A satellite is revolving in a circular orbit of radius R. Ratio of KE and its PE is (only magnitude )

Options

A. `1:4`
B. `2:1`
C. `3:2`
D. `1:2`

Answer

To find the ratio of the kinetic energy (KE) to the potential energy (PE) of a satellite revolving in a circular orbit of radius \( R \), we can follow these steps:

### Step 1: Write the expression for gravitational potential energy (PE)
The gravitational potential energy of a satellite of mass \( m \) at a distance \( R \) from the center of a planet of mass \( M \) is given by the formula:
\[ PE = -\frac{G M m}{R} \]
where \( G \) is the gravitational constant.

### Step 2: Write the expression for kinetic energy (KE)
The kinetic energy of the satellite moving with orbital velocity \( v_0 \) is given by:
\[ KE = \frac{1}{2} m v_0^2 \]
To find \( v_0 \), we can use the formula for orbital velocity:
\[ v_0 = \sqrt{\frac{G M}{R}} \]
Substituting this expression for \( v_0 \) into the kinetic energy formula gives:
\[ KE = \frac{1}{2} m \left(\sqrt{\frac{G M}{R}}\right)^2 = \frac{1}{2} m \frac{G M}{R} \]

### Step 3: Calculate the ratio of KE to PE
Now we have both expressions:
- \( KE = \frac{1}{2} m \frac{G M}{R} \)
- \( PE = -\frac{G M m}{R} \)

To find the ratio of the magnitudes of KE to PE, we compute:
\[ \text{Ratio} = \frac{KE}{|PE|} = \frac{\frac{1}{2} m \frac{G M}{R}}{\frac{G M m}{R}} \]

### Step 4: Simplify the ratio
When we simplify the ratio:
\[ \text{Ratio} = \frac{\frac{1}{2} m \frac{G M}{R}}{\frac{G M m}{R}} = \frac{1}{2} \]

### Final Answer
Thus, the ratio of the kinetic energy to the potential energy (only in magnitude) is:
\[ \text{Ratio} = \frac{1}{2} \]