Question:
The escape velocity of an object from a planet is \(16\,\text{km s}^{-1}\). If the escape velocity of the object from another planet having twice the density and three times the radius of the planet is \(V\sqrt{2}\,\text{m s}^{-1}\), then the value of \(V\) is
The escape velocity of an object from a planet is \(16\,\text{km s}^{-1}\). If the escape velocity of the object from another planet having twice the density and three times the radius of the planet is \(V\sqrt{2}\,\text{m s}^{-1}\), then the value of \(V\) is
Show Hint
For a spherical planet,
\[
v_e=\sqrt{\frac{2GM}{R}}
\]
and since
\[
M=\frac{4}{3}\pi R^3\rho,
\]
we get
\[
v_e\propto R\sqrt{\rho}.
\]
Updated On: Jun 18, 2026
- \(12\)
- \(48\)
- \(18\)
- \(36\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Write the formula for escape velocity.
Escape velocity from a planet is \[ v_e=\sqrt{\frac{2GM}{R}}. \] For a spherical planet, \[ M=\frac{4}{3}\pi R^3\rho. \] Substituting this in the formula, \[ v_e=\sqrt{\frac{2G}{R}\cdot \frac{4}{3}\pi R^3\rho} \] \[ v_e=\sqrt{\frac{8}{3}\pi G\rho R^2} \] Therefore, \[ v_e\propto R\sqrt{\rho}. \]
Step 2: Compare escape velocities of the two planets.
Let the original planet have radius \(R\) and density \(\rho\).
The new planet has radius \[ R'=3R \] and density \[ \rho'=2\rho. \] Therefore, \[ \frac{v'_e}{v_e} = \frac{R'\sqrt{\rho'}}{R\sqrt{\rho}} \] \[ = \frac{3R\sqrt{2\rho}}{R\sqrt{\rho}} \] \[ =3\sqrt{2}. \]
Step 3: Find the escape velocity from the second planet.
Given, \[ v_e=16\,\text{km s}^{-1}. \] Thus, \[ v'_e=3\sqrt2\times 16 \] \[ v'_e=48\sqrt2\,\text{km s}^{-1}. \] Comparing with \[ V\sqrt2, \] we get \[ V=48. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{48} \]
Escape velocity from a planet is \[ v_e=\sqrt{\frac{2GM}{R}}. \] For a spherical planet, \[ M=\frac{4}{3}\pi R^3\rho. \] Substituting this in the formula, \[ v_e=\sqrt{\frac{2G}{R}\cdot \frac{4}{3}\pi R^3\rho} \] \[ v_e=\sqrt{\frac{8}{3}\pi G\rho R^2} \] Therefore, \[ v_e\propto R\sqrt{\rho}. \]
Step 2: Compare escape velocities of the two planets.
Let the original planet have radius \(R\) and density \(\rho\).
The new planet has radius \[ R'=3R \] and density \[ \rho'=2\rho. \] Therefore, \[ \frac{v'_e}{v_e} = \frac{R'\sqrt{\rho'}}{R\sqrt{\rho}} \] \[ = \frac{3R\sqrt{2\rho}}{R\sqrt{\rho}} \] \[ =3\sqrt{2}. \]
Step 3: Find the escape velocity from the second planet.
Given, \[ v_e=16\,\text{km s}^{-1}. \] Thus, \[ v'_e=3\sqrt2\times 16 \] \[ v'_e=48\sqrt2\,\text{km s}^{-1}. \] Comparing with \[ V\sqrt2, \] we get \[ V=48. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{48} \]
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