A light planet is revolving around a massive star in a circular orbit of radius R with a period of revolution T. If the force of attraction between planet and star is proportional to R−3/2 then choose the correct option :T2∝R5/2T2∝R7/2T2∝R3/2T2∝R3
Question
A light planet is revolving around a massive star in a circular orbit of radius R with a period of revolution T. If the force of attraction between planet and star is proportional to R−3/2 then choose the correct option :T2∝R5/2T2∝R7/2T2∝R3/2T2∝R3
Solution
The force of attraction between the planet and the star is given by F = GMm/R^2, where G is the gravitational constant, M is the mass of the star, m is the mass of the planet, and R is the distance between them.
Given that the force of attraction is proportional to R^-3/2, we can write F = kR^-3/2, where k is a constant of proportionality.
The centripetal force required to keep the planet in its circular orbit is given by F = mRω^2, where ω is the angular velocity of the planet.
Setting these two expressions for the force equal to each other, we get:
GMm/R^2 = mRω^2 => GM = R^3ω^2 => ω^2 = GM/R^3
The period of revolution T is related to the angular velocity by ω = 2π/T.
Substituting this into the equation above, we get:
(2π/T)^2 = GM/R^3 => T^2 = (4π^2/GM)R^3
Since GM and 4π^2 are constants, we can see that T^2 is proportional to R^3. Therefore, the correct option is T^2∝R^3.
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