Question:

A satellite is revolving round the earth with orbital speed $v_0$. If it stops suddenly, the speed with which it will strike the surface of earth would be ($v_e= $ escape velocity of a body on earth's surface)

Updated On: Jul 31, 2023

$\frac{v_{e}^{2}}{v_{0}}$
$v_0$
$\sqrt{v_{e}^{2} -2v_{0}^{2}}$
$\sqrt{v_{e}^{2} -v_{0}^{2}}$

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The Correct Option is C

Solution and Explanation

Let $v$ be the velocity with which the satellite strikes the surface of the earth. According to law of conservation of total mechanical energy, we get $- \frac{GMm}{R+h} = \frac{1}{2}mv^{2} - \frac{GMm}{R} $ $v^{2} = \frac{2GM}{R} - \frac{2GM}{R+h}$ $ \because v_{0} = \sqrt{\frac{GM}{R+h}}, v_{e} = \sqrt{\frac{2GM}{R}} $ $ \therefore v^{2} = \sqrt{v_{e}^{2} -2v_{0}^{2}}$

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Newton’s Law of Gravitation

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

F ∝ (M₁M₂) . . . . (1)
(F ∝ 1/r²) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M₁M₂/r²

F = G × [M₁M₂]/r² . . . . (7)

Or, f(r) = GM₁M₂/r²

The dimension formula of G is [M^-1L³T^-2].

A satellite is revolving round the earth with orbital speed $v_0$. If it stops suddenly, the speed with which it will strike the surface of earth would be ($v_e= $ escape velocity of a body on earth's surface)

The Correct Option is C

Solution and Explanation

Top Questions on Gravitation

Concepts Used:

Gravitation

Newton’s Law of Gravitation