Question:
A comet at infinity starts moving towards the earth and passes the earth at a distance of 4R from the center of the earth, where R is the radius of the earth. If escape velocity of a body from the surface of earth is $11.2~kms^{-1}$, then the maximum velocity of the comet is}
A comet at infinity starts moving towards the earth and passes the earth at a distance of 4R from the center of the earth, where R is the radius of the earth. If escape velocity of a body from the surface of earth is $11.2~kms^{-1}$, then the maximum velocity of the comet is}
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Velocity in gravitational field varies as $1/\sqrt{r}$ when starting from infinity.
Updated On: Jun 22, 2026
- $5.6~kms^{-1}$
- $2.8~kms^{-1}$
- $11.2~kms^{-1}$
- $22.4~kms^{-1}$ \bigskip
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The Correct Option is A
Solution and Explanation
Concept:
A comet falling from infinity converts gravitational potential energy into kinetic energy. Conservation of mechanical energy is used in gravitational field:
\[
v^2 = v_\infty^2 + \frac{2GM}{r}
\]
Step 1: Use escape velocity relation.
\[ v_e = \sqrt{\frac{2GM}{R}} = 11.2~kms^{-1} \Rightarrow 2GM = v_e^2 R \]
Step 2: Velocity at distance $r = 4R$.
\[ v^2 = 0 + \frac{2GM}{4R} \] Substitute: \[ v^2 = \frac{v_e^2 R}{4R} = \frac{v_e^2}{4} \]
Step 3: Compute velocity.
\[ v = \frac{v_e}{2} = \frac{11.2}{2} = 5.6~kms^{-1} \] Final Answer: \[ (A)\ 5.6~kms^{-1} \]
Step 1: Use escape velocity relation.
\[ v_e = \sqrt{\frac{2GM}{R}} = 11.2~kms^{-1} \Rightarrow 2GM = v_e^2 R \]
Step 2: Velocity at distance $r = 4R$.
\[ v^2 = 0 + \frac{2GM}{4R} \] Substitute: \[ v^2 = \frac{v_e^2 R}{4R} = \frac{v_e^2}{4} \]
Step 3: Compute velocity.
\[ v = \frac{v_e}{2} = \frac{11.2}{2} = 5.6~kms^{-1} \] Final Answer: \[ (A)\ 5.6~kms^{-1} \]
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