Question

Geostationary orbit is at a height of
(2) 1000 k.m.
(4) 36000 k.m.
(1) 6 k.m.
(3) 3600 k.m.​

Answer 1

Answer:

the correct answer is option 4.

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Answer 2

Answer:

36000 Km

Explanation:

T = 2 π r / v

T = 2 π r / $\sqrt{(GM/r)}$

T = (2 π / $\sqrt{(GM)}$) $r\sqrt{r}$

$T^{2}$ =  $r^{3}$ (4 $(pi)^{2}$) / GM

$r^{3}$  = $T^{2}$ GM / 4 $(pi)^{2}$

Here T = 24 hours = 86400 s (because time period in Geostationary orbit is equal to one rotation period of earth), G = 6.67 x $10^{-11}$ N $(m/kg)^{2}$, M = 5.972 x $10^{24}$ Kg, pi = 3.1415

Substituting these values, we get R = $10^{7}$

Where R is the distance from the center of the earth

But we need to find the height from surface

R = radius of earth + height from surface

$10^{7}$ = 6.4 x $10^{6}$ + height from surface

Therefore, Height of Geostationary orbit is 3.6 x $10^{6}$  m = 36000 Km