ExamBro
ExamBro
KCET · Physics · Gravitation

The total energy of a satellite in a circular orbit at a distance \((\mathrm{R}+\mathrm{h})\) from the centre of the Earth varies as [ \(R\) is the radius of the Earth and \(h\) is the height of the oribit from Earth's surface]

  1. A \(-\frac{1}{(\mathrm{R}+\mathrm{h})}\)
  2. B \(\frac{1}{(\mathrm{R}+\mathrm{h})^2}\)
  3. C \(-\frac{1}{(\mathrm{R}+\mathrm{h})^2}\)
  4. D \(\frac{1}{(\mathrm{R}+\mathrm{h})}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(-\frac{1}{(\mathrm{R}+\mathrm{h})}\)

Step-by-step Solution

Detailed explanation

Total Energy (E) of a Satellite:
\(E=-\frac{G M m}{2 r}\)
Where:
- \(G\) is the gravitational constant,
- \(M\) is the mass of the Earth,
- \(m\) is the mass of the satellite,
- \(r=R+h\) is the distance from the center of the Earth (i.e., Earth's radius \(R\) plus height \(h\) above the surface).
Therefore:
\(E \propto-\frac{1}{r}=-\frac{1}{R+h}\)
Final Answer:
The total energy of a satellite in a circular orbit varies inversely with the distance from the center of the Earth, i.e.,
\(E \propto-\frac{1}{R+h}\)