JEE Main · 2016 · 03 Apr 2016 · Offline · Gravitation

A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth

R

;

h ≪ R

). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to: (Neglect the effect of atmosphere.)

(A) $gR$
(B) $gR /2$
(C) $gR (2 - 1)$ ✓ Correct
(D) $2 gR$

Answer: (C)

Solution

At the low orbit, with

h ≪ R

R + h \approx R

, the circular speed is

v_{0} = \frac{GM}{R + h} \approx \frac{GM}{R} = g R

, using

g = \frac{GM}{R ^{2}}

. To break free the satellite needs the escape speed from the same place,

v_{e} = \frac{2 GM}{R + h} \approx 2 g R

. The extra speed required is the difference,

Δ v = 2 g R - g R = g R (2 - 1)

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option

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