A spaceship is orbiting the earth in a circular orbit at a height equal to radius of the earth `(R_(c) = 6400 km)` from the surface of the earth. An astronaut is on a space walk outside the spaceship. He is at a distance of l = 200 m from the ship and is connected to it with a simple cable which can sustain a maximum tension of10 N. Assume that the centre of the earth, the spaceship and the astronaut are in a line. Mass of astronaut along with all his accessories is 100 kg. Do you think that a weak cable that can only take a load of 10 N, can prevent him from drifting in space ? Make a guess. Estimate the tension in the cable. [Acceleration due to gravity on the surface of`[ Earth =9.8m//s^(2)`]

Answer

To determine whether the cable can prevent the astronaut from drifting in space, we need to calculate the tension in the cable when the astronaut is at a distance of 200 m from the spaceship.

### Step-by-Step Solution:

1. **Identify the total distance from the center of the Earth to the astronaut**:
- The radius of the Earth \( R_c = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \).
- The height of the spaceship from the Earth's surface is equal to the radius of the Earth, so the distance from the center of the Earth to the spaceship \( R \) is:
\[ R = R_c + R_c = 2R_c = 2 \times 6400 \times 10^3 \, \text{m} = 12800 \times 10^3 \, \text{m} \]
- The distance from the spaceship to the astronaut is \( l = 200 \, \text{m} \).
- Therefore, the total distance from the center of the Earth to the astronaut \( R' \) is:
\[ R' = R + l = 12800 \times 10^3 + 200 = 12800.2 \times 10^3 \, \text{m} \]

2. **Calculate the gravitational force acting on the astronaut**:
- The gravitational force \( F_g \) acting on the astronaut due to the Earth can be calculated using Newton's law of gravitation:
\[ F_g = \frac{G M m}{R'^2} \]
- Where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the astronaut (100 kg).

3. **Determine the centripetal force required for the astronaut**:
- The centripetal force \( F_c \) required for the astronaut to maintain circular motion is given by:
\[ F_c = m \omega^2 R' \]
- Here, \( \omega \) is the angular velocity of the spaceship, which can be related to the gravitational force acting on the spaceship.

4. **Calculate the tension in the cable**:
- The tension \( T \) in the cable will be the difference between the gravitational force and the centripetal force:
\[ T = F_g - F_c \]

5. **Estimate the values**:
- We can simplify our calculations by noting that \( R' \) is very close to \( R \) because \( l \) is much smaller than \( R \). Therefore, we can approximate:
\[ R' \approx R \]
- This leads to:
\[ T \approx \frac{G M m}{R^2} - m \omega^2 R \]

6. **Check if the tension is within limits**:
- If \( T \) is less than or equal to 10 N, the cable can hold the astronaut. If \( T \) exceeds 10 N, the cable will break.

### Final Calculation:
Using the values of \( G \), \( M \), and \( R \), we can calculate \( T \) and compare it with 10 N.

### Conclusion:
1. If \( T < 10 \, \text{N} \), the cable can prevent the astronaut from drifting.
2. If \( T > 10 \, \text{N} \), the cable will break, and the astronaut will drift away.