The total energy of a circularly orbiting satellite is

Options

A. a.Negative
B. b.Positive
C. c.Zero
D. d.Either (1) or (2)

Answer

To find the total energy of a circularly orbiting satellite, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Satellite**:
- A satellite in circular orbit experiences two forces: gravitational force and centripetal force. The gravitational force provides the necessary centripetal force for the satellite's circular motion.

2. **Set Up the Equations**:
- The centripetal force (\(F_c\)) required for circular motion is given by:
\[ F_c = \frac{m v^2}{r} \]
where \(m\) is the mass of the satellite, \(v\) is its orbital velocity, and \(r\) is the radius of the orbit.
- The gravitational force (\(F_g\)) acting on the satellite is given by:
\[ F_g = \frac{G M m}{r^2} \]
where \(M\) is the mass of the planet and \(G\) is the gravitational constant.

3. **Equate the Forces**:
- For a satellite in stable orbit, the centripetal force equals the gravitational force:
\[ \frac{m v^2}{r} = \frac{G M m}{r^2} \]

4. **Simplify the Equation**:
- Cancel out \(m\) from both sides (assuming \(m \neq 0\)):
\[ \frac{v^2}{r} = \frac{G M}{r^2} \]
- Rearranging gives:
\[ v^2 = \frac{G M}{r} \]

5. **Calculate Kinetic Energy**:
- The kinetic energy (\(KE\)) of the satellite is given by:
\[ KE = \frac{1}{2} m v^2 \]
- Substituting for \(v^2\):
\[ KE = \frac{1}{2} m \left(\frac{G M}{r}\right) = \frac{G M m}{2r} \]

6. **Calculate Potential Energy**:
- The gravitational potential energy (\(PE\)) of the satellite is given by:
\[ PE = -\frac{G M m}{r} \]

7. **Find Total Energy**:
- The total energy (\(E\)) of the satellite is the sum of its kinetic and potential energy:
\[ E = KE + PE \]
- Substituting the expressions for \(KE\) and \(PE\):
\[ E = \frac{G M m}{2r} - \frac{G M m}{r} \]
- Combine the terms:
\[ E = \frac{G M m}{2r} - \frac{2G M m}{2r} = -\frac{G M m}{2r} \]

8. **Conclusion**:
- The total energy of a circularly orbiting satellite is:
\[ E = -\frac{G M m}{2r} \]
- Since \(G\), \(M\), and \(m\) are all positive, the total energy \(E\) is negative.

### Final Answer:
The total energy of a circularly orbiting satellite is negative.

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