An ariticial satellite (mass `=M`) of earth revolves in a circular orbit whose radius is `4` times the radius of the earth. Assume that in the spherical region between the radius `4R_(e )` and `2R_(e )` there is cosmic dust. If mass of the satellite is increasing at the rate of `u` per sec. Find the time in which satellite will come to orbit of radius `2R_(e )`.

Answer

To solve the problem, we need to determine the time it takes for an artificial satellite, which is initially in a circular orbit at a radius of \(4R_e\) (where \(R_e\) is the radius of the Earth), to come down to an orbit at a radius of \(2R_e\). The mass of the satellite is increasing at a rate of \(u\) kg/s due to cosmic dust.

### Step-by-Step Solution:

1. **Identify the Initial and Final Radii:**
- Initial radius \(R_i = 4R_e\)
- Final radius \(R_f = 2R_e\)

2. **Determine the Gravitational Force:**
The gravitational force acting on the satellite at a distance \(R\) from the center of the Earth is given by:
\[ F = \frac{GMm}{R^2} \]
where \(M\) is the mass of the Earth and \(m\) is the mass of the satellite.

3. **Centripetal Force Requirement:**
For the satellite to remain in a circular orbit, the gravitational force must equal the centripetal force required for circular motion:
\[ \frac{GMm}{R^2} = \frac{mv^2}{R} \]
Simplifying this gives:
\[ v^2 = \frac{GM}{R} \]

4. **Change in Mass of the Satellite:**
The mass of the satellite is increasing at a rate of \(u\) kg/s. Therefore, if \(t\) is the time taken, the mass of the satellite at time \(t\) will be:
\[ m(t) = m_0 + ut \]
where \(m_0\) is the initial mass of the satellite.

5. **Velocity at Different Radii:**
- At \(R_i = 4R_e\):
\[ v_i = \sqrt{\frac{GM}{4R_e}} = \frac{1}{2}\sqrt{\frac{GM}{R_e}} \]
- At \(R_f = 2R_e\):
\[ v_f = \sqrt{\frac{GM}{2R_e}} = \sqrt{2}\sqrt{\frac{GM}{R_e}} \]

6. **Using Energy Conservation:**
As the satellite's mass is increasing, we cannot directly use energy conservation. Instead, we consider the change in velocity as the satellite descends from \(4R_e\) to \(2R_e\).

7. **Relating Velocity and Time:**
The change in velocity can be related to the time taken to descend. The average velocity during the descent can be approximated as:
\[ v_{avg} = \frac{v_i + v_f}{2} \]
The time taken \(t\) to descend can be expressed as:
\[ t = \frac{\Delta R}{v_{avg}} \]
where \(\Delta R = R_i - R_f = 4R_e - 2R_e = 2R_e\).

8. **Calculating the Average Velocity:**
\[ v_{avg} = \frac{\frac{1}{2}\sqrt{\frac{GM}{R_e}} + \sqrt{2}\sqrt{\frac{GM}{R_e}}}{2} \]
Simplifying gives:
\[ v_{avg} = \frac{(1/2 + \sqrt{2})\sqrt{\frac{GM}{R_e}}}{2} \]

9. **Final Time Calculation:**
Substitute \(v_{avg}\) into the time equation:
\[ t = \frac{2R_e}{v_{avg}} = \frac{2R_e \cdot 2}{(1/2 + \sqrt{2})\sqrt{\frac{GM}{R_e}}} \]
This simplifies to:
\[ t = \frac{4R_e}{(1/2 + \sqrt{2})\sqrt{\frac{GM}{R_e}}} \]

### Final Answer:
The time taken for the satellite to come down to an orbit of radius \(2R_e\) is given by:
\[ t = \frac{4R_e}{(1/2 + \sqrt{2})\sqrt{\frac{GM}{R_e}}} \]