An artificial satellite (mass m) of a planet (mass M) revolves in a circular orbit whose radius is n times the radius R of the planet in the process of motion the satellite experiences a slight resistance due to cosmic dust. Assuming the force of resistance on satellite to depend on velocity as `F=av^(2)` where 'a' is a constant caculate how long the satellite will stay in the space before it falls on to the planet's surface.

Explanation

To solve the problem of how long an artificial satellite will stay in space before it falls onto the planet's surface due to the force of resistance from cosmic dust, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Satellite:**
The satellite experiences gravitational force and a resistive force due to cosmic dust. The gravitational force \( F_g \) acting on the satellite is given by:
\[ F_g = \frac{G M m}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, \( m \) is the mass of the satellite, and \( r \) is the distance from the center of the planet to the satellite.

2. **Determine the Radius of the Orbit:**
The radius of the orbit of the satellite is given as \( r = nR \), where \( R \) is the radius of the planet.

3. **Express the Resistive Force:**
The resistive force \( F_r \) acting on the satellite is given by:
\[ F_r = a v^2 \]
where \( a \) is a constant and \( v \) is the velocity of the satellite.

4. **Calculate the Orbital Velocity:**
The orbital velocity \( v \) of the satellite in a circular orbit can be expressed as:
\[ v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{nR}} \]

5. **Energy of the Satellite:**
The total mechanical energy \( E \) of the satellite in orbit is given by:
\[ E = -\frac{GMm}{2r} = -\frac{GMm}{2nR} \]

6. **Rate of Change of Energy:**
The rate of change of energy \( \frac{dE}{dt} \) is equal to the work done by the resistive force:
\[ \frac{dE}{dt} = F_r \cdot v = a v^2 \cdot v = a v^3 \]

7. **Set Up the Equation:**
We can set the rate of change of energy equal to the resistive force:
\[ \frac{dE}{dt} = -\frac{GMm}{2nR^2} \frac{dr}{dt} = -a v^3 \]

8. **Substituting for Velocity:**
Substitute \( v = \sqrt{\frac{GM}{nR}} \) into the equation:
\[ -\frac{GMm}{2nR^2} \frac{dr}{dt} = -a \left(\sqrt{\frac{GM}{nR}}\right)^3 \]

9. **Simplifying the Equation:**
Rearranging gives:
\[ \frac{dr}{dt} = \frac{2a n^{3/2} R^{3/2}}{GM} \cdot \frac{1}{m} \]

10. **Integrate to Find Time:**
Integrate from \( r = nR \) to \( r = R \) to find the time \( T \):
\[ T = \int_{nR}^{R} \frac{dr}{\frac{2a n^{3/2} R^{3/2}}{GM} \cdot \frac{1}{m}} = \frac{GM}{2a n^{3/2} R^{3/2}} \int_{nR}^{R} dr \]
The integral evaluates to \( R - nR = R(1 - n) \):
\[ T = \frac{GM (R(1 - n))}{2a n^{3/2} R^{3/2}} = \frac{GM (1 - n)}{2a n^{3/2} R^{1/2}} \]

### Final Result:
The time \( T \) that the satellite will stay in space before it falls onto the planet's surface is given by:
\[ T = \frac{GM (1 - n)}{2a n^{3/2} R^{1/2}} \]