A planet in a distant solar system is 10 times more  massive than the earth and its radius is  10 times smaller Given that the escape velocity from the earth is  11 km s-1, the escape velocity from the  surface of the planet would be :

  • Option 1)

    0.11 km s-1

  • Option 2)

    1.1 km s-1

  • Option 3)

    11 km s-1

  • Option 4)

    110 km s-1

 

Answers (1)

 

As we learnt in

Escape velocity - Minimum velocity which is required to escape the body from earth's gravitational pull

- wherein it is independent of the mass and direction of projection of body.

 

 V_{e}=\sqrt{\frac{2GM}{R}} for\: the \:earth

V_{e}=11kms^{-1}

mass of the planet =10Me

Radius of the planet= \frac{R}{10}

V_{e}=\sqrt{\frac{2GM\times 10}{\frac{R}{10}}}

= 10X11= 110kms^{-1}

 


Option 1)

0.11 km s-1

Incorrect option

Option 2)

1.1 km s-1

Incorrect option

Option 3)

11 km s-1

Incorrect option

Option 4)

110 km s-1

Incorrect option

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