SATHEE: Gravitation Question 258

Gravitation Question 258

Question: Given radius of Earth ?R? and length of a day ?T? the height of a geostationary satellite is [G?Gravitational Constant, M?Mass of Earth][MP PMT 2002]

Options:

A) $[{(\frac{4 π^{2} G M}{T^{2}})}^{1 / 3}]$

B) $[{(\frac{4 π G M}{R^{2}})}^{1 / 3} - R]$

C) $[{(\frac{G M T^{2}}{4 π^{2}})}^{1 / 3} - R]$

D) $[{(\frac{G M T^{2}}{4 π^{2}})}^{1 / 3} + R]$

Show Answer

Answer:

Correct Answer: C

Solution:

$[T = 2 π \sqrt{\frac{r^{3}}{G M}}, \Rightarrow, T^{2} = \frac{4 π^{2}}{G M} {(R + h)}^{3}]$

$[\Rightarrow R + h = {[\frac{G M T^{2}}{4 π^{2}}]}^{1 / 3} \Rightarrow, h = {[\frac{G M T^{2}}{4 π^{2}}]}^{\frac{1}{3}} - R]$

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Gravitation Question 258

Question: Given radius of Earth ?R? and length of a day ?T? the height of a geostationary satellite is [G?Gravitational Constant, M?Mass of Earth][MP PMT 2002]

Options:

Answer:

Solution:

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