Question:
A test particle is moving in circular orbit in the gravitational field produced by a mass density $r(r)=\frac{\mathrm{K}}{r^{2}}$. Identify the
correct relation between the radius $R$ of the particle's orbit and its period $\mathrm{T}$ :
Correct Option: 1
Solution:
(1) $F=\frac{G M m}{r}=\int a \frac{\rho(d V) m}{r^{2}}$
$=m G \int_{0}^{R} \frac{k}{r^{2}} \frac{4 \pi r^{2} d r}{r^{2}}$
$=-4 \pi k G m\left(\frac{1}{r}\right)_{0}^{R}$
$=-\frac{4 \pi k G m}{R}$
Using Newton's second law, we have
$\frac{m v_{0}^{2}}{R}=\frac{4 \pi k G m}{R}$
or $v_{0}=\mathrm{C}$ (const.)
Time period, $T=\frac{2 \pi R}{v_{0}}=\frac{2 \pi R}{C}$
or $\frac{T}{R}=$ cons
$\rightarrow \quad \rightarrow \quad \rightarrow \quad \rightarrow$