g \propto 1/r^2$, doubling the distance reduces $g$ to one-fourth.

Problem 4:

An object is $2 \times 10^8 \, \text{m}$ away from a star with a mass of $2 \times 10^{30} \, \text{kg}$. Calculate the gravitational force acting on the object.

• a) $-3.35 \times 10^{-6} \, \text{N}$
  
• b) $-4.92 \times 10^{-12} \, \text{N}$
  
• c) $-5.98 \times 10^{-9} \, \text{N}$
  
• d) $-6.67 \times 10^{-11} \, \text{N}$
  

Answer: d) $-6.67 \times 10^{-11} \, \text{N}$

Explanation:

$$F = G \frac{m_1 m_2}{r^2} = \frac{(6.67 \times 10^{-11})(2 \times 10^{30})(1)}{(2 \times 10^8)^2} = 6.67 \times 10^{-11} \, \text{N}$$