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]
- A \(-\frac{1}{(\mathrm{R}+\mathrm{h})}\)
- B \(\frac{1}{(\mathrm{R}+\mathrm{h})^2}\)
- C \(-\frac{1}{(\mathrm{R}+\mathrm{h})^2}\)
- D \(\frac{1}{(\mathrm{R}+\mathrm{h})}\)
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.,
\(E \propto-\frac{1}{R+h}\)
\(E=-\frac{G M m}{2 r}\)
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.,
\(E \propto-\frac{1}{R+h}\)
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