A projectile of mass m is thrown vertically up with an initial velocity v from the surface of earth (mass of earth = M). If it comes to rest at a height h, the change in its potential energy is

Options

A. `GMmh//R(R+h)`
B. `GMmh^(2)//R(R+h)^(2)`
C. `GMmhR//R(R+h)`
D. `GMm//hR(R+h)`

Answer

To find the change in potential energy of a projectile of mass \( m \) thrown vertically upward from the surface of the Earth to a height \( h \), we can follow these steps:

### Step 1: Understand the Initial Potential Energy
The initial potential energy \( U_i \) of the projectile at the surface of the Earth can be expressed using the formula for gravitational potential energy:
\[ U_i = -\frac{G M m}{R} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the projectile, and \( R \) is the radius of the Earth.

### Step 2: Determine the Final Potential Energy
When the projectile reaches a height \( h \), its final potential energy \( U_f \) can be calculated as:
\[ U_f = -\frac{G M m}{R + h} \]
Here, \( R + h \) is the distance from the center of the Earth to the projectile at height \( h \).

### Step 3: Calculate the Change in Potential Energy
The change in potential energy \( \Delta U \) is given by the difference between the final and initial potential energies:
\[ \Delta U = U_f - U_i \]
Substituting the expressions for \( U_f \) and \( U_i \):
\[ \Delta U = \left(-\frac{G M m}{R + h}\right) - \left(-\frac{G M m}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{G M m}{R + h} + \frac{G M m}{R} \]
\[ \Delta U = G M m \left(\frac{1}{R} - \frac{1}{R + h}\right) \]

### Step 4: Simplify the Expression
To further simplify:
\[ \Delta U = G M m \left(\frac{(R + h) - R}{R(R + h)}\right) \]
\[ \Delta U = G M m \left(\frac{h}{R(R + h)}\right) \]

### Final Result
Thus, the change in potential energy when the projectile reaches height \( h \) is:
\[ \Delta U = \frac{G M m h}{R(R + h)} \]