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Escape velocity on a planet from its mass, radius or density compared with the earth

This type of question, asked 4 times: 4 May 2023 S2, 9 May 2023 S2, 10 May 2023 S1, 13 Apr 2026 S2. All 4 questions below, oldest first.

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  1. Question 1

    4 May 2023 S2

    47
    The escape velocity from earth surface is 11 km/s. The escape velocity from a planet having twice the radius and same mean density as earth is

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  2. Question 2

    9 May 2023 S2

    46
    The radius of the planet is double that of the earth, but their average densities are same. VPV_P and VEV_E are the escape velocities of planet and earth respectively. If VPVE=x\frac{V_P}{V_E} = x, the value of 'x' is

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  3. Question 3

    10 May 2023 S1

    8
    The radius and mean density of the planet are four times as that of the earth. The ratio of escape velocity at the earth to the escape velocity at a planet is

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  4. Question 4

    13 Apr 2026 S2

    109
    If earth has a mass nine times and radius twice to the planet P. Then ve3x ms−1\dfrac{v_e}{3}\sqrt{x}\ \mathrm{ms^{-1}} will be the minimum velocity required by a rocket to pull out of gravitational force of P, where vev_e is escape velocity on earth. The value of xx is

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