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

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.

admin:

View Comments (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

Related Questions

  1. Energy stored in the coil of a self inductance 40 mH carrying a steady current of 2A
  2. Radiofrequency choke uses core of
  3. The molar specific heats of an ideal gas at constant pressure and volume are denoted
  4. Electrical force is acting between two charges kept in vacuum. A copper plate
  5. A boat taken 2h to travel 8km and back in still water.If the velocity of water 4 km/h