Work Done Calculation for Raising a Mass

This solution explains how to calculate the work done to move a mass from the Earth's surface to a specific height. We'll break down the physics concepts and perform the calculation step-by-step.

Understanding Gravitational Force and Potential Energy

When an object is moved away from the Earth, the gravitational force exerted by the Earth on the object changes. Unlike calculations near the surface where we can assume a constant force ($F = mg$), moving to significant heights requires considering the inverse square nature of the gravitational force.

• The gravitational force between the Earth (mass M) and the object (mass m) at a distance r from the center of the Earth is given by Newton's Law of Universal Gravitation:

  \(\vec{F} = -\frac{GMm}{r^2} \hat{r}\)

  where G is the gravitational constant and \(\hat{r}\) is the unit vector pointing radially outwards. The negative sign indicates the force is attractive (towards the Earth).
• Work done is the integral of force over distance. To find the work done against gravity, we need to integrate the magnitude of the gravitational force from the initial position to the final position.
• The acceleration due to gravity at the surface, g, is related to G, M, and the Earth's radius R by:

  \(g = \frac{GM}{R^2}\)

  This implies \(GM = gR^2\). We will use this relationship later.

Step-by-Step Work Done Calculation

We need to find the work done (W) to raise a mass m from the Earth's surface (distance R from the center) to a height h above the surface. The question states that the height h is equal to twice the radius of the Earth, so \(h = 2R\).

1. Initial and Final Positions:
   • The initial distance from the center of the Earth is \(r_1 = R\).
   • The final distance from the center of the Earth is \(r_2 = R + h = R + 2R = 3R\).
2. Calculating Work Done: The work done against the gravitational force is calculated by integrating the force from the initial radius to the final radius. The force acting on the mass is \(F(r) = \frac{GMm}{r^2}\) (magnitude).

   \(W = \int_{r_1}^{r_2} F(r) dr\)

   \(W = \int_{R}^{3R} \frac{GMm}{r^2} dr\)

3. Integration:

   \(W = GMm \int_{R}^{3R} r^{-2} dr\)

   The integral of \(r^{-2}\) is \(-r^{-1}\).

   \(W = GMm \left[ -\frac{1}{r} \right]_{R}^{3R}\)

4. Evaluating the Integral:

   \(W = GMm \left( (-\frac{1}{3R}) - (-\frac{1}{R}) \right)\)

   \(W = GMm \left( -\frac{1}{3R} + \frac{1}{R} \right)\)

   \(W = GMm \left( \frac{-1 + 3}{3R} \right)\)

   \(W = GMm \left( \frac{2}{3R} \right)\)

   \(W = \frac{2}{3} \frac{GMm}{R}\)

5. Substituting GM: Now, substitute \(GM = gR^2\) into the expression for work done.

   \(W = \frac{2}{3} \frac{(gR^2)m}{R}\)

   \(W = \frac{2}{3} \frac{gR^2 m}{R}\)

   Simplify by cancelling one R:

   \(W = \frac{2}{3} mgR\)

Final Result Explanation

The calculation shows that the work done to raise a mass m from the Earth's surface to a height equal to twice the Earth's radius (\(h = 2R\)) is \(\frac{2}{3}mgR\). This value represents the change in potential energy of the mass.