Question

Given radius of earth ‘R’ and length of a day ‘T’ the height of a geostationary satellite is

[G – Gravitational constant, M – Mass of earth]

• A. $\left( \frac{4\pi^{2}GM}{T^{2}} \right)^{1/3}$
• B. $\left( \frac{4\pi GM}{R^{2}} \right)^{1/3} ⥂ - R$
• C. $\left( \frac{GMT^{2}}{4\pi^{2}} \right)^{1/3} ⥂ - R$
• D. $\left( \frac{GMT^{2}}{4\pi^{2}} \right)^{1/3} ⥂ + R$

Answer 1

Answer: (C)

$\left( \frac{GMT^{2}}{4\pi^{2}} \right)^{1/3} ⥂ - R$

Answer 2

From the expression $h = \left( \frac{T^{2}gR^{2}}{4\pi^{2}} \right)^{1/3} - R$ $\therefore$ $h = \left( \frac{GMT^{2}}{4\pi^{2}} \right)^{1/3} ⥂ - R$ [As $gR^{2} = GM$]