An artificial satellite revolves around the earth in circular orbit of radius `r` with time period `T`. The satellite is made to stop in the orbit which makes it fall onto the earth. Time of fall of the satellite onto the earth is given by

Options

A. `sqrt(3)T/6`
B. `(sqrt(2))/8T`
C. `T/(sqrt(3))`
D. `sqrt(2/3)T/(pi)`

Explanation

To solve the problem of determining the time of fall of an artificial satellite that has been stopped in its orbit and is falling towards the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The satellite is initially in a circular orbit around the Earth with a radius \( r \) and a time period \( T \).

2. **Stopping the Satellite**:
- When the satellite is stopped, it will no longer maintain its circular motion and will begin to fall towards the Earth. The path it takes will be an elliptical trajectory, with the point of stopping being one of the foci of the ellipse.

3. **Determining the Semi-Major Axis**:
- The semi-major axis \( a \) of the elliptical path can be approximated as \( \frac{r}{2} \) since the satellite will fall towards the Earth from its initial height.

4. **Applying Kepler's Third Law**:
- Kepler's Third Law states that the square of the time period \( T \) of an orbit is proportional to the cube of the semi-major axis \( a \) of the orbit:
\[ T^2 \propto r^3 \]
- For the elliptical orbit, we can express this as:
\[ T'^2 \propto \left(\frac{r}{2}\right)^3 \]

5. **Calculating the New Time Period**:
- From the proportionality, we can write:
\[ T'^2 = K \left(\frac{r}{2}\right)^3 = K \frac{r^3}{8} \]
- Since \( T^2 = K r^3 \), we can relate \( T'^2 \) to \( T^2 \):
\[ T'^2 = \frac{T^2}{8} \]
- Taking the square root gives:
\[ T' = \frac{T}{\sqrt{8}} = \frac{T}{2\sqrt{2}} \]

6. **Time of Fall Calculation**:
- The time of fall of the satellite until it reaches the Earth can be approximated as half of the time period of the elliptical orbit:
\[ \text{Time of fall} = \frac{T'}{2} = \frac{T}{2 \cdot 2\sqrt{2}} = \frac{T}{4\sqrt{2}} \]

### Final Answer:
The time of fall of the satellite onto the Earth is given by:
\[ \text{Time of fall} = \frac{T}{4\sqrt{2}} \]