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  • Solve this problem 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 surfa

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

  • Option 1)

    \sqrt{\frac {2GM}{R}}

  • Option 2)

    \sqrt{\frac{2GM}{R^{2}}}

  • Option 3)

    \sqrt{2gR^{2}}

  • Option 4)

    \sqrt{\frac{2GM}{R^{2}}}

 

Answers (1)

best_answer

As we learnt

Relation of escape velocity and orbital velocity -

V=sqrt{frac{GM}{R}}

V_{e}=sqrt{frac{2GM}{R}}

V ightarrow Orbital velocity

V_{e} ightarrow Escape velocity

- wherein

v=frac{V_{e}}{sqrt{2}}

V_{escape}= sqrt{2}V_{orbital}

 

 According to law of conservation of machanical energy

\frac{1}{2}mv^{2}-\frac{Gmm}{R}=0

\frac{1}{2}mv^{2}-\frac{Gmm}{R}=v^{2}=\frac{2Gm}{R}

v=\sqrt{\frac{2Gm}{R}}


Option 1)

\sqrt{\frac {2GM}{R}}

Correct Option

Option 2)

\sqrt{\frac{2GM}{R^{2}}}

Incorrect Option

Option 3)

\sqrt{2gR^{2}}

Incorrect Option

Option 4)

\sqrt{\frac{2GM}{R^{2}}}

Incorrect Option

Posted by

prateek

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