- 1(current)
Q. No.If \(v_e\) is the escape velocity and \(v_0\) is the orbital velocity of a satellite for orbit close to the earth's surface, then these are related by:
| 1. | \(v_o=v_e\) | 2. | \(v_e=\sqrt{2v_o}\) |
| 3. | \(v_e=\sqrt{2}~v_o\) | 4. | \(v_o=\sqrt{2}~v_e\) |
Subtopic: Orbital velocity |
77%
From NCERT
AIPMT - 2012
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A remote sensing satellite of earth revolves in a circular orbit at a height of \(0.25 \times10^6~\text{m}\) above the surface of the earth. If Earth’s radius is \(6.38\times10^6~\text{m}\) and \(g=9.8~\text{ms}^{-2}\), then the orbital speed of the satellite is:
1. \(7.76~\text{kms}^{-1}\)
2. \(8.56~\text{kms}^{-1}\)
3. \(9.13~\text{kms}^{-1}\)
4. \(6.67~\text{kms}^{-1}\)
Subtopic: Orbital velocity |
52%
From NCERT
NEET - 2015
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