Question:

A geostationary satellite is orbiting the earth at a height of $5\,R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2\,R$ from the surface of the earth is

Updated On: May 4, 2024
  • 5
  • 10
  • $6\sqrt2$
  • $\frac {6}{\sqrt2}$
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The Correct Option is C

Solution and Explanation

According to Kepler's third law, $T ^{2} \propto r ^{3}$
$\frac{ T _{2}}{ T _{1}}= \left(\frac{ r _{2}}{ r _{1}}\right)^{3 / 2}$
Given $T _{1}=24\, h$
$r _{1}= R + h _{1}= R +5 R =6 R$
$r _{2}= R + h _{2}= R +2 R =3 R$
On solving we get
$T _{2}=6 \sqrt{2}\, h$
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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].