Question

A satellite of the earth is revolving in a circular orbit with a uniform velocity V. If the gravitational force suddenly disappears, the satellite will
(A) Continue to move with the same velocity in the same orbit
(B) Move tangentially to the original orbit with velocity V.
(C) Fall down with increasing velocity.
(D) Come to a stop somewhere in its original orbit.

Answer 1

Hint : A satellite revolves in a circular orbit because the forces are balanced. The gravitational force of the earth is balanced by the centripetal force due to the motion of the satellite. So by equating the two forces we will get the answer.

Formula Used: The following formulae have been used below.
  $ \Rightarrow {{\text{F}}_{{\text{gravity}}}}{\text{ = G}}\dfrac{{{{\text{M}}_{{\text{Earth}}}}{\text{m}}}}{{{{\text{R}}_{{\text{Earth}}}}^{\text{2}}}} $ where $ {{\text{F}}_{{\text{gravity}}}} $ is the force of gravity, $ {\text{G}} $ is the gravitational constant, $ {{\text{M}}_{{\text{Earth}}}} $ is the mass of the Earth, $ {{\text{R}}_{{\text{Earth}}}} $ is the radius of the Earth, $ {\text{m}} $ is the mass of the satellite.
  $ \Rightarrow {{\text{F}}_{{\text{centripetal}}}}{\text{ = m}}\dfrac{{{{\text{V}}^{\text{2}}}}}{{\text{r}}} $ where $ {{\text{F}}_{{\text{centripetal}}}} $ is the centripetal force, $ {\text{m}} $ is the mass of the satellite, $ {\text{V}} $ is the velocity of the satellite and $ {\text{r}} $ is the radial distance measured from the centre of the Earth.

Complete step by step answer
Satellites orbiting the Earth are close enough to be acted upon by the gravitational force of the Earth. The centripetal force that keeps the satellite in its orbit is provided by the force of gravity of the Earth. In other words, the force of gravity that keeps the satellite in its circular orbit acts normally to the direction of motion of the satellite. This is why; the gravitational force does not do any work on the satellite if it is in a circular orbit.
According to Newton’s Law of Gravitation,
  $ \Rightarrow {{\text{F}}_{{\text{gravity}}}}{\text{ = G}}\dfrac{{{{\text{M}}_{{\text{Earth}}}}{\text{m}}}}{{{{\text{R}}_{{\text{Earth}}}}^{\text{2}}}} $ where $ {{\text{F}}_{{\text{gravity}}}} $ is the force of gravity, $ {\text{G}} $ is the gravitational constant, $ {{\text{M}}_{{\text{Earth}}}} $ is the mass of the Earth, $ {{\text{R}}_{{\text{Earth}}}} $ is the radius of the Earth, $ {\text{m}} $ is the mass of the satellite.
The centripetal force is necessary to keep the satellite moving in a curved path, and it is directed towards the centre of rotation. The centripetal force acting on the satellite is given by,
  $ \Rightarrow {{\text{F}}_{{\text{centripetal}}}}{\text{ = m}}\dfrac{{{{\text{V}}^{\text{2}}}}}{{\text{r}}} $ where $ {{\text{F}}_{{\text{centripetal}}}} $ is the centripetal force, $ {\text{m}} $ is the mass of the satellite, $ {\text{V}} $ is the velocity of the satellite and $ {\text{r}} $ is the radial distance measured from the centre of the Earth.
For a circular orbit, as described in the question,
  $ \Rightarrow {{\text{F}}_{{\text{centripetal}}}} = {{\text{F}}_{{\text{gravity}}}} $
  $ \Rightarrow {\text{m}}\dfrac{{{{\text{V}}^{\text{2}}}}}{{\text{r}}} = {\text{G}}\dfrac{{{{\text{M}}_{{\text{Earth}}}}{\text{m}}}}{{{{\text{R}}_{{\text{Earth}}}}^{\text{2}}}} $ .
If the gravitational pull of the Earth disappears, the centripetal force supplied by the force of gravity of the Earth will also disappear. As a result the satellite moves in a perfectly straight line, in a tangential direction to the circular path. The velocity of revolution will not be affected and its value will be equal to V.
$ \therefore $ The correct option is Option B.

Note
We should always keep it in mind that the satellite only moves in a circular path as long as the centripetal force and gravitational force of the Earth are balanced. The inertia of direction is a major factor in this case.