Problem 68 - Master Resource Book in JEE Main Physics

Question

A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is \(11 \mathrm{kms}^{-1}\), the escape velocity from the surface of the planet would be (a) \(0.11 \mathrm{kms}^{-1}\) (b) \(1.1 \mathrm{kms}^{-1}\) (c) \(11 \mathrm{kms}^{-1}\) (d) \(110 \mathrm{kms}^{-1}\)

Step-by-Step Solution

Answer

The escape velocity from the planet is closer to 49.17 km/s, not matching provided options precisely.

Step 1: Understanding the Escape Velocity Formula

The formula for escape velocity (v_e) is given by \( v_e = \sqrt{\frac{2GM}{R}} \) where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet. We will use this formula to compare Earth's escape velocity and that of the given planet.

Step 2: Comparing Mass and Radius for the Given Planet

We are told that the planet's mass is 10 times Earth's mass (\( M' = 10M_{earth} \)) and its radius is 10 times smaller (\( R' = \frac{R_{earth}}{10} \)). We need to substitute these into the escape velocity formula for the planet.

Step 3: Calculating the Escape Velocity Ratio

Substitute the values into the escape velocity formula to find the new escape velocity ratio: \( v'_e = \sqrt{\frac{2G(10M_{earth})}{\frac{R_{earth}}{10}}} = \sqrt{\frac{20GM_{earth}}{R_{earth}}} = \sqrt{20} \times \sqrt{\frac{2GM_{earth}}{R_{earth}}} = \sqrt{20} \times v_{earth} \).

Step 4: Calculating the New Escape Velocity

Given that \( v_{earth} = 11 \, \mathrm{kms^{-1}} \), we substitute this into the formula: \( v'_e = \sqrt{20} \times 11 \, \mathrm{kms^{-1}} \). \( \sqrt{20} \approx 4.47 \), so \( v'_e \approx 49.17 \, \mathrm{kms^{-1}} \).

Step 5: Analyzing the Options

With \( v'_e \approx 49.17 \, \mathrm{kms^{-1}} \), the closest answer option would be \( 49.17 \, \mathrm{kms^{-1}} \). However, since it isn't an option, there may be an oversight in matching. Upon re-evaluation, note that higher discrepancies often point to higher multiples or close values misaligned.

Key Concepts

Gravitational Constant

The gravitational constant, represented by the symbol \( G \), is a fundamental constant in physics that appears in the law of universal gravitation. It is used to calculate the gravitational force between two masses and is an integral part of many equations relating to gravitational phenomena, such as the escape velocity.
Escape velocity is the minimum speed needed for an object to break free from a celestial body's gravitational pull without any additional propulsion.
The formula for escape velocity \( v_e \) is \( v_e = \sqrt{\frac{2GM}{R}} \), where:

• \( G \) is the gravitational constant.
• \( M \) is the mass of the planet or celestial body.
• \( R \) is the radius of the planet or celestial body from its center to its surface.

This formula shows how the escape velocity depends both on the gravitational constant and on the mass and radius of the celestial body. Thus, understanding the role of \( G \) is crucial in solving problems related to planetary motion and escape velocities.

Mass and Radius Comparison

When comparing two planets or celestial bodies, it is essential to consider both their masses and radii because they directly affect gravitational forces and escape velocities.
In our problem, the planet in question has a mass that is 10 times that of Earth and a radius that is 10 times smaller. These differences lead to significant changes in the calculated escape velocity.
To see how this comparison impacts escape velocity, let’s consider the modifications in the escape velocity formula:- For this new planet, mass \( M' \) is 10 times of Earth’s mass, so \( M' = 10M_{earth} \).- Radius \( R' \) is 10 times smaller, so \( R' = \frac{R_{earth}}{10} \).Once these values are substituted into the escape velocity formula, they lead to a different ratio, demonstrating how mass and radius differences play a pivotal role.

Velocity Ratio Calculation

To find the escape velocity of a new planet relative to Earth, we calculate the ratio of escape velocities using known values. Given Earth's escape velocity is \( 11 \, ext{kms}^{-1} \), we use this value as a reference.
The formula for the planet's escape velocity \( v'_e \) is derived from the modified equation:\[ v'_e = \sqrt{\frac{2G(10M_{earth})}{R_{earth}/10}} = \sqrt{20} \times \sqrt{\frac{2GM_{earth}}{R_{earth}}} = \sqrt{20} \times v_{earth} \]Here:

• \( \sqrt{20} \) is approximately 4.47.
• Thus, \( v'_e = 4.47 \times 11 \text{ kms}^{-1} \), approximately equal to 49.17 \text{ kms}^{-1}.

This calculation illustrates how planetary characteristics directly influence escape velocity. While our answer was close at 49.17 \text{ kms}^{-1}, it highlights the necessity of rechecking options when discrepancies occur, ensuring accuracy in quantitative solutions.