Question
Question: A satellite with kinetic energy E revolving round the earth in a circular orbit. The minimum additio...
A satellite with kinetic energy E revolving round the earth in a circular orbit. The minimum additional kinetic energy required for it to escape into outer space is
A. 2E
B. 2E
C. 2E
D. E
Solution
The satellite orbiting the earth is held in its orbital constantly, due to the gravitational force. We can solve this problem when we find the total mechanical energy possessed by the satellite due to gravitational force so that an equal amount of energy can be supplied to the satellite to nullify the gravitational force and escape the orbit.
Complete step by step answer:
The satellite orbiting around the Earth is held by its gravitational force. Because of the gravitational force, there is a centripetal force acting on the satellite. This centripetal force constantly acting on the satellite, causes it to move in circular orbits at a constant velocity, v.
This velocity is given by –
v=R+hGM
where M is the mass of earth, R is the radius of earth and G is the gravitational constant and h = height at which the satellite is present.
Since, the value of h is very less compared to R, we can say, R+h≈R
Thus,
⇒v=RGM
Due to this velocity, the kinetic energy possessed by the satellite in its orbit is –
E=21mv2
where m = mass of the satellite.
Substituting,
⇒E=21RGMm
Due to Earth’s gravitational field, the potential energy of the satellite is given by –
P=−R+hGmM
Since, R+h≈R
⇒P=−RGmM
The total mechanical energy is the sum of potential energy and the kinetic energy.
Hence, we have
T=P+E
⇒T=21RGMm−RGMm=−21RGMm
⇒T+21RGMm=0
From the equation of kinetic energy, we have –
⇒T+E=0
∴T=−E
So, in order for the satellite to escape, energy equivalent to the total energy i.e. the kinetic energy must be supplied.
Hence, the correct option is Option D.
Note:
The minimum value of the velocity that the object should have, to escape the Earth’s gravitational influence, is termed as the escape velocity. The value of escape velocity for Earth is 11kms−1. This must be the minimum velocity that should be substituted in the above value for this equation to remain correct.