Two satellites `A` and `B` of the same mass are orbiting the earth at altitudes `R` and `3R` respectively, where `R` is the radius of the earth. Taking their orbit to be circular obtain the ratios of their kinetic and potential energies.

Explanation

To solve the problem, we need to find the ratios of kinetic and potential energies of two satellites A and B orbiting the Earth at different altitudes. Let's denote:

- \( R \): Radius of the Earth
- \( m \): Mass of each satellite
- \( M \): Mass of the Earth
- \( G \): Gravitational constant

### Step 1: Determine the orbital velocities of satellites A and B

1. **For Satellite A** (at altitude \( R \)):
- The distance from the center of the Earth is \( R + R = 2R \).
- The orbital velocity \( V_A \) is given by:
\[ V_A = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{2R}} \]

2. **For Satellite B** (at altitude \( 3R \)):
- The distance from the center of the Earth is \( R + 3R = 4R \).
- The orbital velocity \( V_B \) is given by:
\[ V_B = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{4R}} \]

### Step 2: Calculate the kinetic energies of satellites A and B

The kinetic energy \( KE \) of a satellite is given by:
\[ KE = \frac{1}{2} m V^2 \]

1. **Kinetic Energy of Satellite A**:
\[ KE_A = \frac{1}{2} m V_A^2 = \frac{1}{2} m \left(\sqrt{\frac{GM}{2R}}\right)^2 = \frac{1}{2} m \cdot \frac{GM}{2R} = \frac{mGM}{4R} \]

2. **Kinetic Energy of Satellite B**:
\[ KE_B = \frac{1}{2} m V_B^2 = \frac{1}{2} m \left(\sqrt{\frac{GM}{4R}}\right)^2 = \frac{1}{2} m \cdot \frac{GM}{4R} = \frac{mGM}{8R} \]

### Step 3: Find the ratio of kinetic energies

Now, we can find the ratio of the kinetic energies of A and B:
\[ \frac{KE_A}{KE_B} = \frac{\frac{mGM}{4R}}{\frac{mGM}{8R}} = \frac{8R}{4R} = 2 \]

Thus, the ratio of kinetic energies is:
\[ \frac{KE_A}{KE_B} = 2:1 \]

### Step 4: Calculate the potential energies of satellites A and B

The gravitational potential energy \( PE \) of a satellite is given by:
\[ PE = -\frac{GMm}{r} \]

1. **Potential Energy of Satellite A**:
\[ PE_A = -\frac{GMm}{2R} \]

2. **Potential Energy of Satellite B**:
\[ PE_B = -\frac{GMm}{4R} \]

### Step 5: Find the ratio of potential energies

Now, we can find the ratio of the potential energies of A and B:
\[ \frac{PE_A}{PE_B} = \frac{-\frac{GMm}{2R}}{-\frac{GMm}{4R}} = \frac{4R}{2R} = 2 \]

Thus, the ratio of potential energies is:
\[ \frac{PE_A}{PE_B} = 2:1 \]

### Final Results

- The ratio of kinetic energies \( KE_A : KE_B = 2 : 1 \)
- The ratio of potential energies \( PE_A : PE_B = 2 : 1 \)