A particle of mass m is projected with a velocity u = k v_e ( k < 1 ) from the surface of the earth ( v_e = escape velocity). The maximum height above - Physics MCQ NEET 2021 2026 | ExamTest.live

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A particle of mass m is projected with a velocity u = k v_{e} (k < 1) from the surface of the earth (v_{e} = escape velocity). The maximum height above the surface reached by the particle is:

Solution & Step-by-step Explanation

By conservation of energy: T E_{s u r f a ce} = T E_{ma x_h e i g h t} \frac{1}{2} m u^{2} - \frac{GM m}{R} = 0 - \frac{GM m}{R + h} Substitute u = k v_{e} = k \frac{2 GM}{R} : \frac{1}{2} m (k^{2} \frac{2 GM}{R}) - \frac{GM m}{R} = - \frac{GM m}{R + h} \frac{k ^{2}}{R} - \frac{1}{R} = - \frac{1}{R + h} \frac{1 - k ^{2}}{R} = \frac{1}{R + h} R + h = \frac{R}{1 - k ^{2}} h = \frac{R}{1 - k ^{2}} - R = R [\frac{1 - ( 1 - k ^{2} )}{1 - k ^{2}}] = \frac{R k ^{2}}{1 - k ^{2}}

Try it yourself before checking the explanation above.

A particle of mass m is projected with a velocity u = k v_{e} (k < 1) from the surface of the earth (v_{e} = escape velocity). The maximum height above the surface reached by the particle is:

\frac{R}{1 + k}

\frac{R k ^{2}}{1 - k ^{2}}

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

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

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