Question

A satellite is moving with a constant speed ‘v’ in a circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of object is
A. $\dfrac{3}{2}m{{v}^{2}}$
B. $m{{v}^{2}}$
C. $2m{{v}^{2}}$
D. $\dfrac{1}{2}m{{v}^{2}}$

Answer 1

Hint: The speed of the satellite and mass of the object is given. We shall use the orbital velocity formula and compare it with kinetic energy of the object ejected from the satellite. It is in accordance with energy conservation.

Complete step by step answer:
Orbital velocity is described as the velocity with which an object revolves around another object. The objects that travel in a uniform circular path around the Earth are known as orbits. The velocity of each orbit depends upon the distance between the object and the center of Earth. It is the velocity given to satellites so that they can revolve around a planet. Orbital velocity can be expressed as,

${{V}_{orbit}}=\sqrt{\dfrac{GM}{R}}$

Where, ‘G’ is the gravitational constant
‘M’ is the mass the object at center
‘R’ is the radius of the orbit
The unit of orbital velocity is $m{{s}^{-1}}$

For a height ‘r’ from the center of the Earth, the orbital velocity is given as

$\sqrt{\dfrac{GM}{r}}$

In gravitational physics of orbits, there are two main energies that get exchanged. They are gravitational potential energy and kinetic energy.
By energy conservation, we know

K.E of object with mass ‘m’+$\left( -\dfrac{GMm}{r} \right)=0+0$
At infinity, gravitational potential energy = Kinetic Energy = 0
KE of ‘m’ =$\dfrac{GMm}{r}={{\left( \sqrt{\dfrac{GM}{r}} \right)}^{2}}m=m{{v}^{2}}$
Therefore, the correct answer for the given question is option (B).

Note: There is another term known as escape velocity which is the minimum velocity required to overcome the gravitational potential of the object and escape to the infinity. The relation between escape and orbital velocity can be written as ${{V}_{e}}=\sqrt{2}{{V}_{o}}$. Its unit is also $m{{s}^{-1}}$.