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`

Answer

B

Explanation

`a_(1)` acceleration of first
`=(Gm_(1)m_(2))/(r_(2))xx1/m-1=(Gm_2)/(r^(2))=(6.67xx10^-11xx100)/((100)^(2))`
`=6.67xx10^(-13)ms^(-1)`
`a_(2)=` acceleration of second
`=(Gm_(1))/(r^(2))=(6.67xx10^(-11)xx100)/((100)^(2))=6.67xx10^(-13)ms^(-1)`
Net acceleration of approach
`a=a_(1)+a_(2)=2xx6.67xx10^(-13)ms^(-1)`
Now `s=1/2at^(2)`
`1xx10^(-2)=1/2xx6.67xx10^(-13)xx2xxt^(2)`
Solving we get `t=1.41`days.