Question

The expression for critical velocity of a satellite orbiting the earth is derived using the condition
(A) Gravitational force on the satellite = weight of the satellite
(B) Gravitational force on the satellite = Centripetal force
(C) Gravitational force on the satellite =Atmospheric pressure on a given area
 (D) Gravitational force on the satellite = Thrust of the engine in a satellite

Answer 1

Hint Gravitational force on a body is responsible for its rotation around the other body of greater mass. Also, when a body revolves around the other body, it exerts a centripetal force on it which is also directed towards the center.

Complete step by step solution:
Let's take the mass of the satellite as m and mass of the earth as M. If the satellite is thrown from a very high tower of height say R+h, it will follow a horizontal projectile's path. This satellite is bound to land somewhere on the surface of the earth. If its velocity is increased or the height from which it is thrown in increase, it will travel a greater distance. If we keep on increasing the value, a condition will occur, when the satellite travels along the curvature of the earth. After a threshold value, the satellite will make a complete circle around the earth. At this point, the centripetal force acting on the satellite will be equal to the gravitational pull on the satellite from the earth.
So, the correct answer is option B.
Note The minimum velocity v of the satellite at a height h above the surface of the earth will be :
 $$\begin{gathered} {\displaystyle \frac{m{v}^2}{r}}={\displaystyle \frac{GmM}{r^2}} \\ {v}^2={\displaystyle \frac{GM}{r}} \\ v=\sqrt{{\displaystyle \frac{GM}{R+h}}} \end{gathered}$$