A body of mass `m` is raised to a height 10 R from the surface of the earth, where R is the radius of the earth. Find the increase in potential energy. (G = universal constant of gravitational, M = mass of the earth and g= acceleration due to gravity)

Options

A. `(GMm)/(11R)`
B. `(GMm)/(10 R)`
C. `(mgR)/(11G)`
D. `(10 GMm)/(11 R)`

Explanation

To find the increase in potential energy when a body of mass `m` is raised to a height of \(10R\) from the surface of the Earth, we can follow these steps:

### Step 1: Understand the formula for gravitational potential energy
The gravitational potential energy (U) of a mass `m` at a distance `d` from the center of the Earth is given by the formula:
\[ U = -\frac{GMm}{d} \]
where:
- \(G\) is the universal gravitational constant,
- \(M\) is the mass of the Earth,
- \(m\) is the mass of the object,
- \(d\) is the distance from the center of the Earth.

### Step 2: Calculate the initial potential energy (U_initial)
When the body is on the surface of the Earth, the distance from the center of the Earth is equal to the radius of the Earth \(R\). Therefore, the initial potential energy \(U_{\text{initial}}\) is:
\[ U_{\text{initial}} = -\frac{GMm}{R} \]

### Step 3: Calculate the final potential energy (U_final)
When the body is raised to a height of \(10R\), the total distance from the center of the Earth becomes:
\[ d = R + 10R = 11R \]
Thus, the final potential energy \(U_{\text{final}}\) is:
\[ U_{\text{final}} = -\frac{GMm}{11R} \]

### Step 4: Calculate the change in potential energy (ΔU)
The change in potential energy \(\Delta U\) is given by the difference between the final and initial potential energies:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values we found:
\[ \Delta U = \left(-\frac{GMm}{11R}\right) - \left(-\frac{GMm}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{GMm}{11R} + \frac{GMm}{R} \]
To combine these fractions, we find a common denominator:
\[ \Delta U = \frac{GMm}{R} - \frac{GMm}{11R} = \frac{GMm \cdot 11}{11R} - \frac{GMm}{11R} = \frac{10GMm}{11R} \]

### Final Answer
Thus, the increase in potential energy when the body is raised to a height of \(10R\) from the surface of the Earth is:
\[ \Delta U = \frac{10GMm}{11R} \]