Energy required to move a body of mass m from an orbit of radius 2R to 3R is

Explanation

To find the energy required to move a body of mass \( m \) from an orbit of radius \( 2R \) to \( 3R \), we can follow these steps:

### Step 1: Calculate the initial energy in the orbit of radius \( 2R \)

The gravitational potential energy \( U \) of a body in a circular orbit is given by the formula:

\[ U = -\frac{G M m}{r} \]

where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the object,
- \( r \) is the radius of the orbit.

For the initial orbit at radius \( 2R \):

\[ U_{\text{initial}} = -\frac{G M m}{2R} \]

### Step 2: Calculate the final energy in the orbit of radius \( 3R \)

For the final orbit at radius \( 3R \):

\[ U_{\text{final}} = -\frac{G M m}{3R} \]

### Step 3: Calculate the energy required to move from \( 2R \) to \( 3R \)

The energy required \( E \) to move the body from the initial orbit to the final orbit is given by the difference in potential energy:

\[ E = U_{\text{final}} - U_{\text{initial}} \]

Substituting the values from Steps 1 and 2:

\[ E = \left(-\frac{G M m}{3R}\right) - \left(-\frac{G M m}{2R}\right) \]

### Step 4: Simplify the expression

This simplifies to:

\[ E = -\frac{G M m}{3R} + \frac{G M m}{2R} \]

To combine these fractions, we need a common denominator, which is \( 6R \):

\[ E = \left(-\frac{2G M m}{6R} + \frac{3G M m}{6R}\right) \]

\[ E = \frac{(3 - 2) G M m}{6R} \]

\[ E = \frac{G M m}{6R} \]

### Final Answer

Thus, the energy required to move the body from an orbit of radius \( 2R \) to \( 3R \) is:

\[ E = \frac{G M m}{6R} \]

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