The work that must be done in lifting a body of weight P from the surface of the earth to a height h is

Options

A. `(PRh)/(R-h)`
B. `(R+h)/(PRh)`
C. `(PRh)/(R+h)`
D. `(R-h)/(PRh)`

Answer

To find the work done in lifting a body of weight \( P \) from the surface of the Earth to a height \( h \), we can follow these steps:

### Step 1: Understand the Concept of Work Done
Work done in lifting an object against gravity is equal to the change in gravitational potential energy. The gravitational potential energy at a height \( h \) can be calculated using the formula:

\[ U = -\frac{G M m}{r} \]

where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the object, and \( r \) is the distance from the center of the Earth.

### Step 2: Calculate Initial and Final Potential Energy
1. **Initial Potential Energy (at the surface)**:
At the surface of the Earth (height = 0), the potential energy is given by:
\[ U_i = -\frac{G M m}{R} \]
where \( R \) is the radius of the Earth.

2. **Final Potential Energy (at height \( h \))**:
At height \( h \), the potential energy becomes:
\[ U_f = -\frac{G M m}{R + h} \]

### Step 3: Calculate the Change in Potential Energy
The work done \( W \) in lifting the object is equal to the change in potential energy:
\[ W = U_f - U_i \]
Substituting the values we found:
\[ W = \left(-\frac{G M m}{R + h}\right) - \left(-\frac{G M m}{R}\right) \]
\[ W = \frac{G M m}{R} - \frac{G M m}{R + h} \]

### Step 4: Factor Out Common Terms
Factoring out \( G M m \):
\[ W = G M m \left(\frac{1}{R} - \frac{1}{R + h}\right) \]

### Step 5: Simplify the Expression
To simplify:
\[ W = G M m \left(\frac{(R + h) - R}{R(R + h)}\right) \]
\[ W = G M m \left(\frac{h}{R(R + h)}\right) \]

### Step 6: Relate Mass to Weight
Since the weight \( P \) is given by \( P = mg \), we can express \( m \) in terms of \( P \):
\[ m = \frac{P}{g} \]
Substituting this into the work equation:
\[ W = G M \left(\frac{P}{g}\right) \left(\frac{h}{R(R + h)}\right) \]

### Step 7: Substitute \( g \) for \( \frac{GM}{R^2} \)
Using \( g = \frac{GM}{R^2} \):
\[ W = \frac{P h}{g} \cdot \frac{g R^2}{R(R + h)} \]
\[ W = \frac{P h R}{R + h} \]

### Final Result
Thus, the work done in lifting the body is:
\[ W = \frac{P h}{R + h} \]