A spherically symmetric gravitational system of particles has a mass density` rho={(rho_0,for, r,lt,R),(0,for,r,gt,R):}` where`rho_0` is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed v as a function of distahce `r(0ltrltOO)` form the centre of the system is represented by

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Explanation

To solve the problem, we need to analyze the gravitational field created by a spherically symmetric mass distribution and how it affects the motion of a test mass in circular motion. The mass density is given as:

\[ \rho = \begin{cases} \rho_0 & \text{for } r < R \\ 0 & \text{for } r > R \end{cases} \]

where \(\rho_0\) is a constant.

### Step-by-Step Solution:

1. **Understanding the Gravitational Force**:
For a test mass \(m\) undergoing circular motion at a distance \(r\) from the center, the gravitational force provides the necessary centripetal force. The centripetal force is given by:
\[ F_c = \frac{mv^2}{r} \]
where \(v\) is the speed of the test mass.

2. **Calculating the Mass Inside Radius \(r\)**:
The mass \(M\) enclosed within a radius \(r\) (for \(r < R\)) can be calculated using the volume and density:
\[ M = \rho_0 \cdot \text{Volume} = \rho_0 \cdot \frac{4}{3} \pi r^3 \]

3. **Using Newton's Law of Gravitation**:
The gravitational force acting on the test mass \(m\) due to the enclosed mass \(M\) is given by:
\[ F_g = \frac{GMm}{r^2} \]
where \(G\) is the gravitational constant.

4. **Setting the Forces Equal**:
For circular motion, we set the gravitational force equal to the centripetal force:
\[ \frac{mv^2}{r} = \frac{GMm}{r^2} \]
Here, we can cancel \(m\) from both sides:
\[ \frac{v^2}{r} = \frac{GM}{r^2} \]

5. **Substituting for \(M\)**:
Substitute \(M = \frac{4}{3} \pi r^3 \rho_0\) into the equation:
\[ \frac{v^2}{r} = \frac{G \left(\frac{4}{3} \pi r^3 \rho_0\right)}{r^2} \]
This simplifies to:
\[ v^2 = \frac{4}{3} G \pi \rho_0 r \]
Thus, we find:
\[ v \propto \sqrt{r} \]

6. **For \(r > R\)**:
When \(r > R\), the mass \(M\) enclosed is constant (equal to the total mass of the sphere):
\[ M = \frac{4}{3} \pi R^3 \rho_0 \]
The gravitational force acting on the test mass is:
\[ F_g = \frac{GMm}{r^2} \]
Setting the forces equal again:
\[ \frac{mv^2}{r} = \frac{GMm}{r^2} \]
Canceling \(m\) gives:
\[ \frac{v^2}{r} = \frac{G \left(\frac{4}{3} \pi R^3 \rho_0\right)}{r^2} \]
This simplifies to:
\[ v^2 = \frac{4}{3} G \pi R^3 \rho_0 \cdot \frac{1}{r} \]
Thus, we find:
\[ v \propto \frac{1}{\sqrt{r}} \]

### Final Result:
- For \(r < R\): \(v \propto \sqrt{r}\)
- For \(r > R\): \(v \propto \frac{1}{\sqrt{r}}\)