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 (1)

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

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