Question:

The ratio of the energy required to raise a satellite upto a height $h$ above the earth of radius $R$ to that the kinetic energy of the satellite into that orbit is

Updated On: Jun 7, 2022
  • $R : h$
  • $h : R$
  • $R : 2h$
  • $2h : R$
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The Correct Option is D

Solution and Explanation

Energy required to raise the satellite to a height h from surface of earth is given by,
$U = \frac{GMm}{\left(R +h\right)}- \left(-\frac{GMm}{R}\right)= \frac{GMmh}{R\left(R+h\right)}$
Kinetic energy of satellite is given by,
$K = \frac{1}{2}m\upsilon^{2}_{0} = \frac{1}{2}m \frac{GM}{\left(R +h\right)} \:\:\: \therefore \frac{U}{K}=\frac{2h}{R}$
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Concepts Used:

Escape Speed

Escape speed is the minimum speed, which is required by the object to escape from the gravitational influence of a plannet. Escape speed for Earth’s surface is 11,186 m/sec. 

The formula for escape speed is given below:

ve = (2GM / r)1/2 

where ,

ve = Escape Velocity 

G = Universal Gravitational Constant 

M = Mass of the body to be escaped from 

r = Distance from the centre of the mass