Question:

Rohini satellite at a height of $500 \,km$ and INSAT-B at a height of $3600\, km$ from surface of earth, then relation between their orbital velocity $\left(v_{R}, v_{I}\right)$ is

  • $\left(v_{R} > v_{I}\right)$
  • $\left(v_{R}< v_{I}\right)$
  • $\left(v_{R}=v_{I}\right)$
  • No relation
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The Correct Option is A

Solution and Explanation

Orbital velocity of a satellite is given by $v_{o}=\sqrt{\frac{G M_{e}}{r}}$ where $G=$ gravitational constant. $M_{e}=$ mass of the earth $r=$ distance of the satellite from the centre of the earth $\therefore v_{o} \propto \frac{1}{\sqrt{r}}$ As the distance of INSAT-B from the centre of the earth is greater than the distance of Rohini from the centre of the earth $\therefore v_{R} > v_{I}$
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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].