In order to shift a body of mass `m` from a circular orbit of radius `3R` to a higher orbit of radius `5R` around the earth, the work done is

Options

A. `(3GMm)/(5R)`
B. `(GMm)/(2R)`
C. `2/15(GMm)/R`
D. `(GMm)/(5R)`

Explanation

To find the work done in shifting a body of mass \( m \) from a circular orbit of radius \( 3R \) to a higher orbit of radius \( 5R \) around the Earth, we can follow these steps:

### Step 1: Understand the Potential Energy in Circular Orbits
The gravitational potential energy \( U \) of a body of mass \( m \) in a circular orbit around the Earth is given by the formula:
\[ U = -\frac{GMm}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the radius of the orbit.

### Step 2: Calculate Initial Potential Energy
For the initial orbit at radius \( 3R \):
\[ U_i = -\frac{GMm}{3R} \]

### Step 3: Calculate Final Potential Energy
For the final orbit at radius \( 5R \):
\[ U_f = -\frac{GMm}{5R} \]

### Step 4: Calculate the Change in Potential Energy
The work done \( W \) by an external agent to shift the body from the initial orbit to the final orbit is equal to the change in potential energy:
\[ W = U_f - U_i \]
Substituting the values we calculated:
\[ W = \left(-\frac{GMm}{5R}\right) - \left(-\frac{GMm}{3R}\right) \]
\[ W = -\frac{GMm}{5R} + \frac{GMm}{3R} \]

### Step 5: Simplify the Expression
To simplify the expression, we need a common denominator. The least common multiple of \( 5R \) and \( 3R \) is \( 15R \):
\[ W = \left(-\frac{3GMm}{15R}\right) + \left(\frac{5GMm}{15R}\right) \]
\[ W = \frac{2GMm}{15R} \]

### Final Answer
Thus, the work done to shift the body from a circular orbit of radius \( 3R \) to a higher orbit of radius \( 5R \) is:
\[ W = \frac{2GMm}{15R} \]

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