Two point like objects each with mass m are connected by a massless rope of length `l` the object are suspended vertically near the surface of earth, so that one object hanging below the other then the objects are relased. Show that the tension in the rope is `T=(GMml)/(R^(3))` where M is the mass of the earth and R is its radius `|l lt lt R|`

Explanation

To solve the problem, we need to analyze the forces acting on the two masses connected by a rope and derive the expression for the tension in the rope.

### Step-by-Step Solution:

1. **Identify the System**: We have two masses \( m \) connected by a massless rope of length \( l \). The masses are suspended vertically, with one mass hanging below the other.

2. **Understand the Gravitational Acceleration**: The gravitational acceleration \( g \) at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

3. **Consider the Height Change**: When the two masses are at a distance \( l \) apart, the upper mass experiences a slightly reduced gravitational acceleration due to the distance from the center of the Earth. The gravitational acceleration at height \( l \) can be approximated as:
\[ g' = g \left(1 - \frac{l}{R}\right) \approx g - \frac{gl}{R} \]
(using the binomial approximation for small \( l \) compared to \( R \)).

4. **Set Up the Force Equations**:
- For the lower mass (mass \( m \)):
\[ mg - T = ma \quad \text{(1)} \]
- For the upper mass (mass \( m \)):
\[ T - mg' = ma \quad \text{(2)} \]

5. **Substituting \( g' \)**: Substitute \( g' \) into equation (2):
\[ T - \left(g - \frac{gl}{R}\right)m = ma \]
Simplifying this gives:
\[ T = mg - ma + \frac{mgl}{R} \]

6. **Equate the Two Equations**: From equations (1) and (2), we can equate the expressions for \( T \):
\[ mg - ma = T = mg - ma + \frac{mgl}{R} \]
Rearranging gives:
\[ 2T = mg - mg' = mg - \left(g - \frac{gl}{R}\right)m = \frac{mgl}{R} \]

7. **Solve for Tension \( T \)**:
\[ T = \frac{mgl}{2R} \]

8. **Substituting for \( g \)**: Now substitute \( g = \frac{GM}{R^2} \) into the equation for \( T \):
\[ T = \frac{m \left(\frac{GM}{R^2}\right) l}{2R} = \frac{GMml}{2R^3} \]

9. **Final Expression**: Thus, we arrive at the final expression for the tension in the rope:
\[ T = \frac{GMml}{R^3} \]