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Gravitation question

2011 · Q102
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Gravitation question

2011 · Q102

NEETPhysicsGravitationMCQ+4 / −1
A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is
  1. A
    2GMR2\sqrt {{{2GM} \over {{R^2}}}}R22GM​​
  2. B
    2GMR\sqrt {{{2GM} \over R}}R2GM​​
  3. C
    2gMR2\sqrt {{{2gM} \over {{R^2}}}}R22gM​​
  4. D
    2gR2\sqrt {2g{R^2}}2gR2​
View written solutionFree

Correct answer: B

According to law of conservation of mechanical energy

12mu2−GMmR=0{1 \over 2}m{u^2} - {{GMm} \over R} = 021​mu2−RGMm​=0

⇒\Rightarrow⇒ u2=2GMR{u^2} = {{2GM} \over R}u2=R2GM​

∴\therefore∴ u=2GMR=2gRu = \sqrt {{{2GM} \over R}} = \sqrt {2gR} u=R2GM​​=2gR​   (∵g=GMR2)\left(\because {g = {{GM} \over {{R^2}}}} \right)(∵g=R2GM​)

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