At satellite of mass m is launched from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 3R. K.E. of satellite at the time of launching is

Options

A. `(7)/(8) (GMm)/(R )`
B. `(8)/(7)(GMm)/(R )`
C. `(6)/(7)(GMm)/(R )`
D. `(7)/(8)(GMm)/(R^(2))`

Explanation

To find the kinetic energy (K.E.) of the satellite at the time of launching, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters**:
- Mass of the satellite = \( m \)
- Mass of the planet = \( M \)
- Radius of the planet = \( R \)
- Altitude of the satellite = \( 3R \)
- Distance from the center of the planet to the satellite = \( R + 3R = 4R \)

2. **Calculate the gravitational potential energy (U) at the altitude**:
The gravitational potential energy at a distance \( r \) from the center of a planet is given by:
\[ U = -\frac{GMm}{r} \]
For the satellite at a distance of \( 4R \):
\[ U_{\text{orbit}} = -\frac{GMm}{4R} \]

3. **Calculate the gravitational potential energy (U) at the surface of the planet**:
At the surface of the planet (distance \( R \)):
\[ U_{\text{surface}} = -\frac{GMm}{R} \]

4. **Use the conservation of energy principle**:
The total mechanical energy (potential + kinetic) at the surface must equal the total mechanical energy at the orbit:
\[ K.E. + U_{\text{surface}} = U_{\text{orbit}} + K.E._{\text{orbit}} \]
Rearranging gives:
\[ K.E. = U_{\text{orbit}} + K.E._{\text{orbit}} + U_{\text{surface}} \]

5. **Express kinetic energy at the orbit**:
The orbital velocity \( v_0 \) of the satellite can be expressed as:
\[ v_0 = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{4R}} = \frac{1}{2}\sqrt{\frac{GM}{R}} \]
Therefore, the kinetic energy at the orbit is:
\[ K.E._{\text{orbit}} = \frac{1}{2} mv_0^2 = \frac{1}{2} m \left(\frac{GM}{4R}\right) = \frac{GMm}{8R} \]

6. **Substituting values**:
Now substituting the values back into the energy equation:
\[ K.E. = -\frac{GMm}{4R} + \frac{GMm}{8R} - \frac{GMm}{R} \]
Combining these terms:
\[ K.E. = -\frac{2GMm}{8R} + \frac{GMm}{8R} - \frac{8GMm}{8R} = -\frac{2GMm + 8GMm - GMm}{8R} \]
Simplifying this gives:
\[ K.E. = \frac{7GMm}{8R} \]

### Final Answer:
The kinetic energy of the satellite at the time of launching is:
\[ K.E. = \frac{7GMm}{8R} \]