If `w_(1)` joule work is done against gravitational attraction, to carry 10 kg mass from earth's surface to infinity. Then the magnitude of workdone by gravitational attraction in bringing 20 kg mass from infinity to the centre without any acceleration is `w_(2)`. Now if `w_(2)=n xx w_(1)`. Find the value of n.

Answer

To solve the problem, we need to find the value of \( n \) in the equation \( w_2 = n \times w_1 \), where \( w_1 \) is the work done against gravitational attraction to carry a 10 kg mass from the Earth's surface to infinity, and \( w_2 \) is the work done by gravitational attraction in bringing a 20 kg mass from infinity to the center of the Earth.

### Step-by-Step Solution:

1. **Calculate \( w_1 \)**:
- The work done \( w_1 \) to carry a mass \( m \) from the Earth's surface to infinity is given by:
\[ w_1 = m \times \Delta V \]
- Here, \( m = 10 \, \text{kg} \) and \( \Delta V \) is the change in gravitational potential energy. The gravitational potential at the Earth's surface is:
\[ V_i = -\frac{G M}{R} \]
and at infinity,
\[ V_f = 0. \]
- Therefore, the change in potential is:
\[ \Delta V = V_f - V_i = 0 - \left(-\frac{G M}{R}\right) = \frac{G M}{R}. \]
- Substituting this into the equation for \( w_1 \):
\[ w_1 = 10 \times \frac{G M}{R} = \frac{10 G M}{R}. \]

2. **Calculate \( w_2 \)**:
- The work done \( w_2 \) in bringing a 20 kg mass from infinity to the center of the Earth is given by:
\[ w_2 = m \times \Delta V \]
- Here, \( m = 20 \, \text{kg} \). The gravitational potential at the center of the Earth can be derived from the formula for potential inside a uniform sphere:
\[ V = -\frac{3 G M}{2 R}. \]
- At infinity, the potential is \( 0 \), thus:
\[ \Delta V = V_f - V_i = -\frac{3 G M}{2 R} - 0 = -\frac{3 G M}{2 R}. \]
- Therefore, substituting into the equation for \( w_2 \):
\[ w_2 = 20 \times \left(-\frac{3 G M}{2 R}\right) = -\frac{60 G M}{2 R} = -\frac{30 G M}{R}. \]

3. **Relate \( w_2 \) to \( w_1 \)**:
- We have:
\[ w_1 = \frac{10 G M}{R} \]
and
\[ w_2 = -\frac{30 G M}{R}. \]
- To find \( n \), we set up the equation:
\[ w_2 = n \times w_1. \]
- Substituting the values:
\[ -\frac{30 G M}{R} = n \times \frac{10 G M}{R}. \]
- Dividing both sides by \( \frac{G M}{R} \) (assuming \( G, M, R \) are not zero):
\[ -30 = n \times 10. \]
- Solving for \( n \):
\[ n = -3. \]

### Final Answer:
The value of \( n \) is \( -3 \).