A satellite A of mass m is at a distance of r from the centre of the earth. Another satellite B of mass 2m is at distance of 2r from the earth's centre. Their time periode are in the ratio of

Options

A. `1:2`
B. `1:16`
C. `1:32`
D. `1:2sqrt(2)`

Answer

To solve the problem of finding the ratio of the time periods of two satellites A and B, we will follow these steps:

### Step 1: Understand the formula for the time period of a satellite
The time period \( T \) of a satellite in orbit is given by the formula:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]
where:
- \( r \) 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 2: Calculate the time period for satellite A
For satellite A, the distance from the center of the Earth is \( r \). Therefore, the time period \( T_A \) can be calculated as:
\[ T_A = 2\pi \sqrt{\frac{r^3}{GM}} \]

### Step 3: Calculate the time period for satellite B
For satellite B, the distance from the center of the Earth is \( 2r \). Therefore, the time period \( T_B \) can be calculated as:
\[ T_B = 2\pi \sqrt{\frac{(2r)^3}{GM}} = 2\pi \sqrt{\frac{8r^3}{GM}} = 2\pi \cdot 2\sqrt{\frac{r^3}{GM}} = 4\pi \sqrt{\frac{r^3}{GM}} \]

### Step 4: Find the ratio of the time periods
Now, we can find the ratio of the time periods \( \frac{T_A}{T_B} \):
\[ \frac{T_A}{T_B} = \frac{2\pi \sqrt{\frac{r^3}{GM}}}{4\pi \sqrt{\frac{r^3}{GM}}} \]
The \( 2\pi \) and \( \sqrt{\frac{r^3}{GM}} \) terms cancel out:
\[ \frac{T_A}{T_B} = \frac{1}{2} \]

### Step 5: Simplify the ratio
Since we have \( T_A = 2\pi \sqrt{\frac{r^3}{GM}} \) and \( T_B = 4\pi \sqrt{\frac{r^3}{GM}} \), we can express the ratio as:
\[ T_A : T_B = 1 : 2 \]

### Final Answer
Thus, the ratio of the time periods of satellite A to satellite B is:
\[ T_A : T_B = 1 : 2 \]