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Gravitation question

2020 · 4 Sep · Shift 2 · Q63
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Gravitation question

2020 · 4 Sep · Shift 2 · Q63

JEE MainPhysicsGravitationMCQ+4 / −1
A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:
  1. A
    2
  2. B
    1
  3. C
    2\sqrt 22​
  4. D
    12{1 \over {\sqrt 2 }}2​1​
View written solutionFree

Correct answer: D

  1. Speed in a low circular orbit

For a circular orbit of radius rrr, the orbital speed is vo=GMrv_o = \sqrt{\frac{GM}{r}}vo​=rGM​​ Here the orbit is very close to the planet, so we take r=Rr = Rr=R Thus, vo=GMRv_o = \sqrt{\frac{GM}{R}}vo​=RGM​​

  1. Escape velocity from the planet

Escape velocity from the surface of a planet is ve=2GMRv_e = \sqrt{\frac{2GM}{R}}ve​=R2GM​​

  1. Required ratio

We need vove=GM/R2GM/R\frac{v_o}{v_e} = \frac{\sqrt{GM/R}}{\sqrt{2GM/R}}ve​vo​​=2GM/R​GM/R​​

Cancelling the common factor, vove=12=12\frac{v_o}{v_e} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}ve​vo​​=21​​=2​1​

  1. Match with options

The correct option is 12\boxed{\frac{1}{\sqrt{2}}}2​1​​ which is Option D.

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