Question

At what distance above earth would a satellite have a period of 125 min?

Answer 1

Answer : $h = 2.02\times 10^{6}\ m$

Explanation :  We know that the time period

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

angular acceleration $\omega$

Now,  we know

Centripetal acceleration = gravitational acceleration

$\dfrac{GM}{r^{2}} = \dfrac {4\pi^{2}r}{T^{2}}$

$r ^{3} = \dfrac{GMT^{2}}{4\pi^{2}}$

Now, put the value of G, M and T

$r^{3} = \dfrac{6.67\times10^{-11}\times5.9\times10^{24}\times(7500)^{2}}{4\times(3.14)^{2}}$

$r^{3} = 5.6128\times10^{20}\ m^{3}$

$r = 8.25\times 10^{6}\ m$

Now, we know the height of satellite from the earth surface

$r = R+ h$

$h = 8.25\times10^{6} - 6.385\times10^{6}$

$h = 2.02\times 10^{6}\ m$

Where, h is the distance of the satellite above earth surface.

Answer 2

Rw^2 = GmM/r^2 
 Leads to 
 w^2 r^3 = GM 
 (2pi /T) ^2 r^3 = GM 
 4pi^2 r^3 = GM T^2 
 r^3 = GM T^2 / 4pi^2 
 Work out r^3 then r. 
 T = 125 min = 125(60) = 7500 s 
 R = 6.38E6 m 
 m = 5.97E24 kg 
 G = 6.673E-11 
 r= 8279791.78 m
 Since r = radius R of Earth + height above urface,h 
 h = r - R =  8279791.78 - 6.38E6 =  1899791.78 m
 h= 1899.79178 Km