A particle of mass m is under the influence of the gravitational field of a body of mass M(≫ m). The particle is moving in a circular orbit of radius r₀ with time period T₀ around the mass M. Then, the particle is subjected to an additional central force, corresponding to the potential energy V_c(r) = mα/r³, where α is a positive constant of suitable dimensions and r is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius r₀ in the combined gravitational potential due to M and V_c(r), but with a new time period T₁, then$\left({\mathrm{T}}_{1}^{2}-{\mathrm{T}}_{0}^{2}/{\mathrm{T}}_{1}^{2}\right)$ is given by
[G is the gravitational constant.]

Question

YESPYQ · Accepted Answer

Correct answer: a) $\frac{3\alpha }{{\mathrm{GMr}}_{0}^{2}}$. $\frac{\mathrm{GMm}}{{\mathrm{r}}_{0}^{2}}={\mathrm{mω}}^{2}{\mathrm{r}}_{0}$ $\omega =\sqrt{\frac{\mathrm{GM}}{{\mathrm{r}}_{0}^{3}}}$ ${\mathrm{T}}_{0}^{2}=4{\pi }^{2}(\frac{{\mathrm{r}}_{0}^{3}}{\mathrm{GM}})$ .....(1) $\mathrm{F}=−\frac{{\mathrm{dv}}_{\mathrm{c}}(\mathrm{r})}{\mathrm{dr}}$ $\mathrm{F}=\frac{3\mathrm{mα}}{{\mathrm{r}}^{4}}$ Net force $\frac{\mathrm{GMm}}{{\mathrm{r}}_{0}^{2}}−\frac{3\mathrm{mα}}{{\mathrm{r}}_{0}^{4}}={\mathrm{mω}}^{2}{\mathrm{r}}_{0}$ ${\omega }^{2}=\frac{\mathrm{GM}}{{\mathrm{r}}_{0}^{3}}−\frac{3\alpha }{{\mathrm{r}}_{0}^{5}}$ We know $\omega =\frac{2\pi }{{\mathrm{T}}_{1}}$ $\frac{4{\pi }^{2}}{{\mathrm{T}}_{1}^{2}}=\frac{\mathrm{GM}}{{\mathrm{r}}_{0}^{3}}−\frac{3\alpha }{{\mathrm{r}}_{0}^{5}}$ ${\mathrm{T}}_{1}^{2}=\frac{4{\pi }^{2}}{\frac{\mathrm{GM}}{{\mathrm{r}}_{0}^{3}}−\frac{3\alpha }{{\mathrm{r}}_{0}^{5}}}$ ......(2) From equation (1) and (2) $\frac{{\mathrm{T}}_{1}^{2}−{\mathrm{T}}_{0}^{2}}{{\mathrm{T}}_{1}^{2}}=\frac{3\alpha }{{\mathrm{GMr}}_{0}^{2}}$

A particle of mass m is under the influence of the gravitational field of a body of mass M(≫ m). The particle is moving …

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