Question

If the length of the day is $T$, the height of that TV satellite above the earth's surface which always appears stationary from earth, will be : $( R =$ radius of earth, $M =$ Mass of earth, $G =$ Gravitational constant)

• A. $h =\left[\frac{4 \pi^{2} GM }{ T ^{2}}\right]^{1 / 3}$
• B. $h=\left[\frac{4 \pi GM }{ T ^{2}}\right]^{1 / 2}- R$
• C. $h=\left[\frac{G M T^{2}}{4 \pi^{2}}\right]^{1 / 3}-R$
• D. $h=\left[\frac{G M T^{2}}{4 \pi^{2}}\right]^{1 / 3}+R$

Answer 1

Answer: (C)

The time period must be $T$.
Let the height be $h$ radius of the orbit $a=R$ + h
From the kepler third law
$T ^{2}=\frac{( a )^{3}\left(4 \pi^{2}\right)}{ GM }$
$T ^{2}=\frac{( R + h )^{3}\left(4 \pi^{2}\right)}{ GM }$
$h =\left[\frac{ T ^{2} GM }{4 \pi^{2}}\right]^{1 / 3}- R$