Question

If suddenly the gravitational force of attraction between the earth and a satellite revolving around it becomes zero, then the satellite will
A) Continue to move in its orbit with the same velocity.
B) Move tangentially to the orbit and escape away from its orbit.
C) Become stationary in its orbit.
D) Move towards the earth.

Answer 1

Hint: When a satellite revolves around the earth then a centrifugal force is created due to the motion of the satellite which is directed outward from the center of the earth. But the earth’s gravitational force acts towards the center of the earth and opposes the satellite to fly away. At a stable orbit, these two forces balance each other.

Formula used:
At orbit forces on the satellite is given by
$
  \dfrac{{m{v^2}}}{R} = mg \\
  \therefore \dfrac{{{v^2}}}{R} = g \\
 $ …………………….. (1)
Where,
$m$ is the mass of the satellite,
$v$ is its linear velocity at any particular moment,
$R$ is the radius of the orbit,
$g$ is the acceleration due to gravity.
Here, the direction of the velocity v of the satellite is always tangential to the orbit of it (see the picture).

Complete step by step answer:
As the formula says, at equilibrium i.e. when the satellite revolves around the earth in a stable orbit the centrifugal force and the gravitational force of the earth balance each other.

Now, if suddenly the gravitational force of earth becomes zero then the RHS of eq.(1) turns 0. Hence, the term $\dfrac{{{v^2}}}{R}$ must be zero. Now, $v$ can’t be 0 because no external force is applied on the satellite to stop its motion. Hence, $R$ must go to infinity. So, the satellite starts moving out at a velocity v which is tangential to the orbit towards infinity.

The satellite will Move tangentially to the original orbit with the same velocity. Hence option (B) is correct.

Note:
Many students might get confused as the relation in eq.(1) can be written in terms of the angular velocity $\omega $ as $m{\omega ^2}R = mg$. Now, making RHS of this equation 0 implies $R$ must be 0. But this form of the eq.(1) is misleading as $\omega $itself is inversely proportional to $R$. Hence, this conclusion is not true.