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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)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A uranium nucleus ₉₂ U ²³⁸ emits an α-particle and a β-particle in succession.
- A solid sylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 ms⁻¹
- The magnetic susceptibility does not depend upon the temperature in
- Photoelectric emision occurs only when the incident light has more than a certain minimum
- The radius of curvature of a convave mirror is 24 cm and the imager is magnified
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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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