JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Gravitation

A satellite is moving with a constant speed

v

in circular orbit around the earth. An object of mass '

m

' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is:

(A) $2 m v^{2}$
(B) $m v^{2}$ ✓ Correct
(C) $\frac{1}{2} m v^{2}$
(D) $\frac{3}{2} m v^{2}$

Answer: (B)

Solution

A satellite in a circular orbit obeys

v^{2} = G M_{e} / r

, so gravity exactly supplies the centripetal pull. The question is how much kinetic energy the ejected object needs to just escape from that same distance, and just escape means its total mechanical energy there is zero. Writing that,

K E - \frac{G M _{e} m}{r} = 0

, gives

K E = \frac{G M _{e} m}{r}

, and substituting

G M_{e} / r = v^{2}

collapses it to

m v^{2}

. Notice this is twice the object's original orbital kinetic energy

\frac{1}{2} m v^{2}

, which is just the familiar statement that escape speed is

2

times orbital speed.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Gravitation

Solution

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