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 solve the problem of finding 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: Understand the expressions for KE and PE
The kinetic energy (KE) of a satellite in orbit is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the satellite and \( v \) is its orbital velocity.

The potential energy (PE) of the satellite in the gravitational field is given by:
\[ PE = -\frac{G M m}{r} \]
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 orbital velocity to gravitational force
For a satellite in a stable circular orbit, the gravitational force provides the necessary centripetal force. Thus, we can equate the gravitational force to the centripetal force:
\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]
From this equation, we can solve for \( v^2 \):
\[ v^2 = \frac{G M}{r} \]

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

### Step 4: Substitute \( r \) for the radius of the orbit
In this case, if the satellite is at a height \( h \) above the Earth's surface, the radius \( r \) can be expressed as:
\[ r = R + h \]
where \( R \) is the radius of the Earth. For simplicity, we can consider \( r = R \) if the height \( h \) is negligible.

### Step 5: Calculate the PE
Now, substituting \( r \) into the potential energy formula:
\[ PE = -\frac{G M m}{r} = -\frac{G M m}{R} \]

### Step 6: Find the ratio of KE to PE
Now we can find the ratio of KE to PE:
\[ \frac{KE}{PE} = \frac{\frac{G M m}{2R}}{-\frac{G M m}{R}} = \frac{1}{2} \cdot (-1) = -\frac{1}{2} \]
Taking the magnitude, we have:
\[ \left|\frac{KE}{PE}\right| = \frac{1}{2} \]

### Step 7: Express the ratio in terms of 1:2
Thus, the ratio of the magnitudes of kinetic energy to potential energy is:
\[ \frac{KE}{PE} = \frac{1}{2} \implies 1 : 2 \]

### Final Answer
The ratio of the kinetic energy to the potential energy (in terms of magnitude) is \( 1 : 2 \).

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