A planet of mass `M` has uniform density in a spherical volume of radius `R`. Calculate the work done by the external agent to deassemble the planet in eight identical spherical part against gravitational pull amongst its constitute particle.

Answer

To solve the problem of calculating the work done by an external agent to deassemble a planet of mass \( M \) and radius \( R \) into eight identical spherical parts against gravitational pull, we can follow these steps:

### Step 1: Determine the mass of each smaller sphere
Since the planet is divided into 8 identical spheres, the mass of each smaller sphere \( m \) can be calculated as:
\[ m = \frac{M}{8} \]

### Step 2: Calculate the radius of each smaller sphere
The volume of the original planet is given by:
\[ V = \frac{4}{3} \pi R^3 \]
The volume of each smaller sphere is:
\[ v = \frac{4}{3} \pi r^3 \]
Since the total volume remains the same, we have:
\[ \frac{4}{3} \pi R^3 = 8 \left(\frac{4}{3} \pi r^3\right) \]
This simplifies to:
\[ R^3 = 8r^3 \implies r = \frac{R}{2} \]

### Step 3: Calculate the gravitational potential energy of the original planet
The gravitational potential energy \( U \) of a uniform sphere is given by:
\[ U = -\frac{3}{5} \frac{GM^2}{R} \]

### Step 4: Calculate the gravitational potential energy of the 8 smaller spheres
For each smaller sphere, the gravitational potential energy is:
\[ U' = -\frac{3}{5} \frac{G m^2}{r} = -\frac{3}{5} \frac{G \left(\frac{M}{8}\right)^2}{\frac{R}{2}} = -\frac{3}{5} \frac{G \frac{M^2}{64}}{\frac{R}{2}} = -\frac{3GM^2}{160R} \]
Thus, for 8 spheres, the total gravitational potential energy \( U_{total} \) is:
\[ U_{total} = 8 \left(-\frac{3GM^2}{160R}\right) = -\frac{3GM^2}{20R} \]

### Step 5: Calculate the work done by the external agent
The work done \( W \) by the external agent to deassemble the planet can be calculated using the change in gravitational potential energy:
\[ W = U_{final} - U_{initial} \]
Where:
- \( U_{initial} = -\frac{3GM^2}{5R} \)
- \( U_{final} = -\frac{3GM^2}{20R} \)

Now substituting these values:
\[ W = \left(-\frac{3GM^2}{20R}\right) - \left(-\frac{3GM^2}{5R}\right) \]
Converting \( -\frac{3GM^2}{5R} \) to a common denominator:
\[ -\frac{3GM^2}{5R} = -\frac{12GM^2}{20R} \]
Thus:
\[ W = -\frac{3GM^2}{20R} + \frac{12GM^2}{20R} = \frac{9GM^2}{20R} \]

### Final Answer
The work done by the external agent to deassemble the planet into eight identical spherical parts is:
\[ W = \frac{9GM^2}{20R} \]