Let `S` be an imaginary closed surface enclosing mass `m`. Let `dvecS` be an element of area on `S`, the direction of `dvecS` being outward from `S`. Let `vecE` be the gravitational intensity at `dvecS`. We define `phi=oint_(s)E*dS`, the integration being carried out over the entire surface `S`.

Options

A. `phi=-Gm`
B. `phi=-4piGm`
C. `phi=-(Gm)/(4pi)`
D. No relation of the type `(a)` , `(b)` or `(c )` can exist

Answer

To solve the problem, we need to calculate the gravitational flux through a closed surface \( S \) that encloses a mass \( m \). The gravitational intensity \( \vec{E} \) at any point on the surface is directed towards the mass \( m \), while the area vector \( d\vec{S} \) is directed outward from the surface.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a closed surface \( S \) enclosing a mass \( m \).
- The gravitational field \( \vec{E} \) at any point on the surface is directed inward towards the mass.

2. **Expressing the Flux**:
- The gravitational flux \( \Phi \) through the surface \( S \) is defined as:
\[ \Phi = \oint_S \vec{E} \cdot d\vec{S} \]
- Here, \( d\vec{S} \) is the outward area vector of the surface element.

3. **Using the Dot Product**:
- The dot product \( \vec{E} \cdot d\vec{S} \) can be expressed as:
\[ \vec{E} \cdot d\vec{S} = |\vec{E}| |d\vec{S}| \cos \theta \]
- Since \( \vec{E} \) is directed inward and \( d\vec{S} \) is directed outward, the angle \( \theta \) between them is \( 180^\circ \) (or \( \pi \) radians), which gives \( \cos \theta = -1 \).

4. **Calculating the Magnitude of \( \vec{E} \)**:
- The gravitational field \( \vec{E} \) at a distance \( r \) from the mass \( m \) is given by:
\[ |\vec{E}| = \frac{Gm}{r^2} \]
- Where \( G \) is the gravitational constant.

5. **Integrating Over the Surface**:
- The total surface area \( A \) of a sphere of radius \( r \) is \( 4\pi r^2 \).
- Therefore, the flux becomes:
\[ \Phi = \oint_S \vec{E} \cdot d\vec{S} = \oint_S \left(-\frac{Gm}{r^2}\right) dS \]
- Substituting \( dS = 4\pi r^2 \):
\[ \Phi = -\frac{Gm}{r^2} \cdot 4\pi r^2 \]

6. **Simplifying the Expression**:
- The \( r^2 \) terms cancel out:
\[ \Phi = -4\pi Gm \]

7. **Final Result**:
- Thus, the gravitational flux through the closed surface \( S \) is:
\[ \Phi = -4\pi Gm \]