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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Physics
Chapter
Gravitation

JEE Advanced 2011 · Paper 2 · Physics · Gravitation

A satellite is moving with a constant speed v in a circular orbit...

A satellite is moving with a constant speed vv in a circular orbit about the earth. An object of mass mm 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

Options

  1. A12mv2\frac{1}{2} m v^2
  2. Bmv2m v^2
  3. C32mv2\frac{3}{2} m v^2
  4. D2mv22 m v^2

Correct answer

B. mv2m v^2

Step-by-step solution

In circular orbit of a satellite, potential energy =−2×( kinetic energy) =−2×12mv2=−mv2\begin{aligned}& =-2 \times(\text{ kinetic energy) }\\ & =-2 \times \frac{1}{2}m v²=-m v² \end{aligned}Justtoescapefromthegravitationalpull,itstotalmechanicalenergyshouldbezero.Therefore,itskineticenergyshouldbe+mv2to escape from the gravitational pull, its total mechanical energy should be zero. Therefore, its kinetic energy should be+m v^2. ∴\therefore Correct answer is (b). Analysis of Question (i) Question is moderately difficult. (ii) In circular orbit of satellite U=−GMmrandK=GMm2rorU=−2KU=-\frac{G M m}{r} and K=\frac{G M m}{2 r} or U=-2 K Here, UU is potential energy and $K is kinetic energy.
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