Question:

Assume that the earth moves around the sun in a circular orbit of radius R and there exists a planet which also moves around the sun in circular orbit with an angular speed twice as large as that of the earth. The radius of the orbit of the planet is

Updated On: May 29, 2024
  • $2-\frac{2}{3}R$
  • $2\frac{2}{3}R$
  • $2\frac{-1}{3}R$
  • $\frac{R}{\sqrt{2}}$
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The Correct Option is A

Solution and Explanation

$T^{2} \propto r^{3},\frac{rE}{rP}=\left(\frac{T_{E}}{T_{P}}\right)^{\frac{2}{3}} =\left(\frac{\omega_{P}}{\omega_{E}}\right)^{\frac{2}{3}}$ $\frac{R}{r_{p}} = \left(2\right)^{\frac{2}{3}} , r_{p} =2^{-\frac{2}{3}R}$
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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].