Potential energy in circular orbits

Classical Mechanics Level 3

Two particles of identical mass are moving in circular orbits under a potential given by V ( r ) = K r − n V(r) = Kr^{-n} , where K K is a constant.

If the radii of their orbits are r 1 , r 2 r_1,r_2 and their speeds are v 1 , v 2 v_1, v_2 , respectively, then which of the following equations must be true?

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v 1 2 r 1 n = v 2 2 r 2 n v_1 ^2 r_1 ^n = v_2 ^2 r_2 ^n v 1 2 r 1 − n = v 2 2 r 2 − n v_1 ^2 r_1 ^{-n} = v_2 ^2 r_2 ^{-n } v 1 2 r 1 = v 2 2 r 2 v_1 ^2 r_1 = v_2 ^2 r_2 v 1 2 r 1 2 − n = v 2 2 r 2 2 − n v_1 ^2 r_1 ^{2-n} = v_2 ^2 r_2 ^{2-n}

Md Zuhair by Md Zuhair , 20, India

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1 solution

Aryaman Maithani
Jun 30, 2018

Given: V ( r ) = K r − n V(r) = Kr^{-n}

Using F ( r ) = − d V d r F(r) = -\frac{dV}{dr}

F ( r ) = K n r − n − 1 F(r) = Knr^{-n-1}

When in orbit, the centripetal force is provided by the above force, so equating the two:

m v 2 r = K n r − n − 1 \frac{mv^2}{r} = Knr^{-n-1}

⟹ v 2 r n = K n m \implies v^2r^n = \frac{Kn}{m}

As both particles have the same mass, v 2 r n v^2r^n must be equal for both.

∴ v 1 2 r 1 n = v 2 2 r 2 n \therefore \boxed{ v_1 ^2 r_1 ^n = v_2 ^2 r_2 ^n}

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The given thing is Potential, Not potential energy. As constant will cancel out, thats why you came to the answer I guess.

Md Zuhair - 2 years, 11 months ago

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