JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 10)
A test particle is moving in a circular orbit in
the gravitational field produced by a mass
density $$\rho (r) = {K \over {{r^2}}}$$ . Identify the correct relation
between the radius R of the particle's orbit and
its period T
T2/R3 is a constant
TR is a constant
T/R2 is a constant
T/R is a constant
Explanation
For circular motion of particle:
$${{m{V^2}} \over r} = mE$$
$$ = m\left( {{{GM} \over {{r^2}}}} \right)$$
Where $$M = \int\limits_0^r {\left( {4\pi {x^2}dx} \right)\left( {{k \over {{x^2}}}} \right)} = 4\pi kr$$
$$ \Rightarrow {{m{V^2}} \over r} = m\left( {{{G\left( {4\pi k} \right)} \over r}} \right)$$
$$ \Rightarrow $$ V = constant
$$T = {{2\pi r} \over V}$$
$$ \Rightarrow $$ $${T \over R}$$ = Constant
$${{m{V^2}} \over r} = mE$$
$$ = m\left( {{{GM} \over {{r^2}}}} \right)$$
Where $$M = \int\limits_0^r {\left( {4\pi {x^2}dx} \right)\left( {{k \over {{x^2}}}} \right)} = 4\pi kr$$
$$ \Rightarrow {{m{V^2}} \over r} = m\left( {{{G\left( {4\pi k} \right)} \over r}} \right)$$
$$ \Rightarrow $$ V = constant
$$T = {{2\pi r} \over V}$$
$$ \Rightarrow $$ $${T \over R}$$ = Constant
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