At the surface of earth, potential energy of a particle is U and potential is V.Change in potential energy and potential at height h=R are suppose `DeltaU" and " Delta V`.Then

Options

A. `DeltaU=-U//2`
B. `DeltaU=U//2`
C. `DeltaV=V//2`
D. `DeltaV=-V//2`

Answer

To solve the problem, we need to find the change in potential energy (ΔU) and the change in potential (ΔV) of a particle when it is moved from the surface of the Earth to a height equal to the radius of the Earth (h = R).

### Step-by-Step Solution

1. **Identify the Given Variables:**
- At the surface of the Earth:
- Potential energy (U) = \(-\frac{GMm}{R}\)
- Potential (V) = \(-\frac{GM}{R}\)
- At height \(h = R\):
- The distance from the center of the Earth to the particle is \(R + R = 2R\).

2. **Calculate Potential Energy at Height \(h = R\):**
- The potential energy at height \(h\) (U1) is given by:
\[ U_1 = -\frac{GMm}{2R} \]

3. **Calculate the Change in Potential Energy (ΔU):**
- The change in potential energy is given by:
\[ \Delta U = U_1 - U \]
- Substituting the values:
\[ \Delta U = \left(-\frac{GMm}{2R}\right) - \left(-\frac{GMm}{R}\right) \]
- Simplifying:
\[ \Delta U = -\frac{GMm}{2R} + \frac{GMm}{R} = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R} \]
- Thus:
\[ \Delta U = -\frac{U}{2} \]

4. **Calculate Potential at Height \(h = R\):**
- The potential at height \(h\) (V1) is given by:
\[ V_1 = -\frac{GM}{2R} \]

5. **Calculate the Change in Potential (ΔV):**
- The change in potential is given by:
\[ \Delta V = V_1 - V \]
- Substituting the values:
\[ \Delta V = \left(-\frac{GM}{2R}\right) - \left(-\frac{GM}{R}\right) \]
- Simplifying:
\[ \Delta V = -\frac{GM}{2R} + \frac{GM}{R} = \frac{GM}{R} - \frac{GM}{2R} = \frac{GM}{2R} \]
- Thus:
\[ \Delta V = -\frac{V}{2} \]

### Final Results:
- Change in potential energy: \(\Delta U = -\frac{U}{2}\)
- Change in potential: \(\Delta V = -\frac{V}{2}\)