Two identical satellites A and B are circulating round the earth at the height of R and 2 R respectively, (where R is radius of the earth). The ratio of kinetic energy of A to that of B is

Options

A. `1/2`
B. `2/3`
C. `2`
D. `3/2`

Answer

To find the ratio of the kinetic energy of satellites A and B, we can follow these steps:

### Step 1: Understand the heights of the satellites
- Satellite A is at a height of \( R \) (the radius of the Earth).
- Satellite B is at a height of \( 2R \).

### Step 2: Determine the distances from the center of the Earth
- The distance of satellite A from the center of the Earth, \( r_A \), is:
\[ r_A = R + R = 2R \]
- The distance of satellite B from the center of the Earth, \( r_B \), is:
\[ r_B = R + 2R = 3R \]

### Step 3: Write the formula for kinetic energy of a satellite
The kinetic energy (KE) of a satellite in orbit is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the satellite and \( v \) is its orbital velocity.

### Step 4: Determine the orbital velocity of the satellites
The orbital velocity \( v \) of a satellite is given by:
\[ v = \sqrt{\frac{GM}{r}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

#### For satellite A:
\[ v_A = \sqrt{\frac{GM}{r_A}} = \sqrt{\frac{GM}{2R}} \]

#### For satellite B:
\[ v_B = \sqrt{\frac{GM}{r_B}} = \sqrt{\frac{GM}{3R}} \]

### Step 5: Calculate the kinetic energies
#### Kinetic energy of satellite A:
\[ KE_A = \frac{1}{2} m v_A^2 = \frac{1}{2} m \left(\sqrt{\frac{GM}{2R}}\right)^2 = \frac{1}{2} m \frac{GM}{2R} = \frac{GMm}{4R} \]

#### Kinetic energy of satellite B:
\[ KE_B = \frac{1}{2} m v_B^2 = \frac{1}{2} m \left(\sqrt{\frac{GM}{3R}}\right)^2 = \frac{1}{2} m \frac{GM}{3R} = \frac{GMm}{6R} \]

### Step 6: Find the ratio of kinetic energies
Now, we can find the ratio of the kinetic energies of A to B:
\[ \frac{KE_A}{KE_B} = \frac{\frac{GMm}{4R}}{\frac{GMm}{6R}} = \frac{6}{4} = \frac{3}{2} \]

### Final Answer:
The ratio of the kinetic energy of satellite A to that of satellite B is:
\[ \frac{KE_A}{KE_B} = \frac{3}{2} \]