Question:

The mean radius of earth is $R$, its angular speed on its own axis is $\omega$ and the acceleration due to gravity at earth's surface is $g$. What will be the radius of the orbit of a geostationary satellite ?

Updated On: Jun 27, 2024
  • $(R^2 g / {\omega}^2 ) ^{1/3}$
  • $(R g / {\omega}^2 ) ^{1/3}$
  • $(R^2 {\omega}^2 /g ) ^{1/3}$
  • $(R^2 g / \omega) ^{1/3}$
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The Correct Option is A

Solution and Explanation

$ \frac{ GM m }{r^2} = m {\omega}^2 r \Rightarrow r^3 = \frac{GM}{ {\omega}^2} = \frac{ g R^2}{{\omega}^2}$
$\therefore \, \, \, \, r = ( g R^2 / {\omega}^2 ) ^{1/3}$
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Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

Newton’s Law of Gravitation

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

  • F ∝ (M1M2) . . . . (1)
  • (F ∝ 1/r2) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2

The dimension formula of G is [M-1L3T-2].