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

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
$T^2=\frac{(a{)}^3\left(4{\pi }^2\right)}{GM}$
$T^2=\frac{(R+h{)}^3\left(4{\pi }^2\right)}{GM}$
$h={\left[\frac{T^{2}GM}{4{\pi }^2}\right]}^{1/3}-R$