A satellite of mass `m` is orbiting the earth in a circular orbit of radius `r`. It starts losing energy due to small air resistance at the rate of `C J//s`. Then the time teken for the satellite to reach the earth is...........

Explanation

To solve the problem of a satellite losing energy due to air resistance while orbiting the Earth, we will follow these steps:

### Step 1: Understand the Energy Loss
The satellite is losing energy at a rate of \( C \) joules per second. This means that over time, the total energy lost can be expressed as:
\[ \text{Energy lost} = C \times t \]
where \( t \) is the time in seconds.

### Step 2: Calculate the Initial and Final Energies
The gravitational potential energy \( U \) of the satellite at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{G M m}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the satellite.

The kinetic energy \( K \) of the satellite in orbit is given by:
\[ K = \frac{1}{2} m v^2 \]
For a satellite in a circular orbit, the orbital speed \( v \) can be derived from the gravitational force:
\[ F = \frac{G M m}{r^2} = \frac{m v^2}{r} \implies v^2 = \frac{G M}{r} \]
Thus, the kinetic energy becomes:
\[ K = \frac{1}{2} m \left(\frac{G M}{r}\right) = \frac{G M m}{2r} \]

### Step 3: Total Energy of the Satellite
The total mechanical energy \( E \) of the satellite in orbit is the sum of its kinetic and potential energy:
\[ E = K + U = \frac{G M m}{2r} - \frac{G M m}{r} = -\frac{G M m}{2r} \]

### Step 4: Energy at the Surface of the Earth
When the satellite reaches the Earth's surface (at radius \( r_e \)), its total energy will be:
\[ E_{surface} = -\frac{G M m}{2r_e} \]

### Step 5: Calculate the Change in Energy
The change in energy \( \Delta E \) as the satellite moves from radius \( r \) to radius \( r_e \) is:
\[ \Delta E = E_{surface} - E = -\frac{G M m}{2r_e} - \left(-\frac{G M m}{2r}\right) = \frac{G M m}{2r} - \frac{G M m}{2r_e} \]

### Step 6: Relate Energy Loss to Time
Setting the energy lost equal to the energy change:
\[ C \times t = \frac{G M m}{2} \left(\frac{1}{r} - \frac{1}{r_e}\right) \]
From this, we can solve for \( t \):
\[ t = \frac{G M m}{2C} \left(\frac{1}{r} - \frac{1}{r_e}\right) \]

### Final Result
Thus, the time taken for the satellite to reach the Earth is:
\[ t = \frac{G M m}{2C} \left(\frac{1}{r} - \frac{1}{r_e}\right) \]