A satellite of mass m is orbiting around the earth (mass M, radius R) in a circular orbital of radius 4R. It starts losing energy slowly at a constant `-(dE)/(dt)=eta`due to friction. Find the time (t) in which the satellite will spiral down to the surface of the earth.

Answer

To solve the problem of a satellite spiraling down to the surface of the Earth due to energy loss, we can follow these steps:

### Step 1: Determine the Initial Energy of the Satellite
The gravitational potential energy \( E_i \) of the satellite in orbit at radius \( 4R \) is given by the formula:
\[ E_i = -\frac{GMm}{2r} \]
Substituting \( r = 4R \):
\[ E_i = -\frac{GMm}{2 \times 4R} = -\frac{GMm}{8R} \]

### Step 2: Determine the Final Energy of the Satellite
When the satellite spirals down to the Earth's surface (radius \( R \)), the final energy \( E_f \) is:
\[ E_f = -\frac{GMm}{2R} \]

### Step 3: Calculate the Loss of Energy
The loss of energy \( \Delta E \) as the satellite spirals down is given by:
\[ \Delta E = E_i - E_f \]
Substituting the values we found:
\[ \Delta E = \left(-\frac{GMm}{8R}\right) - \left(-\frac{GMm}{2R}\right) \]
This simplifies to:
\[ \Delta E = -\frac{GMm}{8R} + \frac{4GMm}{8R} = \frac{3GMm}{8R} \]

### Step 4: Relate Energy Loss to Time
The satellite is losing energy at a constant rate of \( -\frac{dE}{dt} = \eta \). Therefore, we can set up the equation:
\[ \Delta E = \eta t \]
From this, we can express time \( t \) as:
\[ t = \frac{\Delta E}{\eta} \]

### Step 5: Substitute the Loss of Energy
Substituting the expression for \( \Delta E \):
\[ t = \frac{\frac{3GMm}{8R}}{\eta} \]
Thus, the time \( t \) in which the satellite will spiral down to the surface of the Earth is:
\[ t = \frac{3GMm}{8R\eta} \]

### Final Answer
The time \( t \) for the satellite to spiral down to the surface of the Earth is:
\[ t = \frac{3GMm}{8R\eta} \]