An object is taken to height 2R above the surface of earth , the increase in potential energy is [ R is radius of earth ]

Options

A. `(mgR)/2 `
B. `(mgR)/3 `
C. `(2mgR)/3`
D. 2mgR

Answer

To find the increase in potential energy when an object is taken to a height of \(2R\) above the surface of the Earth, we can follow these steps:

### Step 1: Understand the Potential Energy Formula
The gravitational potential energy (\(U\)) between two masses is given by the formula:
\[ U = -\frac{G m_1 m_2}{r} \]
where:
- \(G\) is the gravitational constant,
- \(m_1\) is the mass of the Earth,
- \(m_2\) is the mass of the object,
- \(r\) is the distance between the centers of the two masses.

### Step 2: Calculate Initial Potential Energy (\(U_i\))
When the object is at the surface of the Earth (height = 0), the distance \(r\) from the center of the Earth to the object is equal to the radius of the Earth \(R\). Thus, the initial potential energy (\(U_i\)) is:
\[ U_i = -\frac{G m M}{R} \]
where \(m\) is the mass of the object and \(M\) is the mass of the Earth.

### Step 3: Calculate Final Potential Energy (\(U_f\))
When the object is raised to a height of \(2R\) above the surface of the Earth, the total distance \(r\) from the center of the Earth to the object becomes \(R + 2R = 3R\). Therefore, the final potential energy (\(U_f\)) is:
\[ U_f = -\frac{G m M}{3R} \]

### Step 4: Calculate the Increase in Potential Energy
The increase in potential energy (\(\Delta U\)) is given by:
\[ \Delta U = U_f - U_i \]
Substituting the expressions for \(U_f\) and \(U_i\):
\[ \Delta U = \left(-\frac{G m M}{3R}\right) - \left(-\frac{G m M}{R}\right) \]
This simplifies to:
\[ \Delta U = -\frac{G m M}{3R} + \frac{G m M}{R} \]
Factoring out the common terms:
\[ \Delta U = G m M \left(-\frac{1}{3R} + \frac{1}{R}\right) \]
\[ \Delta U = G m M \left(\frac{-1 + 3}{3R}\right) = G m M \left(\frac{2}{3R}\right) \]

### Step 5: Substitute for \(G\)
Using the relation \(g = \frac{G M}{R^2}\), we can express \(G M\) in terms of \(g\):
\[ G M = g R^2 \]
Substituting this back into the equation for \(\Delta U\):
\[ \Delta U = \frac{2}{3} \cdot g R^2 \cdot \frac{m}{R} = \frac{2}{3} g m R \]

### Final Answer
Thus, the increase in potential energy when the object is taken to a height of \(2R\) above the surface of the Earth is:
\[ \Delta U = \frac{2}{3} g m R \]