The ratio of the time periods of two satellites at heights `h_1,h_2` from the surface of the earth.

Options

A. `(h_1/h_2)^(3//2)`
B. `((R+h_1)/(R+h_2))^(3//2)`
C. `((R+h_2)/(R+h_1))^(3//2)`
D. `(h_2/h_1)^(3//2)`

Answer

To find the ratio of the time periods of two satellites at heights \( h_1 \) and \( h_2 \) from the surface of the Earth, we can follow these steps:

### Step 1: Understand the relationship between time period and radius
According to Kepler's third law, the square of the time period \( T \) of a satellite is directly proportional to the cube of the radius \( r \) of its orbit. This can be expressed mathematically as:
\[ T^2 \propto r^3 \]
Thus, we can write:
\[ T = k \cdot r^{3/2} \]
where \( k \) is a constant.

### Step 2: Define the radius for each satellite
For two satellites located at heights \( h_1 \) and \( h_2 \) above the Earth's surface, the effective radius \( r \) for each satellite can be defined as:
- For the first satellite at height \( h_1 \):
\[ r_1 = R + h_1 \]
- For the second satellite at height \( h_2 \):
\[ r_2 = R + h_2 \]
where \( R \) is the radius of the Earth.

### Step 3: Write the time periods for each satellite
Using the relationship established in Step 1, we can express the time periods \( T_1 \) and \( T_2 \) for the two satellites:
\[ T_1 = k \cdot (R + h_1)^{3/2} \]
\[ T_2 = k \cdot (R + h_2)^{3/2} \]

### Step 4: Find the ratio of the time periods
Now, we can find the ratio of the time periods \( T_1 \) and \( T_2 \):
\[ \frac{T_1}{T_2} = \frac{(R + h_1)^{3/2}}{(R + h_2)^{3/2}} = \left( \frac{R + h_1}{R + h_2} \right)^{3/2} \]

### Final Answer
Thus, the ratio of the time periods of the two satellites is:
\[ \frac{T_1}{T_2} = \left( \frac{R + h_1}{R + h_2} \right)^{3/2} \]