The magnitude of binding energy of the satellite is E and kinetic energy is K. The ratio E/K is

Options

A. 2
B. `1/4`
C. `1`
D. `1/2`

Answer

To find the ratio of the binding energy (E) of a satellite to its kinetic energy (K), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Binding Energy**:
The binding energy (E) is the energy required to separate the satellite from the gravitational influence of the planet. It is equal to the negative of the total mechanical energy (TME) of the satellite.

\[ E = -TME \]

2. **Total Mechanical Energy**:
The total mechanical energy (TME) of the satellite is the sum of its kinetic energy (K) and potential energy (U).

\[ TME = K + U \]

Therefore, we can express the binding energy as:

\[ E = - (K + U) \]

3. **Expressing E in terms of K and U**:
Rearranging the equation gives:

\[ E = -K - U \]

4. **Finding the Ratio E/K**:
We want to find the ratio \( \frac{E}{K} \). From the expression for E, we can substitute:

\[ \frac{E}{K} = \frac{-K - U}{K} = -1 - \frac{U}{K} \]

5. **Finding Potential Energy (U)**:
For a satellite of mass \( m \) orbiting a planet of mass \( M \) at a distance \( r \):

\[ U = -\frac{GMm}{r} \]

where \( G \) is the gravitational constant.

6. **Finding Kinetic Energy (K)**:
The kinetic energy of the satellite in orbit is given by:

\[ K = \frac{1}{2} mv^2 \]

Using the gravitational force as the centripetal force, we have:

\[ \frac{mv^2}{r} = \frac{GMm}{r^2} \]

This simplifies to:

\[ v^2 = \frac{GM}{r} \]

Thus, substituting for K gives:

\[ K = \frac{1}{2} m \left(\frac{GM}{r}\right) = \frac{GMm}{2r} \]

7. **Finding the Ratio U/K**:
Now we can find the ratio \( \frac{U}{K} \):

\[ \frac{U}{K} = \frac{-\frac{GMm}{r}}{\frac{GMm}{2r}} = -2 \]

8. **Substituting back into the E/K ratio**:
Now substituting \( \frac{U}{K} = -2 \) into the ratio \( \frac{E}{K} \):

\[ \frac{E}{K} = -1 - (-2) = -1 + 2 = 1 \]

### Final Answer:
The ratio of the binding energy to the kinetic energy is:

\[ \frac{E}{K} = 1 \]