A satellite is revolving around earth in its equatorial plane with a period `T`. If the radius of earth suddenly shrinks to half without change in the mass. Then, the new period of revolution will be

Options

A. `8 T`
B. `2sqrt(2) T`
C. `2T`
D. `T`

Explanation

To solve the problem, we need to understand how the period of a satellite in orbit is affected by the radius of the Earth. The period \( T \) of a satellite in a circular orbit is given by the formula:

\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]

where:
- \( T \) is the orbital period,
- \( r \) is the distance from the center of the Earth to the satellite,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth.

### Step-by-Step Solution:

1. **Initial Conditions**:
Let the initial radius of the orbit be \( r \) (which is the radius of the Earth plus the altitude of the satellite). The period of the satellite is given as \( T \).

2. **Change in Radius**:
According to the problem, the radius of the Earth shrinks to half. Therefore, the new radius \( r' \) of the orbit will be:
\[ r' = \frac{r}{2} \]

3. **New Period Calculation**:
We need to find the new period \( T' \) using the same formula:
\[ T' = 2\pi \sqrt{\frac{(r')^3}{GM}} \]
Substituting \( r' = \frac{r}{2} \):
\[ T' = 2\pi \sqrt{\frac{\left(\frac{r}{2}\right)^3}{GM}} = 2\pi \sqrt{\frac{\frac{r^3}{8}}{GM}} = 2\pi \sqrt{\frac{r^3}{8GM}} \]

4. **Relating New Period to Old Period**:
We know from the original period:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]
Now, we can express \( T' \) in terms of \( T \):
\[ T' = 2\pi \sqrt{\frac{1}{8}} \sqrt{\frac{r^3}{GM}} = \frac{T}{\sqrt{8}} = \frac{T}{2\sqrt{2}} \]

5. **Final Answer**:
Therefore, the new period of revolution when the radius of the Earth shrinks to half is:
\[ T' = \frac{T}{2\sqrt{2}} \]