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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
- The current in the windings of a toroid is 2.0 A. There are 400 turns
- The distance travelled by an object along a straight line in time t is given by
- A linear aperture whose width is 0.02 cm is placed immediately in front of a lens
- Young’s modulus of the material of a wire is 18×10¹¹ dyne cm⁻².its value in SI is
- If the radius of star is R and it acts as a black body, what would be the temperature
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