Question

Two identical 100 kg objects (of negligible size) are in space 100 m away from each other. The force of gravity causes both objects to accelerate towards one another. Find the time it takes for the objects to collide.

Answer 1

m1=m2 = 100 kg
x = distance between the objects = 100m
F = Gm1m2/x^2
therefore
F = 6.6 x 10-11 * 100 * 100 / 100^2 = 6.6 x 10-11
also acceleration a = F/m = 6.6 x 10-11/100=6.6 x 10-9
so both objects are gonna move towards each other with an accelertaion a
so they both will cover x/2 distance in same time
and collide
so we need to find that time
in which each object covers a distance of x/2
s = ut + 1/2at^2
100/2 = 1/2 6.6 x 10-9 t^2
implies
t^2 = 100/ 6.6 x 10-9
t = (150 * 10-8)^1/2
t = 12.2 * 10-4 seconds
Keep me posted if it is right answer, I m not sure abt its correctness :)

Answer 2

I don't believe that is right. Because they are constantly moving toward each other, acceleration is changing (always increasing).

Answer 3

eh yes , missed that point totally

Answer 4

ok wait it cud be solved by Integration or something
F = Integrating x from 100 m to 0m (Gmm/x^2) as initial distance is 100 and when they collide x becomes 0 .
Lemme Integrate

Answer 5

The integral of the acceleration will yield velocity; the integral of the velocity will give the distance.

I'm figuring it out, but I can't figure out how to account for both of them moving. We are only looking at the force on one. Should we treat the distance as 50?

Answer 6

yeah we shud treat the distance as 50 , as both the objects will move equal distance as they r movin with equal acceleration

Answer 7

got the soution yet?

Answer 8

may be we can use conservation of energy, at any time where object have moved x
\[-G\frac{m^2}{x_0}=2\frac{mv^2}{2}-G\frac{m^2}{x_0-2x}\]
thus velocity of the object:
\[v=\sqrt{Gm(\frac{1}{x_0-2x}-\frac{1}{x_0})}=\sqrt{GM\frac{2x}{x_0-2x}}\]
then distance travelled in time 'dt' can be written as
\[dx=vdt\]
moving velocity to left side,integrating both sides
\[\int_0^{x_0/2}\frac{dx}{\sqrt{GM\frac{2x}{x_0-2x}}}=\int_0^tdt\]

does it make sense?

Answer 9

in second line I missed one 'x_0' thus
\[t=\frac{1}{\sqrt{GM}}\int_0^{x_0/2}\Big(\sqrt{\frac{x_0^2}{2x}-x_0}\Big)dx\]
this is
\[t=\frac{x_0}{\sqrt{GM}}\int_0^{x_0/2}\Big(\sqrt{\frac{1}{2x}-\frac{1}{x_0}}\Big)dx\]