Which of the following is true for a satellite orbiting close to the Earth's surface?

The correct answer is option 4) i.e. All of the above

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:

\(\Rightarrow v_0 =\sqrt{\frac{GM}{R+h}}\)

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:

• The orbital velocity is inversely proportional to the distance from the Earth. Hence, the speed becomes maximum as the satellite gets closer to Earth.
• The total energy of the satellite has an inverse relationship with the distance from Earth.

i.e. \(TE \propto \frac{-1}{r}\).

• Hence, as the satellite orbits are nearer to Earth, the total energy of the earth plus satellite is minimum (negative value).
• As the satellite orbits are closer to the Earth, it will be able to complete a revolution in less time period. Hence, its time period becomes minimum.
• Hence, all the above statements are correct.