JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 7)
The mass density of a spherical galaxy varies
as
$${K \over r}$$ over a large distance ‘r’ from its centre.
In that region, a small star is in a circular orbit
of radius R. Then the period of revolution, T
depends on R as :
T2 $$ \propto $$ R
T2 $$ \propto $$ R3
T $$ \propto $$ R
T2 $$ \propto $$ $${1 \over {{R^3}}}$$
Explanation
_2nd_September_Morning_Slot_en_7_2.png)
dm = $$\rho $$dv
$$ \Rightarrow $$ dm = $$\left( \frac{k}{r} \right)$$ (4$$\pi $$r2dr)
$$ \Rightarrow $$ dm = 4$$\pi $$krdr
M = $$\int\limits^{R}_{0} dm$$ = $$\int\limits^{R}_{0} 4\pi krdr$$
$$ \Rightarrow $$ M = $$4\pi k\left[ \frac{r^{2}}{2} \right]^{R}_{0} $$
$$ \Rightarrow $$ M = 2$$\pi $$kR2
For circular motion gravitational force will provide required centripetal force.
$$\frac{GMm}{R^{2}} $$ = $$\frac{mv^{2}}{R} $$
$$ \Rightarrow $$ $$\frac{G\left( 2\pi kR^{2}\right) m}{R^{2}} $$ = $$\frac{mv^{2}}{R} $$
$$ \Rightarrow $$ v = $$\sqrt{2\pi GkR} $$
Time period, T = $$\frac{2\pi R}{v} $$
= $$\frac{2\pi R}{\sqrt{2\pi GkR} } $$ $$ \propto $$ $$\sqrt{R} $$
$$ \Rightarrow $$ T2 $$ \propto $$ R
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