Let gravitation field in a space be given as `E = - (k//r)` If the reference point is a distance d where potential is `V_(1)` then relation for potential is .

Options

A. `V = kl n(1)/(V_(i)) + 0`
B. `V =k ln (r/d) + Vi`
C. `V = "ln" (r)/(d_(i)) + KV`
D. `V = "ln" (r)/(d_(i)) + (V_(i))/(k)` .

Answer

To solve the problem, we need to find the relation for gravitational potential \( V \) in terms of the gravitational field \( E \) and the reference potential \( V_1 \) at a distance \( d \).

### Step-by-Step Solution:

1. **Understand the Given Information**:
We are given the gravitational field \( E = -\frac{k}{r} \), where \( k \) is a constant, and \( r \) is the distance from the mass creating the gravitational field.

2. **Relate Gravitational Field to Potential**:
The gravitational field \( E \) is related to the gravitational potential \( V \) by the equation:
\[ E = -\frac{dV}{dr} \]
Substituting the expression for \( E \):
\[ -\frac{k}{r} = -\frac{dV}{dr} \]
This simplifies to:
\[ \frac{dV}{dr} = \frac{k}{r} \]

3. **Separate Variables**:
We can separate the variables to integrate:
\[ dV = \frac{k}{r} dr \]

4. **Integrate Both Sides**:
Now we integrate both sides. The limits for \( r \) will be from \( d \) to \( r \) and for \( V \) from \( V_1 \) to \( V \):
\[ \int_{V_1}^{V} dV = \int_{d}^{r} \frac{k}{r} dr \]

5. **Perform the Integration**:
The left side integrates to:
\[ V - V_1 \]
The right side integrates to:
\[ k \int_{d}^{r} \frac{1}{r} dr = k [\ln(r) - \ln(d)] = k \ln\left(\frac{r}{d}\right) \]

6. **Combine the Results**:
Equating both sides gives us:
\[ V - V_1 = k \ln\left(\frac{r}{d}\right) \]

7. **Rearranging the Equation**:
Finally, we can express \( V \) in terms of \( V_1 \):
\[ V = V_1 + k \ln\left(\frac{r}{d}\right) \]

### Final Result:
The relation for potential \( V \) is:
\[ V = V_1 + k \ln\left(\frac{r}{d}\right) \]