A satellite of mass m is orbiting the earth in a circular orbit at a height 2R abvoe earth surface where R is the radius of the earth. If it starts losing energy at a constant rate `beta`, then it will fall on the earth surface after the time `t=(GMm)/(nbetaR)`. Assuming that the satellite is approximately in a circular orbit at all times, find the value of n.

Answer

To solve the problem, we need to analyze the energy changes of the satellite as it loses energy at a constant rate and eventually falls to the Earth's surface. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the initial conditions
The satellite is at a height of \(2R\) above the Earth's surface, which means its distance from the center of the Earth is \(R + 2R = 3R\), where \(R\) is the radius of the Earth.

### Step 2: Calculate the initial energy of the satellite
The total mechanical energy \(E\) of the satellite in orbit is given by the sum of its kinetic energy (KE) and gravitational potential energy (PE).

1. **Kinetic Energy (KE)**:
The centripetal force required for circular motion is provided by the gravitational force. Thus, we have:
\[ \frac{mv^2}{r} = \frac{GMm}{r^2} \]
Rearranging gives:
\[ mv^2 = \frac{GMm}{r} \]
Therefore, the kinetic energy is:
\[ KE = \frac{1}{2} mv^2 = \frac{1}{2} \cdot \frac{GMm}{r} = \frac{GMm}{2r} \]

2. **Potential Energy (PE)**:
The gravitational potential energy at a distance \(d\) from the center of the Earth is:
\[ PE = -\frac{GMm}{d} = -\frac{GMm}{3R} \]

3. **Total Energy (E)**:
Thus, the total energy \(E\) at the height of \(2R\) is:
\[ E = KE + PE = \frac{GMm}{2R} - \frac{GMm}{3R} \]
To combine these, we find a common denominator:
\[ E = \frac{3GMm}{6R} - \frac{2GMm}{6R} = \frac{GMm}{6R} \]

### Step 3: Energy loss rate
The satellite loses energy at a constant rate \(\beta\). Therefore, the rate of change of energy is:
\[ \frac{dE}{dt} = -\beta \]

### Step 4: Integrate the energy loss
We can set up the equation for energy loss over time:
\[ E_f - E_i = -\beta t \]
Where \(E_i\) is the initial energy and \(E_f\) is the final energy when the satellite reaches the Earth's surface (at distance \(R\)):
\[ E_f = -\frac{GMm}{R} \]

### Step 5: Substitute the energies into the equation
Substituting the energies into the equation gives:
\[ -\frac{GMm}{R} - \frac{GMm}{6R} = -\beta t \]
This simplifies to:
\[ -\frac{GMm}{R} + \frac{GMm}{6R} = -\beta t \]
Combining the terms:
\[ -\frac{6GMm}{6R} + \frac{GMm}{6R} = -\beta t \]
\[ -\frac{5GMm}{6R} = -\beta t \]
Thus, we have:
\[ \beta t = \frac{5GMm}{6R} \]

### Step 6: Solve for \(t\)
Rearranging gives:
\[ t = \frac{5GMm}{6\beta R} \]

### Step 7: Compare with the given formula
We are given that:
\[ t = \frac{GMm}{n\beta R} \]
By comparing both expressions for \(t\):
\[ \frac{5GMm}{6\beta R} = \frac{GMm}{n\beta R} \]
Cancelling \(GMm\) and \(\beta R\) from both sides gives:
\[ \frac{5}{6} = \frac{1}{n} \]
Thus, solving for \(n\) gives:
\[ n = \frac{6}{5} \]

### Final Answer
The value of \(n\) is \(6/5\).