The minimum and maximum distance of a satellite from the centre of the earth are `2R` and `4R` respectively, where `R` is the radius of earth and `M` is the mass of the earth. The radius of curvature at the point of maximum distance is

Options

A. `(8R)/(3)`
B. `(4R)/(3)`
C. `(3R)/(8)`
D. `(3R)/(4)`

Explanation

To find the radius of curvature at the point of maximum distance of a satellite from the center of the Earth, we can follow these steps:

### Step 1: Understand the Problem
We are given that the minimum distance of the satellite from the center of the Earth is \(2R\) and the maximum distance is \(4R\). We need to find the radius of curvature at the maximum distance.

### Step 2: Identify Forces Acting on the Satellite
At the maximum distance \(4R\), the gravitational force acting on the satellite provides the necessary centripetal force for circular motion. The gravitational force \(F\) can be expressed as:
\[ F = \frac{GMm}{(4R)^2} \]
where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(m\) is the mass of the satellite.

### Step 3: Write the Expression for Centripetal Force
The centripetal force required for circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where \(v\) is the velocity of the satellite at distance \(4R\) and \(r\) is the radius of curvature at that point.

### Step 4: Set Gravitational Force Equal to Centripetal Force
At the maximum distance, we equate the gravitational force to the centripetal force:
\[ \frac{GMm}{(4R)^2} = \frac{mv^2}{4R} \]
We can cancel \(m\) from both sides:
\[ \frac{GM}{(4R)^2} = \frac{v^2}{4R} \]

### Step 5: Solve for Velocity \(v\)
Rearranging the equation gives:
\[ v^2 = \frac{GM \cdot 4R}{(4R)^2} = \frac{GM}{4R} \]

### Step 6: Substitute \(v^2\) into the Centripetal Force Equation
Now, substituting \(v^2\) back into the centripetal force equation:
\[ \frac{GMm}{(4R)^2} = \frac{m \cdot \frac{GM}{4R}}{4R} \]
This simplifies to:
\[ \frac{GM}{16R^2} = \frac{GM}{16R^2} \]
This confirms our calculations are consistent.

### Step 7: Find the Radius of Curvature
The radius of curvature \(r\) at the maximum distance can be found from the centripetal force equation:
\[ r = \frac{v^2}{g} \]
where \(g\) is the acceleration due to gravity at distance \(4R\):
\[ g = \frac{GM}{(4R)^2} = \frac{GM}{16R^2} \]
Thus, substituting \(v^2\):
\[ r = \frac{\frac{GM}{4R}}{\frac{GM}{16R^2}} = \frac{16R^2}{4R} = 4R \]

### Final Answer
The radius of curvature at the point of maximum distance \(4R\) is:
\[ r = 4R \]