If `W` is the weight of a satellite on the surface of the earth, then the energy required to lauch that satellite from the surface of earth into a circular orbit of radius `3R` is (here `R` is the radius of the earth)

Options

A. `5 WR//6`
B. `6WR//5`
C. `2WR//3`
D. `3WR//2`

Explanation

To solve the problem of finding the energy required to launch a satellite from the surface of the Earth to a circular orbit of radius \(3R\), where \(R\) is the radius of the Earth, we can follow these steps:

### Step 1: Define the variables
Let:
- \(m\) = mass of the satellite
- \(M\) = mass of the Earth
- \(g\) = acceleration due to gravity at the surface of the Earth
- \(W\) = weight of the satellite on the surface of the Earth, which is given by \(W = mg\)

### Step 2: Calculate the gravitational potential energy at the surface
The gravitational potential energy \(U\) of the satellite at the surface of the Earth (radius \(R\)) is given by:
\[ U_{\text{surface}} = -\frac{GMm}{R} \]
where \(G\) is the universal gravitational constant.

### Step 3: Calculate the gravitational potential energy in the orbit
The gravitational potential energy \(U\) of the satellite in a circular orbit at a radius \(3R\) is given by:
\[ U_{\text{orbit}} = -\frac{GMm}{3R} \]

### Step 4: Calculate the total energy in the orbit
The total mechanical energy \(E\) in a circular orbit is given by:
\[ E_{\text{orbit}} = U + K \]
where \(K\) is the kinetic energy. The total energy in a circular orbit can also be expressed as:
\[ E_{\text{orbit}} = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R} \]

### Step 5: Calculate the kinetic energy in the orbit
From the total energy expression, we can rearrange to find the kinetic energy:
\[ K_{\text{orbit}} = E_{\text{orbit}} - U_{\text{orbit}} = -\frac{GMm}{6R} + \frac{GMm}{3R} \]
\[ K_{\text{orbit}} = -\frac{GMm}{6R} + \frac{2GMm}{6R} = \frac{GMm}{6R} \]

### Step 6: Calculate the energy required to launch the satellite
The energy required to launch the satellite is the difference in total energy from the surface to the orbit:
\[ \Delta E = E_{\text{orbit}} - U_{\text{surface}} = \left(-\frac{GMm}{6R}\right) - \left(-\frac{GMm}{R}\right) \]
\[ \Delta E = -\frac{GMm}{6R} + \frac{GMm}{R} = \frac{GMm}{R} - \frac{GMm}{6R} = \frac{5GMm}{6R} \]

### Step 7: Substitute \(W\) into the equation
Since \(W = mg\) and \(g = \frac{GM}{R^2}\), we can express \(GM\) in terms of \(W\):
\[ GM = gR^2 = \frac{W R^2}{m} \]
Substituting this into the energy equation gives:
\[ \Delta E = \frac{5}{6} \cdot \frac{W R^2}{R} = \frac{5}{6} W R \]

### Final Answer
Thus, the energy required to launch the satellite into a circular orbit of radius \(3R\) is:
\[ \Delta E = \frac{5}{6} W R \]