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Given radius of earth ‘R’ and length of a day ‘T’ the height of a geostationary satellite is [G = gravitational constant, M = mass of the earth]

Option: 1

\left ( \frac{4\pi ^{2}GM}{T^{2}} \right )^{\frac{1}{3}}


Option: 2

\left ( \frac{4\pi GM}{T^{2}} \right )^{\frac{1}{3}}-R


Option: 3

(\frac{GMT^2}{4\pi^2})^{\frac{1}{3}}-R


Option: 4

\left ( \frac{GMT^{2}}{4\pi^{2}} \right )+R


Answers (1)

best_answer

As we had learnt in

Height of geostationery satellite -

T=2\pi \sqrt{\frac{r^{3}}{GM}}\simeq 24hr

r=R+h

r\rightarrow direction of satellite from centre of planet

R\rightarrow Radius of planet

T= time period

- wherein

height of geostationery  satellite from surface of earth is h=6R=36000Km

 

 

T=2\pi \sqrt{\frac{r^{3}}{GM}}

T=2\pi \sqrt{\frac{(r+h)^{3}}{GM}}

  h=(\frac{GMT^2}{4\pi^2})^{\frac{1}{3}}-R

Posted by

HARSH KANKARIA

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