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  • Solve! The escape velocity from the earth’s surface is 11 km/sec. A certain planet has a radius twice that of the earth but its mean density is the same as that of the earth. The value of the escape velocity from this planet would be

The escape velocity from the earth’s surface is 11 km/sec. A certain planet has a radius twice that of the earth but its mean density is the same as that of the earth. The value of the escape velocity from this planet would be

  • Option 1)

    22 km/sec

  • Option 2)

    11 km/sec

  • Option 3)

    5.5 km/sec

  • Option 4)

    16.5 km/sec

 

Answers (2)

As we learnt in 

Escape velocity ( in terms of radius of planet) -

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

V_{c}=\sqrt{2gR}

V_{c}\rightarrow Escape velocity

R\rightarrowRadius of earth

- wherein

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

 

 V_{e}=\sqrt{\frac{2GM}{R}}\left ( escape\ velocity \right )

=\sqrt{\frac{2G}{R}}.\rho.\frac{4\pi}{3}R^{3}

=\sqrt{\frac{8\pi}{3}G\rho R^{2}}

\frac{V_{p}}{V_{e}}=\frac{R_{p}}{R_{e}}=2\ \: \: \: \: \Rightarrow V_{p}=22 km/sec

 


Option 1)

22 km/sec

This is correct option

Option 2)

11 km/sec

This is incorrect option

Option 3)

5.5 km/sec

This is incorrect option

Option 4)

16.5 km/sec

This is incorrect option

Posted by

Vakul

View full answer

Answer 1- 22km/sec

Posted by

Ankita Singh

View full answer

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