Question:
Escape velocity from earth's surface is 11 km/s. If the radius of a planet is double that of earth and the density is the same as that of earth, then the escape velocity from this planet will be:
Escape velocity from earth's surface is 11 km/s. If the radius of a planet is double that of earth and the density is the same as that of earth, then the escape velocity from this planet will be:
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Escape velocity depends on both the mass and the radius of a planet. When the radius is doubled and the density remains constant, the escape velocity increases by a factor of \( \sqrt{2} \).
Updated On: Apr 22, 2026
- 5.5 km/s
- 11 km/s
- 16.5 km/s
- 22 km/s
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The Correct Option is D
Solution and Explanation
Step 1: Formula for Escape Velocity.
The escape velocity \( v_e \) is given by the formula: \[ v_e = \sqrt{\frac{2GM}{R}} \] where: - \( G \) is the gravitational constant, - \( M \) is the mass of the planet, and - \( R \) is the radius of the planet.
Step 2: Understand the Given Data.
We are told that: - The radius of the new planet is double that of Earth, so \( R_{\text{new}} = 2R_{\text{earth}} \). - The density of the new planet is the same as Earth, meaning the mass of the new planet \( M_{\text{new}} \) will be 8 times the mass of Earth. This is because the mass of a planet depends on its volume, and volume scales with the cube of the radius. Therefore, if the radius is doubled, the volume (and thus the mass) increases by a factor of \( 2^3 = 8 \).
Step 3: Relation for New Planet.
The escape velocity \( v_e \) depends on both the mass \( M \) and the radius \( R \) of the planet. The formula for the escape velocity can be written for the new planet as: \[ v_{e,\text{new}} = \sqrt{\frac{2GM_{\text{new}}}{R_{\text{new}}}} \] Substitute \( M_{\text{new}} = 8M_{\text{earth}} \) and \( R_{\text{new}} = 2R_{\text{earth}} \) into the equation: \[ v_{e,\text{new}} = \sqrt{\frac{2G(8M_{\text{earth}})}{2R_{\text{earth}}}} \]
Step 4: Simplifying the Expression.
Now simplify the expression: \[ v_{e,\text{new}} = \sqrt{\frac{8GM_{\text{earth}}}{2R_{\text{earth}}}} \] \[ v_{e,\text{new}} = \sqrt{4 \cdot \frac{2GM_{\text{earth}}}{R_{\text{earth}}}} \]
Step 5: Compare with Escape Velocity from Earth.
We know that the escape velocity from Earth is: \[ v_{e,\text{earth}} = \sqrt{\frac{2GM_{\text{earth}}}{R_{\text{earth}}}} = 11 \, \text{km/s} \] Thus, we can write: \[ v_{e,\text{new}} = 2 \cdot v_{e,\text{earth}} = 2 \cdot 11 \, \text{km/s} = 22 \, \text{km/s} \]
Step 6: Conclusion.
Therefore, the escape velocity from the new planet is \( 22 \, \text{km/s} \).
The escape velocity \( v_e \) is given by the formula: \[ v_e = \sqrt{\frac{2GM}{R}} \] where: - \( G \) is the gravitational constant, - \( M \) is the mass of the planet, and - \( R \) is the radius of the planet.
Step 2: Understand the Given Data.
We are told that: - The radius of the new planet is double that of Earth, so \( R_{\text{new}} = 2R_{\text{earth}} \). - The density of the new planet is the same as Earth, meaning the mass of the new planet \( M_{\text{new}} \) will be 8 times the mass of Earth. This is because the mass of a planet depends on its volume, and volume scales with the cube of the radius. Therefore, if the radius is doubled, the volume (and thus the mass) increases by a factor of \( 2^3 = 8 \).
Step 3: Relation for New Planet.
The escape velocity \( v_e \) depends on both the mass \( M \) and the radius \( R \) of the planet. The formula for the escape velocity can be written for the new planet as: \[ v_{e,\text{new}} = \sqrt{\frac{2GM_{\text{new}}}{R_{\text{new}}}} \] Substitute \( M_{\text{new}} = 8M_{\text{earth}} \) and \( R_{\text{new}} = 2R_{\text{earth}} \) into the equation: \[ v_{e,\text{new}} = \sqrt{\frac{2G(8M_{\text{earth}})}{2R_{\text{earth}}}} \]
Step 4: Simplifying the Expression.
Now simplify the expression: \[ v_{e,\text{new}} = \sqrt{\frac{8GM_{\text{earth}}}{2R_{\text{earth}}}} \] \[ v_{e,\text{new}} = \sqrt{4 \cdot \frac{2GM_{\text{earth}}}{R_{\text{earth}}}} \]
Step 5: Compare with Escape Velocity from Earth.
We know that the escape velocity from Earth is: \[ v_{e,\text{earth}} = \sqrt{\frac{2GM_{\text{earth}}}{R_{\text{earth}}}} = 11 \, \text{km/s} \] Thus, we can write: \[ v_{e,\text{new}} = 2 \cdot v_{e,\text{earth}} = 2 \cdot 11 \, \text{km/s} = 22 \, \text{km/s} \]
Step 6: Conclusion.
Therefore, the escape velocity from the new planet is \( 22 \, \text{km/s} \).
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