A satallite of mass `m`, initally at rest on the earth, is launched into a circular orbit at a height equal to the the radius of the earth. The minimum energy required is

Options

A. `sqrt3/4` mgR
B. `1/2` mgR
C. `1/4` mgR
D. `3/4` mgR

Explanation

To find the minimum energy required to launch a satellite of mass \( m \) into a circular orbit at a height equal to the radius of the Earth, we can follow these steps:

### Step 1: Understand the Problem
The satellite is launched from the surface of the Earth to a height equal to the radius of the Earth. Therefore, the distance from the center of the Earth to the satellite in orbit is \( R + R = 2R \), where \( R \) is the radius of the Earth.

### Step 2: Calculate the Gravitational Potential Energy (U)
The gravitational potential energy of the satellite at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{G M m}{r} \]
At the surface of the Earth (initial position, \( r = R \)):
\[ U_1 = -\frac{G M m}{R} \]
At the height of \( R \) (final position, \( r = 2R \)):
\[ U_2 = -\frac{G M m}{2R} \]

### Step 3: Calculate the Change in Gravitational Potential Energy
The change in gravitational potential energy (\( \Delta U \)) as the satellite moves from the surface to the orbit is:
\[ \Delta U = U_2 - U_1 = \left(-\frac{G M m}{2R}\right) - \left(-\frac{G M m}{R}\right) \]
\[ \Delta U = -\frac{G M m}{2R} + \frac{G M m}{R} = \frac{G M m}{R} - \frac{G M m}{2R} = \frac{G M m}{2R} \]

### Step 4: Calculate the Kinetic Energy (K) in Orbit
The satellite in a circular orbit has a centripetal force provided by gravity. The gravitational force acting on the satellite is:
\[ F = \frac{G M m}{(2R)^2} = \frac{G M m}{4R^2} \]
The centripetal force required for circular motion is:
\[ F_c = \frac{m v^2}{2R} \]
Setting these equal gives:
\[ \frac{m v^2}{2R} = \frac{G M m}{4R^2} \]
Cancelling \( m \) and rearranging gives:
\[ v^2 = \frac{G M}{2R} \]

The kinetic energy (\( K \)) of the satellite in orbit is:
\[ K = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{G M}{2R}\right) = \frac{G M m}{4R} \]

### Step 5: Calculate the Total Energy in Orbit
The total mechanical energy (\( E \)) in orbit is the sum of kinetic and potential energy:
\[ E = K + U_2 = \frac{G M m}{4R} - \frac{G M m}{2R} \]
\[ E = \frac{G M m}{4R} - \frac{2G M m}{4R} = -\frac{G M m}{4R} \]

### Step 6: Calculate the Minimum Energy Required
The minimum energy required to launch the satellite is the change in total energy from the initial state to the final state:
\[ \Delta E = E - U_1 = \left(-\frac{G M m}{4R}\right) - \left(-\frac{G M m}{R}\right) \]
\[ \Delta E = -\frac{G M m}{4R} + \frac{G M m}{R} = \frac{G M m}{R} - \frac{G M m}{4R} = \frac{3G M m}{4R} \]

### Step 7: Express in Terms of \( mg \)
Using \( g = \frac{G M}{R^2} \), we can express \( G M \) as \( g R^2 \):
\[ \Delta E = \frac{3}{4} \cdot \frac{g R^2 m}{R} = \frac{3}{4} g m R \]

### Final Answer
Thus, the minimum energy required to launch the satellite is:
\[ \Delta E = \frac{3}{4} g m R \]

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