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 of determining the speed \( v \) of a test mass undergoing circular motion in a spherically symmetric gravitational system, we will analyze two cases based on the distance \( r \) from the center of the system.

### Step-by-Step Solution

**Step 1: Define the mass density and gravitational field**

Given the mass density:
\[ \rho = \begin{cases} \rho_0 & \text{for } r < R \\ 0 & \text{for } r \geq R \end{cases} \]

The gravitational field \( g \) at a distance \( r \) from the center can be derived using Gauss's law for gravity.

**Step 2: Case 1: For \( r \geq R \)**

When \( r \geq R \), the gravitational field \( g \) is due to the entire mass enclosed within radius \( R \). The mass \( M \) within radius \( R \) can be calculated as:
\[ M = \rho_0 \cdot \frac{4}{3} \pi R^3 \]

The gravitational field \( g \) at distance \( r \) (where \( r \geq R \)) is given by:
\[ g = \frac{GM}{r^2} = \frac{G \left( \rho_0 \cdot \frac{4}{3} \pi R^3 \right)}{r^2} \]

**Step 3: Relate gravitational force to centripetal force**

For circular motion, the centripetal force required is provided by the gravitational force:
\[ \frac{mv^2}{r} = mg \]
Substituting for \( g \):
\[ \frac{mv^2}{r} = m \cdot \frac{G \left( \rho_0 \cdot \frac{4}{3} \pi R^3 \right)}{r^2} \]
Cancelling \( m \) from both sides:
\[ \frac{v^2}{r} = \frac{G \left( \rho_0 \cdot \frac{4}{3} \pi R^3 \right)}{r^2} \]
Rearranging gives:
\[ v^2 = \frac{G \left( \rho_0 \cdot \frac{4}{3} \pi R^3 \right)}{r} \]
Thus, we find:
\[ v \propto \frac{1}{\sqrt{r}} \]

**Step 4: Case 2: For \( r < R \)**

For \( r < R \), the gravitational field \( g \) is due only to the mass within radius \( r \):
\[ M' = \rho_0 \cdot \frac{4}{3} \pi r^3 \]
Thus, the gravitational field \( g \) is:
\[ g = \frac{GM'}{r^2} = \frac{G \left( \rho_0 \cdot \frac{4}{3} \pi r^3 \right)}{r^2} = \frac{G \rho_0 \cdot \frac{4}{3} \pi r}{3} \]

**Step 5: Relate gravitational force to centripetal force**

Again, using the centripetal force:
\[ \frac{mv^2}{r} = mg \]
Substituting for \( g \):
\[ \frac{mv^2}{r} = m \cdot \frac{G \rho_0 \cdot \frac{4}{3} \pi r}{3} \]
Cancelling \( m \):
\[ \frac{v^2}{r} = \frac{G \rho_0 \cdot \frac{4}{3} \pi r}{3} \]
Rearranging gives:
\[ v^2 = \frac{G \rho_0 \cdot \frac{4}{3} \pi}{3} r^2 \]
Thus, we find:
\[ v \propto r \]

### Conclusion

The speed \( v \) as a function of distance \( r \) from the center of the system is represented by:
- For \( r < R \): \( v \propto r \) (linear relationship)
- For \( r \geq R \): \( v \propto \frac{1}{\sqrt{r}} \) (inverse square root relationship)