AP EAMCET · PHYSICS · Gravitation
If the escape velocity of a body from the surface of the earth is \(11.2 \mathrm{~km} \mathrm{~s}^{-1}\), then the orbital velocity of a satellite in an orbit which is at a height equal to the radius of the earth is
- A \(11.2 \mathrm{~km} \mathrm{~s}^{-1}\)
- B \(2.8 \mathrm{~km} \mathrm{~s}^{-1}\)
- C \(22.4 \mathrm{~km} \mathrm{~s}^{-1}\)
- D \(5.6 \mathrm{~km} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(D) \(5.6 \mathrm{~km} \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
\(v_e = \sqrt{\frac{2GM}{R}}\) \(v_o = \sqrt{\frac{GM}{R+h}}\) For \(h=R\): \(v_o = \sqrt{\frac{GM}{2R}} = \frac{1}{2}\sqrt{\frac{2GM}{R}}\) \(v_o = \frac{v_e}{2}\) \(v_o = \frac{11.2 \mathrm{~km} \mathrm{~s}^{-1}}{2}\) \(v_o = 5.6 \mathrm{~km} \mathrm{~s}^{-1}\)
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