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A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is :
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- 5
- 10
- 6 √2
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6 √2
Correct Option: C
According to Kelpner’s law of period T2 ∝ R3
| = | = | = 8 | ||||
| T22 | R23 | (3R)3 |
| = 8 | ||
| T22 |
| T22 = | = 72 = 36 × 2 | |
| 8 |
T2 = 6 √2
