Q. Two astronauts have deserted their space ships 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?
Explanation
(b) Here, $m_{1} = m_{2} = 100$ kg ; $r = 100$m
Acceleration of first astronaut,
$a_{1} = \frac{Gm_{1}m_{2}}{r^{2}} \times \frac{1}{m_{1}} = \frac{Gm_{1}}{r^{2}}$
Acceleration of second astronaut,
$a_{2} = \frac{Gm_{1}m_{2}}{r^{2}} \times \frac{1}{m_{2}} = \frac{Gm_{2}}{r^{2}}$
Net acceleration of approach
$a = a_{1} + a_{2} = \frac{Gm_{2}}{r^{2}} + \frac{Gm_{1}}{r^{2}} = \frac{2Gm_{1}}{r^{2}}$
$= \frac{2 \times (6.67 \times 10^{- 11}) \times 100}{(100)^{2}}$
$= 2 \times 6.67 \times 10^{- 13}ms^{- 2}$
As $s = \frac{1}{2}at^{2}$
$\therefore t = \left( \frac{2s}{a} \right)^{1/2} = \left\lbrack \frac{2 \times (1/100)}{2 \times 6.67 \times 10^{- 13}} \right\rbrack^{1/2}$second
On solving we get $t = 1.41$ days
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.