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The ratio of escape velocity of earth \left( {\upsilon _\text{e} } \right)  to the escape velocity at a planet \left( {\upsilon _\text{p} } \right) whose radius and mean density are twice as that of earth is:

  • Option 1)

    1 : 2

  • Option 2)

    1:2\sqrt 2

  • Option 3)

    1 : 4

  • Option 4)

    1:\sqrt 2

 

Answers (1)

best_answer

 

Escape velocity ( in terms of radius of planet) -

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

V_{c}=sqrt{2gR}

V_{c} ightarrow Escape velocity

R ightarrowRadius of earth

- wherein

  • depends on the reference body
  • greater the value of frac{M}{R} or left ( gR ight ) greater will be the escape velocity V_{e}=11.2Km/s  For earth

 

 escape velocity=\frac{\sqrt{2GM}}{R}

\frac{\sqrt{2G.(\int .\frac{4\pi }{3}R^{3})}}{R}=\sqrt{\frac{8\pi }{3}G\int R^{2}}

\frac{V_{p}}{V_{e}}=\frac{\sqrt{\int _{2}.R_{p}^{2}}}{\sqrt{\int }_{e}.R_{e}^{2}}=2\sqrt{2}

\frac{V_{e}}{V_{p}}=\frac{1}{2\sqrt{2}}


Option 1)

1 : 2

This option is incorrect 

Option 2)

1:2\sqrt 2

This option is correct 

Option 3)

1 : 4

This option is incorrect 

Option 4)

1:\sqrt 2

This option is incorrect 

Posted by

Plabita

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