A satellite of earth of mass 'm' is taken from orbital radius 2R to 3R, then minimum work done is :-

Options

A. `(GMm)/(6R)`
B. `(GMm)/(12R)`
C. `(GMm)/(24 R)`
D. `(GMm)/(3R)`

Answer

To find the minimum work done when a satellite of mass 'm' is moved from an orbital radius of 2R to 3R, we can follow these steps:

### Step 1: Understand the Potential Energy in Orbit
The gravitational potential energy (U) of a satellite in orbit 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 gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the Earth.

### Step 2: Calculate Initial Potential Energy (U_i)
For the initial position of the satellite at an orbital radius of \( 2R \):
\[ U_i = -\frac{G M m}{2R} \]

### Step 3: Calculate Final Potential Energy (U_f)
For the final position of the satellite at an orbital radius of \( 3R \):
\[ U_f = -\frac{G M m}{3R} \]

### Step 4: Calculate the Change in Potential Energy (ΔU)
The work done (W) in moving the satellite is equal to the change in potential energy:
\[ W = U_f - U_i \]
Substituting the values we found:
\[ W = \left(-\frac{G M m}{3R}\right) - \left(-\frac{G M m}{2R}\right) \]

### Step 5: Simplify the Expression
Now, we can simplify the expression:
\[ W = -\frac{G M m}{3R} + \frac{G M m}{2R} \]
\[ W = \frac{G M m}{R} \left(-\frac{1}{3} + \frac{1}{2}\right) \]

### Step 6: Find a Common Denominator
To combine the fractions, we find a common denominator (which is 6):
\[ -\frac{1}{3} = -\frac{2}{6} \]
\[ \frac{1}{2} = \frac{3}{6} \]
Thus:
\[ W = \frac{G M m}{R} \left(\frac{-2 + 3}{6}\right) \]
\[ W = \frac{G M m}{R} \left(\frac{1}{6}\right) \]

### Step 7: Final Result
So the minimum work done is:
\[ W = \frac{G M m}{6R} \]

### Conclusion
The minimum work done in moving the satellite from an orbital radius of 2R to 3R is:
\[ W = \frac{G M m}{6R} \]

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