KVPY SA Exam 2016 Question Paper

Question : 29 of 80

Marks: +1, -0

Two particles of identical mass are moving in circular orbits under a potential given by

V(r) = Kr^{-n}

, where K is a constant. If the radii of their orbits are

r_1, r_2

and their speeds are

v_1, v_2

, respectively, then

Solution:

V(r)=Kr^{-n}

gravitational field = E

=-{dV}/{dr}

=(-K)d/{dr}(r^{-n})

=(-K)(-n)r^{-n-1}={Kn}/r^{n+1}

force on mass = E × M, where M = mass of body

∴ME_1={MV_1^2}/r_1,

ME_2={MV_2^2}/r_2

∴V_1^2/V_2^2={r_1E_1 }/{r_2E_2 }

=r_1/r_2 .{Kn}/r_1^{n+1}. r_2^{n+1}/{Kn}

=r_2^n/r_1^n

⇒V_1^2r_1^n=V_2^2r_2^n

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