InterviewSolution
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InterviewSolution
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A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is …… hours. |
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Answer» As `T^(2)= k r^(3)` or `T prop r^(3//2)` `:. (T_(1))/(T_(2)) = ((r_(2))/(r_(1)))^(3//2)` or `T_(2) = T_(1) ((r_(2))/(r_(1)))^(3//2) = 24 ((2.5 R + R)/(6R + R))^(3//2)` `= 24 ((1)/(2))^(3//2) = 6 sqrt(2) hour` |
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