What is the change in potential energy of a body of mass 10kg when it is taken to a height of 2R from the earth's surface ? G given

Explanation

To find the change in potential energy of a body of mass 10 kg when it is taken to a height of 2R from the Earth's surface, we can follow these steps:

### Step 1: Understand the Initial and Final Positions
- The initial position of the body is at the Earth's surface, which is at a distance of \( R \) from the center of the Earth.
- The final position of the body is at a height of \( 2R \) above the Earth's surface, making the total distance from the center of the Earth \( R + 2R = 3R \).

### Step 2: Write the Formula for Gravitational Potential Energy
The gravitational potential energy (U) of a mass \( m \) at a distance \( r \) from the center of the Earth is given by the formula:
\[ U = -\frac{G M m}{r} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the object,
- \( r \) is the distance from the center of the Earth.

### Step 3: Calculate Initial Potential Energy
For the initial position at the Earth's surface (distance \( r = R \)):
\[ U_{\text{initial}} = -\frac{G M m}{R} \]

### Step 4: Calculate Final Potential Energy
For the final position at a height of \( 2R \) (distance \( r = 3R \)):
\[ U_{\text{final}} = -\frac{G M m}{3R} \]

### Step 5: Calculate Change in Potential Energy
The change in potential energy (\( \Delta U \)) is given by:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values we calculated:
\[ \Delta U = \left(-\frac{G M m}{3R}\right) - \left(-\frac{G M m}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{G M m}{3R} + \frac{G M m}{R} \]
\[ \Delta U = \frac{G M m}{R} - \frac{G M m}{3R} \]
\[ \Delta U = \frac{G M m}{R} \left(1 - \frac{1}{3}\right) = \frac{G M m}{R} \cdot \frac{2}{3} \]
\[ \Delta U = \frac{2 G M m}{3R} \]

### Step 6: Substitute the Values
Now, we substitute the known values:
- Mass of the body \( m = 10 \, \text{kg} \)
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- Mass of the Earth \( M = 6 \times 10^{24} \, \text{kg} \)
- Radius of the Earth \( R = 6.4 \times 10^6 \, \text{m} \)

Calculating:
\[ \Delta U = \frac{2 \cdot (6.67 \times 10^{-11}) \cdot (6 \times 10^{24}) \cdot (10)}{3 \cdot (6.4 \times 10^6)} \]

### Step 7: Final Calculation
Calculating the above expression:
\[ \Delta U \approx \frac{2 \cdot 6.67 \cdot 6 \cdot 10^{13}}{3 \cdot 6.4} \approx \frac{80.04 \times 10^{13}}{19.2} \approx 4.18 \times 10^6 \, \text{J} \]

### Final Answer
The change in potential energy when the body is taken to a height of \( 2R \) from the Earth's surface is approximately \( 4.18 \times 10^6 \, \text{J} \).

---