If the gravitational field intensity at a point is given by `g = (GM)/(r^(2.5))`. Then, the potential at a distance `r` is

Options

A. `(-2GM)/(3r^(1.5))`
B. `(-GM)/(r^(2.5))`
C. `(2GM)/(3r^(1.5))`
D. `(GM)/(r^(3.5))`

Explanation

To find the gravitational potential \( V \) at a distance \( r \) given the gravitational field intensity \( g = \frac{GM}{r^{2.5}} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between gravitational field intensity and potential**:
The gravitational potential \( V \) at a distance \( r \) from a mass \( M \) can be calculated using the relationship:
\[ V = -\int E \, dr \]
where \( E \) is the gravitational field intensity.

2. **Set up the integral**:
Given \( g = \frac{GM}{r^{2.5}} \), we can substitute \( E \) with \( g \):
\[ V = -\int_{\infty}^{r} \frac{GM}{r^{2.5}} \, dr \]

3. **Change the limits of integration**:
We will evaluate the integral from infinity to \( r \):
\[ V = -\int_{\infty}^{r} \frac{GM}{r^{2.5}} \, dr \]

4. **Evaluate the integral**:
The integral can be simplified as follows:
\[ V = -GM \int_{\infty}^{r} r^{-2.5} \, dr \]
The integral of \( r^{-2.5} \) is:
\[ \int r^{-2.5} \, dr = \frac{r^{-1.5}}{-1.5} = -\frac{2}{3} r^{-1.5} \]
Thus, we have:
\[ V = -GM \left[-\frac{2}{3} r^{-1.5}\right]_{\infty}^{r} \]

5. **Apply the limits**:
Now, we substitute the limits into the integral:
\[ V = -GM \left[-\frac{2}{3} r^{-1.5} - 0\right] \]
Since as \( r \) approaches infinity, \( r^{-1.5} \) approaches 0.

6. **Final expression for potential**:
Therefore, we get:
\[ V = \frac{2GM}{3r^{1.5}} \]

### Conclusion:
The potential at a distance \( r \) is given by:
\[ V = -\frac{2GM}{3r^{1.5}} \]