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 $100 k g$ 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 $1 c m$ closer together?

2.52

days

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1.41

days

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0.70

days

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1.41

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Solution

The correct option is B $1.41$ days
AS both astronauts come closer by 1 cm thus one has to be moved by $\frac{1}{2} c m$ .
Accleration, $a = \frac{G M}{r^{2}}$
Thus using $s = u t + \frac{1}{2} a t^{2}$
$⟹ t^{2} = \frac{2 s r^{2}}{G M}$
Given: $M = 100 k g s = \frac{1}{2} c m r = 100 m u = 0$
$t^{2} = \frac{2 (\frac{1}{200}) (100)^{2}}{6.67 \times 10^{- 11} (100)} ⟹ t = 1.22 \times 10^{5} s$
Thus, $t = 1.41 d a y s$

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The correct option is B 1.41 daysAS both astronauts come closer by 1 cm thus one has to be moved by 12cm.Accleration, a=GMr2 Thus using s=ut+12at2⟹t2=2sr2GMGiven: M=100kgs=12cmr=100mu=0t2=2(1200)(100)26.67×10−11(100)⟹t=1.22×105sThus, t=1.41days