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The escape velocity on the surface of ea...

 The escape velocity on the surface of earth is 11.2 km/s. What will be its value on a planet having double the radius and eight times the mass of earth?

Text Solution

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`implies v_p/v_e=sqrt((2GM_e)/R_p)xx sqrt((R_e)/(GM_e))=sqrt(M_p/M_exxR_e/R_p)`
`v_p/v_e=sqrt(8xx1/2)=sqrt4 =2`
`:. v_p=2v_e=2 (11.2km//s)= 22.4 (km)/s`
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Knowledge Check

  • The escape velocity of the surface of earth is 11.2 km/s. What would be the escape velocity of the surface of another planet of the same mass but (1^(th))/4 times the radius of the earth ?

    A
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    B
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  • The escape velocity from the surface of earth is v_e . The escape velocity from the surface of a planet whose mass and radius are 3 times those of the earth will be ...........

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    `v_e`
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    D
    `27v_e`

  • Escape velocity for earth surface is 11 km/s. If the radius of any planet is two times the radius of the earth but average density is same that of earth, then the escape velocity at the planet will be ........

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