Two artificial satellites of the same mass are moving around the earth in circular orbits of different radii. In comparision to the satellite with lesser orbital radius, the other satellite with higher orbital radius will have:

Options

A. greater kinetic energy
B. less potential energy
C. greater total energy
D. less magnitude of angular momentum, about the centre of the circular orbit.

Explanation

To solve the problem, we need to analyze the properties of two artificial satellites orbiting the Earth at different radii. We will derive the relevant formulas for kinetic energy, potential energy, total energy, and angular momentum to compare the two satellites.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( m \) be the mass of the satellite.
- Let \( M_e \) be the mass of the Earth.
- Let \( r_1 \) be the radius of the orbit of the satellite with lesser orbital radius.
- Let \( r_2 \) be the radius of the orbit of the satellite with higher orbital radius, where \( r_2 > r_1 \).

2. **Kinetic Energy (KE)**:
- The formula for kinetic energy of a satellite in orbit is given by:
\[ KE = \frac{G M_e m}{2r} \]
- For the satellite at radius \( r_1 \):
\[ KE_1 = \frac{G M_e m}{2r_1} \]
- For the satellite at radius \( r_2 \):
\[ KE_2 = \frac{G M_e m}{2r_2} \]
- Since \( r_2 > r_1 \), it follows that \( KE_2 < KE_1 \). Therefore, the satellite with the higher orbital radius has **less kinetic energy**.

3. **Potential Energy (PE)**:
- The formula for gravitational potential energy is given by:
\[ PE = -\frac{G M_e m}{r} \]
- For the satellite at radius \( r_1 \):
\[ PE_1 = -\frac{G M_e m}{r_1} \]
- For the satellite at radius \( r_2 \):
\[ PE_2 = -\frac{G M_e m}{r_2} \]
- Since \( r_2 > r_1 \), it follows that \( |PE_2| < |PE_1| \) (the magnitude of potential energy is less for the satellite at a higher radius). Therefore, the satellite with the higher orbital radius has **greater potential energy** (less negative).

4. **Total Energy (TE)**:
- The total energy of a satellite is the sum of its kinetic and potential energy:
\[ TE = KE + PE \]
- For the satellite at radius \( r_1 \):
\[ TE_1 = KE_1 + PE_1 = \frac{G M_e m}{2r_1} - \frac{G M_e m}{r_1} = -\frac{G M_e m}{2r_1} \]
- For the satellite at radius \( r_2 \):
\[ TE_2 = KE_2 + PE_2 = \frac{G M_e m}{2r_2} - \frac{G M_e m}{r_2} = -\frac{G M_e m}{2r_2} \]
- Since \( r_2 > r_1 \), it follows that \( TE_2 > TE_1 \) (less negative). Therefore, the satellite with the higher orbital radius has **greater total energy**.

5. **Angular Momentum (L)**:
- The angular momentum of a satellite is given by:
\[ L = m v r \]
- The orbital velocity \( v \) can be expressed as:
\[ v = \sqrt{\frac{G M_e}{r}} \]
- Thus, the angular momentum becomes:
\[ L = m \sqrt{\frac{G M_e}{r}} r = m \sqrt{G M_e r} \]
- For the satellite at radius \( r_1 \):
\[ L_1 = m \sqrt{G M_e r_1} \]
- For the satellite at radius \( r_2 \):
\[ L_2 = m \sqrt{G M_e r_2} \]
- Since \( r_2 > r_1 \), it follows that \( L_2 > L_1 \). Therefore, the satellite with the higher orbital radius has **greater angular momentum**.

### Conclusion:
In comparison to the satellite with lesser orbital radius, the satellite with higher orbital radius will have:
- **Less kinetic energy**
- **Greater potential energy**
- **Greater total energy**
- **Greater magnitude of angular momentum**

The correct option from the question is that the satellite with the higher orbital radius has **greater total energy**.