1. R(k/1-k)² 2. R(k/1+k)² 3. R²k/1+k 4. Rk²/1-k²

Answer: 4. Rk²/1-k² Key idea: conserve energy: 1/2mv²-dfracGMmR=-GMm/R+h , with V_e²=dfrac2GMR . 1/2m k²dfrac2GMR-dfracGMmR=-GMm/R+h⇒dfracGMR(1-k²)=GM/R+h R+h=dfrac R1-k²⇒ h=Rk²/1-k² . Option 4.

Topic: Gravitational potential energy and gravitational potential· Gravitation

A particle of mass 'm' is projected with a velocity v = kV_e (k < 1) from the surface of the earth. (V_e = escape velocity)The maximum height above the surface reached by the particle is :

1. \(R\left(\frac{k}{1-k}\right)^2\)

2. \(R\left(\frac{k}{1+k}\right)^2\)

3. \(\frac{R^2k}{1+k}\)

4. \(\frac{Rk^2}{1-k^2}\)

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Answer: 4. \(\frac{Rk^2}{1-k^2}\)

Key idea: conserve energy: \(\tfrac12mv^2-\dfrac{GMm}R=-\dfrac{GMm}{R+h}\), with \(V_e^2=\dfrac{2GM}R\).

\(\tfrac12m k^2\dfrac{2GM}R-\dfrac{GMm}R=-\dfrac{GMm}{R+h}\Rightarrow\dfrac{GM}R(1-k^2)=\dfrac{GM}{R+h}\)

\(R+h=\dfrac R{1-k^2}\Rightarrow h=\dfrac{Rk^2}{1-k^2}\). Option 4.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A particle of mass 'm' is projected with a velocity v = kV_e (k < 1) from the surface of the earth. (V_e = escape velocity)The maximum height above the surface reached by the particle is :

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