If the radius of an orbiting satellite is increased then-

The correct answer is option 2) i.e. its kinetic energy decreases

CONCEPT:

• The time period of the satellite: It is the time taken by the satellite to complete one revolution around the Earth.
• Consider a satellite orbiting the earth at a height h from the surface of the earth of radius R.
• The circumference of orbit of satellite = 2π(R+h)
• The orbital velocity of the satellite at a height h is given by:

\(⇒ v_0 =\sqrt{\frac{GM}{R+h}}\)

Where M is the mass of the Earth.

• The kinetic energy of the satellite of mass m in a circular orbit is

​\(KE = \frac{1}{2}mv^2 = \frac{1}{2}m(\frac{GM}{R+h})\)

• The potential energy at distance (R +h) from the centre of the earth is

​\(PE = -\frac{GmM}{R+h}\)

• The total energy of a circularly orbiting satellite ​is

\(TE = KE +PE =\frac{1}{2}(\frac{GmM}{R+h}) + (-\frac{GmM}{R+h})=-\frac{GmM}{2(R+h)}\)

EXPLANATION:

• Kinetic energy is inversely proportional to the radial distance from the centre of the Earth for an orbiting satellite.
• Hence, when the radius of orbiting is increased for a satellite its kinetic energy decreases.