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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2019 · 11 January · Morning Shift · Physics · Gravitation

A satellite is revolving in a circular orbit at a height h from the...

A satellite is revolving in a circular orbit at a height h from the carth surface, such that h<<R\mathrm{h}< < \mathrm{R} where R\mathrm{R} is the radius of the earth. Assuming that the effect of earth"s atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is

Options

  1. A2gR\sqrt{2 g R}
  2. BgR\sqrt{g R}
  3. CgR2\sqrt{\frac{g R}{2}}
  4. DgR(2−1)\sqrt{g R}(\sqrt{2}-1)

Correct answer

D. gR(2−1)\sqrt{g R}(\sqrt{2}-1)

Step-by-step solution

For a satellite orbiting close to the earth, orbital velocity is given by v0=g(R+h)=gRv_{0}=\sqrt{g(R+h)}=\sqrt{g R} Escape velocity (ve)\left(\mathrm{v}_{\mathrm{e}}\right) is ve=2g(R+h)=2gR[∵h<cRΔv=ve−v0=(2−1)gRv_{e}=\sqrt{2 g(R+h)}=\sqrt{2 g R} \quad[\because h < c R \Delta v=v_{e}-v_{0}=(\sqrt{2}-1) \sqrt{g R}
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