The escape velocity from the earth’s surface is 11 kms⁻¹.A certain planet

The Escape Velocity From The Earths Surface Is 11 Kmsa Physics Question

The escape velocity from the earth’s surface is 11 kms⁻¹.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

Options

(a) 22 kms⁻¹
(b) 11 kms⁻¹
(c) 5.5 kms⁻¹
(d) 16.5 kms⁻¹

Correct Answer:

22 kms⁻¹

Explanation:

No explanation available. Be the first to write the explanation for this question by commenting below.

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Topics: Laws of Motion (103)
Subject: Physics (2479)

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1 Comment on The escape velocity from the earth’s surface is 11 kms⁻¹.A certain planet

  1. Escape velocity is related to the velocity by, v² = 2 GM/r
    where r is the radius of the planet. G is a constant.
    mass density(ρ) = mass/volume
    So mass = Mass Density X volume

    Mass = 4/3 πr³ x ρ

    The escape velocity now is
    v² = 2 G/r x 4/3 πr³ x ρ
    Let Vs be the escape velocity of Earth and Vp be the escape velocity of planet.
    where Rs is the radius of the Earth and Rp is the radius of the planet.
    Also given that ρ is same on both planets.

    Take the ratio
    Vs²/Vp² = Rs²/Rp² as density is same in Earth and the other planet.
    Given that Rp=2 Rs.
    So, Escape velocity will be
    Vp² = Vs² * Rp²/Rs²= 121*4= 484
    => Vp=22 km/s

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