Two satellites are at heights `h_1, h_2`. The ratio of their orbital angular speeds is

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 orbital angular speeds of two satellites at heights \( h_1 \) and \( h_2 \) above the Earth's surface, we can follow these steps:

### Step 1: Define the distances of the satellites from the center of the Earth
Let \( r \) be the radius of the Earth. The distances of the satellites from the center of the Earth are:
- For satellite 1 at height \( h_1 \):
\[ R_1 = r + h_1 \]
- For satellite 2 at height \( h_2 \):
\[ R_2 = r + h_2 \]

### Step 2: Use Kepler's Third Law
According to Kepler's Third Law, the square of the time period \( T \) of a satellite's orbit is directly proportional to the cube of the semi-major axis (which is the distance from the center of the Earth in this case):
\[ T^2 \propto R^3 \]
This can be expressed as:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \]

### Step 3: Relate the time periods to the angular speeds
The angular speed \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Thus, we can express the ratio of angular speeds in terms of the time periods:
\[ \frac{\omega_1}{\omega_2} = \frac{T_2}{T_1} \]

### Step 4: Substitute the time period ratio
From Step 2, we have:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \]
Taking the square root gives:
\[ \frac{T_1}{T_2} = \frac{R_1^{3/2}}{R_2^{3/2}} \]
Thus,
\[ \frac{\omega_1}{\omega_2} = \frac{R_2^{3/2}}{R_1^{3/2}} \]

### Step 5: Substitute the expressions for \( R_1 \) and \( R_2 \)
Now substituting \( R_1 \) and \( R_2 \):
\[ \frac{\omega_1}{\omega_2} = \frac{(r + h_2)^{3/2}}{(r + h_1)^{3/2}} \]

### Final Result
Thus, the ratio of the orbital angular speeds of the two satellites is:
\[ \frac{\omega_1}{\omega_2} = \left(\frac{r + h_2}{r + h_1}\right)^{3/2} \]