A particle of mass 100 g is kept on the surface of a uniform sphere of mass 10 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle away from the sphere.

Explanation

To solve the problem of finding the work done against the gravitational force to take a particle away from a uniform sphere, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the particle, \( m_1 = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Mass of the sphere, \( m_2 = 10 \, \text{kg} \)
- Radius of the sphere, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)

2. **Calculate the Initial Gravitational Potential Energy (PE_initial):**
The formula for gravitational potential energy between two masses is given by:
\[ PE = -\frac{G \cdot m_1 \cdot m_2}{r} \]
Substituting the values:
\[ PE_{\text{initial}} = -\frac{(6.67 \times 10^{-11}) \cdot (0.1) \cdot (10)}{0.1} \]
Simplifying this:
\[ PE_{\text{initial}} = -\frac{6.67 \times 10^{-11} \cdot 1}{0.1} = -6.67 \times 10^{-10} \, \text{J} \]

3. **Determine the Final Gravitational Potential Energy (PE_final):**
When the particle is taken away to infinity, the gravitational potential energy approaches zero:
\[ PE_{\text{final}} = 0 \, \text{J} \]

4. **Calculate the Work Done (W):**
The work done against the gravitational force is equal to the change in potential energy:
\[ W = PE_{\text{final}} - PE_{\text{initial}} \]
Substituting the values:
\[ W = 0 - (-6.67 \times 10^{-10}) = 6.67 \times 10^{-10} \, \text{J} \]

5. **Final Answer:**
The work done against the gravitational force to take the particle away from the sphere is:
\[ W = 6.67 \times 10^{-10} \, \text{J} \]