Answer : [tex]h = 2.02\times 10^{6}\ m[/tex]
Explanation : We know that the time period
[tex]T = 2\pi\sqrt{\dfrac{r^{3}}{GM}}[/tex]
angular acceleration [tex]\omega[/tex]
Now, we know
Centripetal acceleration = gravitational acceleration
[tex]\dfrac{GM}{r^{2}} = \dfrac {4\pi^{2}r}{T^{2}}[/tex]
[tex]r ^{3} = \dfrac{GMT^{2}}{4\pi^{2}}[/tex]
Now, put the value of G, M and T
[tex]r^{3} = \dfrac{6.67\times10^{-11}\times5.9\times10^{24}\times(7500)^{2}}{4\times(3.14)^{2}}[/tex]
[tex]r^{3} = 5.6128\times10^{20}\ m^{3}[/tex]
[tex]r = 8.25\times 10^{6}\ m[/tex]
Now, we know the height of satellite from the earth surface
[tex]r = R+ h[/tex]
[tex]h = 8.25\times10^{6} - 6.385\times10^{6}[/tex]
[tex]h = 2.02\times 10^{6}\ m[/tex]
Where, h is the distance of the satellite above earth surface.
