The PE of a satellite is -U . Its KE is

Options

A. U
B. 2U
C. `-U`
D. U/2

Answer

To find the kinetic energy (KE) of a satellite given that its potential energy (PE) is -U, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Potential Energy (PE) Formula**:
The potential energy (PE) of a satellite in a gravitational field is given by the formula:
\[ PE = -\frac{G M m}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, \( m \) is the mass of the satellite, and \( r \) is the distance from the center of the planet to the satellite.

2. **Set the Given PE**:
According to the problem, the potential energy of the satellite is given as:
\[ PE = -U \]
Therefore, we can equate the two expressions:
\[ -U = -\frac{G M m}{r} \]
From this, we can derive:
\[ U = \frac{G M m}{r} \quad \text{(Equation 1)} \]

3. **Determine the Velocity of the Satellite**:
For a satellite in a circular orbit, the centripetal force required to keep it in orbit is provided by the gravitational force. The formula for the orbital velocity \( v \) of the satellite is:
\[ v = \sqrt{\frac{G M}{r}} \]

4. **Calculate the Kinetic Energy (KE)**:
The kinetic energy (KE) of the satellite can be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the expression for \( v \):
\[ KE = \frac{1}{2} m \left(\sqrt{\frac{G M}{r}}\right)^2 \]
Simplifying this gives:
\[ KE = \frac{1}{2} m \frac{G M}{r} \]

5. **Relate KE to PE**:
From Equation 1, we know that:
\[ U = \frac{G M m}{r} \]
Therefore, we can substitute this into the expression for KE:
\[ KE = \frac{1}{2} \cdot U \]
Thus, we find that:
\[ KE = \frac{U}{2} \]

### Final Answer:
The kinetic energy (KE) of the satellite is:
\[ KE = \frac{U}{2} \]