Problem 39 - Master Resource Book in JEE Main Physics

Question

Mass of moon is \(7.34 \times 10^{22} \mathrm{~kg}\). If the acceleration due to gravity on the moon is \(1.4 \mathrm{~ms}^{-2}\), the radius of the moon is \(\left(G=6.667 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}\right)\) (a) \(0.56 \times 10^{4} \mathrm{~m}\) (b) \(1.87 \times 10^{6} \mathrm{~m}\) (c) \(1.92 \times 10^{6} \mathrm{~m}\) (d) \(1.01 \times 10^{8} \mathrm{~m}\)

Step-by-Step Solution

Answer

The radius of the moon is approximately \(1.87 \times 10^6 \text{ m}\). Option (b) is correct.

Step 1: Identify the Formula

The acceleration due to gravity at the surface of a celestial body is given by the formula: \[ g = \frac{G \cdot M}{R^2} \]where \( g \) is the acceleration due to gravity, \( G \) is the universal gravitational constant, \( M \) is the mass of the body, and \( R \) is the radius of the body.

Step 2: Rearrange the Formula

We need to solve for \( R \), the radius of the moon. Start by rearranging the formula to solve for \( R^2 \): \[ R^2 = \frac{G \cdot M}{g} \]

Step 3: Substitute Given Values

Substitute the given values into the formula: \[ R^2 = \frac{(6.667 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \cdot (7.34 \times 10^{22} \, \text{kg})}{1.4 \, \text{ms}^{-2}} \]

Step 4: Calculate \( R^2 \)

Perform the calculation to find \( R^2 \):\[ R^2 = \frac{4.888078 \times 10^{12}}{1.4} \approx 3.4914843 \times 10^{12} \]

Step 5: Solve for \( R \)

Take the square root of both sides to solve for \( R \):\[ R = \sqrt{3.4914843 \times 10^{12}} \approx 1.868 \times 10^6 \text{ m} \]

Step 6: Select the Closest Option

Compare the calculated \( R \) to the provided options. The closest value is:(b) \( 1.87 \times 10^6 \text{ m} \).

Key Concepts

Acceleration due to Gravity

The acceleration due to gravity is a crucial concept in gravitational physics. It is the rate at which an object accelerates downwards due to the gravitational pull of a celestial body, like Earth or the Moon. The acceleration due to gravity on the Moon is given in this exercise as 1.4 m/s². This is significantly less than Earth's gravitational acceleration, which is approximately 9.8 m/s². This difference explains why we feel much lighter on the Moon.

The formula to calculate this acceleration is:

• \( g = \frac{G \cdot M}{R^2} \)

Here, \( g \) is the acceleration due to gravity, \( G \) is the universal gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius. By knowing the values of \( G \), \( M \), and \( R \), we can find the gravitational acceleration on any celestial body's surface.

Universal Gravitational Constant

The universal gravitational constant is a vital element in the study of gravitational physics. Known as \( G \), it appears in the formula for Newton's law of universal gravitation. The value of \( G \) is approximately \( 6.667 \times 10^{-11} \, \mathrm{Nm}^2 / \mathrm{kg}^2 \).

This constant is crucial because it helps quantify the gravitational force between two masses. While it remains constant, the gravitational force is heavily dependent on these two masses' values and the distance between them. The constant \( G \) ensures that the law of universal gravitation applies consistently across different scenarios, whether it be on Earth or for celestial bodies far away.

Given its essential role in gravitational calculations, understanding \( G \) helps us grasp the forces at play in various gravitational interactions throughout the universe.

Mass and Radius of Celestial Bodies

The mass and radius of celestial bodies, like the Moon or Earth, play a significant role in determining the gravitational forces that act on or around them. Mass measures the amount of matter in a celestial object, while the radius is the distance from its center to its edge.

In our example exercise, the Moon's mass is given as \( 7.34 \times 10^{22} \, \mathrm{kg} \). This massive value influences how much gravitational pull the Moon exerts on objects near it. Coupled with its radius, which was calculated to be approximately \( 1.87 \times 10^6 \, \mathrm{m} \), we can understand how these characteristics affect the strength of gravitational acceleration (1.4 m/s² on the Moon).

Larger celestial bodies tend to have stronger gravitational pulls due to their larger masses. Hence, the greater the mass and the smaller the radius, the stronger the gravitational pull exerted at its surface.