IIT-JEE 2011 · Paper 2 · Physics Q22

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PhysicsSingle correct+3 / -1Question 22 of 60

A satellite is moving with a constant speed ' $V$ ' in a circular orbit about 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 its ejection, the kinetic energy of the object is

\frac12mV^2

mV^2

\frac32mV^2

2mV^2

Answer and worked solution

Official answer: B

Orbit. $\dfrac{mV^2}r=\dfrac{GMm}{r^2}$ , so $\dfrac{GM}r=V^2$ .

Escape from the orbit. The object needs total energy zero: $K=\dfrac{GMm}r=mV^2$ .

(B)

Topics:Escape velocity Satellites and orbital motion

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IIT-JEE 2011 · Paper 2, Physics question 22