A test particle is moving in a circular orbit | vedclass

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(D) The gravitational field $E$ at a distance $r$ is given by $E = \frac{GM(r)}{r^2}$,where $M(r)$ is the mass enclosed within radius $r$.Given $\rho(

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$T/R$ is a constant

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(D) The gravitational field $E$ at a distance $r$ is given by $E = \frac{GM(r)}{r^2}$,where $M(r)$ is the mass enclosed within radius $r$.Given $ho(r) = \frac{K}{r^2}$,the mass $M(r)$ is calculated as:$M(r) = \int_0^r ho(x) \cdot 4\pi x^2 dx = \int_0^r \frac{K}{x^2} \cdot 4\pi x^2 dx = \int_0^r 4\pi K dx = 4\pi Kr$.Substituting $M(r)$ into the expression for the gravitational field:$E = \frac{G(4\pi Kr)}{r^2} = \frac{4\pi GK}{r}$.For a particle of mass $m$ in a circular orbit of radius $R$,the centripetal force is provided by the gravitational force:$\frac{mV^2}{R} = mE = m \left( \frac{4\pi GK}{R} ight)$.This simplifies to $V^2 = 4\pi GK$,which means the orbital speed $V$ is a constant.The period $T$ of the orbit is given by $T = \frac{2\pi R}{V}$.Since $V$ is constant,$T \propto R$,which implies $\frac{T}{R} = 	ext{constant}$.

$A$ test particle is moving in a circular orbit in the gravitational field produced by a mass density $\rho(r) = \frac{K}{r^2}$. Identify the correct relation between the radius $R$ of the particle's orbit and its period $T$.

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