Two astronauts have deserted their spaceship in a region of space far from the gravitational attraction of any other body. Each has a mass of `100 kg` and they are `100 m` apart. They are initially at rest relative to one another. How long will it be before the gravitational attraction brings them `1 cm` closer together?

Options

A. `2.52`days
B. `1.41` days
C. `0.70` days
D. `1.41 s`

Explanation

To solve the problem, we need to determine how long it will take for two astronauts, each with a mass of 100 kg and initially 100 m apart, to be brought 1 cm closer together due to their mutual gravitational attraction.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of each astronaut, \( m = 100 \, \text{kg} \)
- Initial distance between the astronauts, \( r = 100 \, \text{m} \)
- Relative displacement to be achieved, \( d = 1 \, \text{cm} = 0.01 \, \text{m} \)

2. **Calculate the Gravitational Force:**
The gravitational force \( F \) between the two astronauts can be calculated using Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{r^2} \]
where \( G \) is the gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).

Substituting the values:
\[ F = \frac{(6.67 \times 10^{-11}) \times (100) \times (100)}{(100)^2} = \frac{6.67 \times 10^{-11} \times 10000}{10000} = 6.67 \times 10^{-11} \, \text{N} \]

3. **Calculate the Acceleration of Each Astronaut:**
The acceleration \( a \) experienced by each astronaut due to the gravitational force is given by:
\[ a = \frac{F}{m} = \frac{6.67 \times 10^{-11}}{100} = 6.67 \times 10^{-13} \, \text{m/s}^2 \]

4. **Calculate the Relative Acceleration:**
Since both astronauts are attracting each other, the relative acceleration \( a_{rel} \) is:
\[ a_{rel} = 2a = 2 \times 6.67 \times 10^{-13} = 1.334 \times 10^{-12} \, \text{m/s}^2 \]

5. **Use the Kinematic Equation:**
We can use the kinematic equation for displacement to find the time \( t \) it takes to achieve the relative displacement of 1 cm:
\[ d = u_{rel} t + \frac{1}{2} a_{rel} t^2 \]
Given that the initial relative velocity \( u_{rel} = 0 \):
\[ d = \frac{1}{2} a_{rel} t^2 \]
Rearranging gives:
\[ t^2 = \frac{2d}{a_{rel}} \implies t = \sqrt{\frac{2d}{a_{rel}}} \]

6. **Substituting the Values:**
\[ t = \sqrt{\frac{2 \times 0.01}{1.334 \times 10^{-12}}} \]
\[ t = \sqrt{\frac{0.02}{1.334 \times 10^{-12}}} \approx \sqrt{1.498 \times 10^{10}} \approx 122.4 \, \text{s} \]

7. **Convert Time to Days:**
To convert seconds to days:
\[ t \approx \frac{122.4}{86400} \approx 0.00142 \, \text{days} \approx 1.41 \, \text{days} \]

### Final Answer:
The time it will take before the gravitational attraction brings them 1 cm closer together is approximately **1.41 days**.