Orbital Speed in Spherically Symmetric Mass Distribution

To find the orbital speed $V$ of a test mass $m$ as a function of the radial distance $r$ from the centre of the spherically symmetric system, we consider the gravitational force acting as the centripetal force required for circular motion:

$\frac{m V^2}{r} = F_{\text{grav}} = \frac{G M(r) m}{r^2}$

which simplifies to:
$V = \sqrt{\frac{G M(r)}{r}}$

where $M(r)$ is the total mass enclosed within a sphere of radius $r$.

Case 1: Inside the sphere ($r \le R$)

Since the mass density is uniform with $\rho = \rho_0$:
$M(r) = \int_0^r \rho_0 (4\pi x^2)\,dx = \frac{4}{3}\pi \rho_0 r^3$

Substituting $M(r)$ into the expression for orbital speed:
$V = \sqrt{\frac{G \left(\frac{4}{3}\pi \rho_0 r^3\right)}{r}} = \sqrt{\frac{4}{3}\pi G \rho_0} \, r$

Thus, inside the sphere, the speed is directly proportional to the radius:
$V \propto r \quad \text{for } r \le R$
This corresponds to a straight line passing through the origin.

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Case 2: Outside the sphere ($r > R$)

Since $\rho = 0$ for $r > R$, the total enclosed mass remains constant and is equal to the total mass of the sphere:
$M(r) = M_{\text{total}} = \frac{4}{3}\pi \rho_0 R^3$

Substituting $M_{\text{total}}$ into the expression for orbital speed:
$V = \sqrt{\frac{G M_{\text{total}}}{r}} \propto \frac{1}{\sqrt{r}} \quad \text{for } r > R$

Thus, for $r > R$, $V$ decreases non-linearly, approaching zero asymptotically as $r \to \infty$.

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Conclusion:

• For $0 < r \le R$, $V$ increases linearly with $r$.
• For $r > R$, $V$ decreases as $\frac{1}{\sqrt{r}}$.

This behavior matches the graph represented in Option (C).