Question

Outside the influence of any other gravitational influence, two 1kg spheres with a radius of 0.05m rest 1m apart. When will they collide?

Answer 1

The question is not clear if the GAP or the Centres are 1m - but the principle is the same

The force between them is Gmm/r^2

This acts on each of them, and F=ma

So for each ball the acceleration = Gm/r^2


but a = d2v/dt^2
hence integrate once to get dv/dt

integrate again to get v (= ds/dt)

Integrate again to get s

at the point of collision s = 500mm (or 490 if the centres are 1 m apart at start)

not sure about the constants of integration... you know s0 and v0

Answer 2

I mean 495 (subtracting radius..)

Also - at the point of collision s= 500 - so you can solve for t

Answer 3

got this wrong - sorry
a=dv/dt
v=ds/dt
a=d2s/dt^2

So there is one less integration to do than I said above - and you have the 2 constants from the initial conditions.

Answer 4

so we are assuming they are 110 centimeters apart?

Answer 5

*their centers of mass

Answer 6

The drawing seems to imply they are 1m apart at centres of mass.

So at collision they will have move 450mm (yet ANOTHER correction to my witterings above...)

Answer 7

Ummm.
The inside edge of both spheres are 0.9m apart. The centers of mass are 1.0m apart.

Answer 8

OK - so use the acceleration formula I gave above.

You need to integrate this to get Velocity. You know that V=0 when t=0 so can get eh constant of integration from that.

Then you need to integrate that again to get the formula for distance (s)

You know that s=0 when t= 0 (assume we take a datum at centre of one of the spheres)

So now you have a formula for s in terms of t

Both spheres will move equally - so they will collide when they have moved 450mm.

Solve you equation for s=450mm and that is the time it takes to collide..

\[\int\limits_{}^{}\frac{ dv }{ dt }dt = v+c\]
v=ds/dt so
\[\int\limits_{}^{} \frac{ds }{ dt }dt= s+c\]

Answer 9

did you get it?

Answer 10

I'll have to read up on integrals.

Answer 11

\[\frac{ 1 }{r^{2} } =r^{-2}\]

So use the normal method for integrating exponents of r

Answer 12

This page has the guidance you need..
http://hyperphysics.phy-astr.gsu.edu/hbase/intpol.html

Answer 13

I was thinking I could do it with\[x=x_0+v_0t+\frac{1}{2}at^2\]

Answer 14

This is quite a sophisticated question - where did you get it? and how come you don't already have knowledge of integrals?

(just curious - not being intrusive I hope..)

Answer 15

Incidentally - the equation you wrote can be derived from the integrals I wrote by starting out with a=k (constant)

Answer 16

You can't do it with that method since that only applies for constant acceleration

In this case the force increase as the spheres approach each other, and the acceleration therefore increases.

I'm afraid there is no way of finding the solution without integrals.

Answer 17

You told me \(a=\frac{GM}{r^2}\)\[a=\dfrac{6.67\times10^{-11}m^3kg^{-1}s^{-2}\cdot1kg}{1m^2}\]\[a=\dfrac{6.67\times10^{-11}m^{\cancel{3}}\cancel{kg^{-1}}s^{-2}\cdot\cancel{1kg}}{\cancel{1m^2}}\]\[a=6.67\times10^{-11}ms^{-2}\]

Answer 18

yeah - but leave the numbers out until you have done the integrals

just work with the algebra

Answer 19

Why would acceleration change? There is no other force acting until they collide?

Answer 20

The Force = GMm/r^2
But as they approach each other r changes, and the force increases - so the acceleration increases

Answer 21

@MrNood You are not being intrusive at all, you are quite patient and helpful actually.
I came up with the question to explain the simplest concept of gravity.
I haven't learned integrals because I haven't learned them yet. There are many other things I haven't learned yet too.

Answer 22

The acceleration that you calculated above is for the INITIAL condition where the distance = 1m

If you put r =0.8m (for example) you will see that the acceleration is greater.

Integration allows you to 'sum' all the instantaneous values of r and hence get the total effect over the range.

Answer 23

it is impossible to solve this without integral :)

Answer 24

If I plot the velocity it looks like this, right?

Answer 25

Yes - something like that

Answer 26

MrNood, do you have a solution?

Answer 27

Wouldn't acceleration look like this?