For a satellite moving in a circular orbit around the earth, the ratio of its potential energy to kinetic energy is

Options

A. 1
B. `-1`
C. `2`
D. `-2`

Explanation

To find the ratio of potential energy (U) to kinetic energy (K) for a satellite moving in a circular orbit around the Earth, we can follow these steps:

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

2. **Potential Energy (U)**: The gravitational potential energy of the satellite is given by:
\[ U = -\frac{G M m}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( r \) is the distance from the center of the Earth to the satellite.

### Step 2: Relate kinetic energy to gravitational force
For a satellite in a stable circular orbit, the gravitational force provides the necessary centripetal force. Thus, we can set the gravitational force equal to the centripetal force:
\[ \frac{G M m}{r^2} = \frac{mv^2}{r} \]
From this equation, we can solve for \( v^2 \):
\[ v^2 = \frac{G M}{r} \]

### Step 3: Substitute \( v^2 \) into the kinetic energy formula
Now, substitute \( v^2 \) into the kinetic energy formula:
\[ K = \frac{1}{2} m \left(\frac{G M}{r}\right) = \frac{G M m}{2r} \]

### Step 4: Calculate the ratio of potential energy to kinetic energy
Now we can find the ratio \( \frac{U}{K} \):
\[ \frac{U}{K} = \frac{-\frac{G M m}{r}}{\frac{G M m}{2r}} \]
Simplifying this gives:
\[ \frac{U}{K} = \frac{-\frac{G M m}{r}}{\frac{G M m}{2r}} = \frac{-1}{\frac{1}{2}} = -2 \]

### Final Answer
The ratio of the potential energy to the kinetic energy for a satellite in a circular orbit around the Earth is:
\[ \frac{U}{K} = -2 \]

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