- 20 May '08 13:04 / 1 edit

Gravitational attraction.*Originally posted by thyme***if they are motionless they will never touch.**

Basically, you need to solve this:

d^2x/dt^2 = -GM/4x^2

where x(0) = d/2 and dx/dt(0) = 0

What is t when x = r?

(x being the distance between the mid-point and the centre of one of the spheres) - 20 May '08 14:55 / 1 edit

Based on that, I worked out:*Originally posted by mtthw***Gravitational attraction.**

Basically, you need to solve this:

d^2x/dt^2 = -GM/4x^2

where x(0) = d/2 and dx/dt(0) = 0

What is t when x = r?

(x being the distance between the mid-point and the centre of one of the spheres)

dx/dt = -4GM int (x^2)

dx/dt = -4GM((x^3) / 3) + k

dx/dt = -(4/3) GM x^3 + k

x = -(1/3) GM x^4 + kx +c

Where k and c are a constants. int is integral.

My integration is rusty. Am I going down the correct path?

EDIT: Oops. Didn't take into account the t. Will try again. - 20 May '08 15:15 / 1 edit

Ah I see. I hadn't though of there being no resistance in space ...*Originally posted by mtthw***Gravitational attraction.**

Basically, you need to solve this:

d^2x/dt^2 = -GM/4x^2

where x(0) = d/2 and dx/dt(0) = 0

What is t when x = r?

(x being the distance between the mid-point and the centre of one of the spheres) - 20 May '08 16:59

OK.. so what you are saying is that I have to solve the integral from d to r for t of d^2x/dt^2 = -GM/4x^2 to arrive at the solution?*Originally posted by mtthw***Gravitational attraction.**

Basically, you need to solve this:

d^2x/dt^2 = -GM/4x^2

where x(0) = d/2 and dx/dt(0) = 0

What is t when x = r?

(x being the distance between the mid-point and the centre of one of the spheres)

but that seems the wrong way round... if I solve that I'm solving for x aren't I?

God it is a long time ago.. i used to be really good at this

d^2x/dt^2 = -GM/4x^2 is d^2x/dt^2 = -GM/4 * (x^-2) because -GM/4 is constant, G being 6.67 * 10^-11 and with M = 1 kg (yes?)

so the only thing to worry about is x^-2

dx/dt = GM/2 * (x^-1) +C

x/t = GM/2 *ln|x| +Cx +D

ehm.. t = GM/2 *xln|x| +Cx^2 +Dx

and x = 45 .. calculator... assume C and D are 0 (what are C and D?)

..i'm lost... I have no idea if this gibberish makes sense. - 20 May '08 18:51
*Originally posted by mtthw*

Basically, you need to solve this:

d^2x/dt^2 = -GM/4x^2

where x(0) = d/2 and dx/dt(0) = 0

What is t when x = r?

(x being the distance between the mid-point and the centre of one of the spheres)

t=0 x=0.5 dx/dt=0

We need to find t when x=0.05

We know d^2x/dt^2 = (G/4) x^(-2)

So we have to integrate once to get dx/dt and use fact that dx/dt = 0

Then integrate again to get x in terms of t

Then solve for x=0.05

25 years ago I could do this!!

- 20 May '08 19:12

Remember, we're trying to integrate d/dt(dx/dt), where x is a function of t. You can't just integrate with respect to x. Just going to look up my old differential equations notes and see if I can solve this one.*Originally posted by thyme***OK.. so what you are saying is that I have to solve the integral from d to r for t of d^2x/dt^2 = -GM/4x^2 to arrive at the solution?**

but that seems the wrong way round... if I solve that I'm solving for x aren't I?

God it is a long time ago.. i used to be really good at this

d^2x/dt^2 = -GM/4x^2 is d^2x/dt^2 = -GM/4 * (x^-2) because -GM/4 is constan ...[text shortened]... and D are 0 (what are C and D?)

..i'm lost... I have no idea if this gibberish makes sense. - 20 May '08 19:17 / 2 edits

People are trying integration 'n' stuff here while I consider the elementary school way much more simple, though longer.*Originally posted by David113***Two spheres of 10-cm diameter and 1-kg mass are out in space, motionless with respect to each other. Their centers are one meter apart. The only force on each sphere is the gravitational pull of the other sphere. How long will it be until they touch?**

First of all we have to calculate the gravitational force between the spheres. F = (G * m1 * m2) / R ^ 2; (G is the gravitational constant) F = (6.7 * 10^-11 * 1 * 1) / 1 ^ 2 = 6.7 * 10^-11 N. The distance between surfaces is 1 - 2 * 0.5 = 0.9 m. Next step is to calculate the acceleration. F = m * a => a = F/m; a = 6.7 * 10^-11 / 2 = 3.35 * 10^-11 m/s^2. Now we have to look for a formula from kinematics from which we can get the time if we have distance and acceleration. S = v0 * t + (a * t^2) / 2 where v0 is the speed at the beginning of event which therefore can be crossed out, so S = (a * t^2) / 2 => t = sqrt((2 * s) / a); t = sqrt((2 * 0.9) / 3.35 * 10^-11) which is approximately 231800 s.

So the answer I propose is 64 hours, 23 min, 20 s.

P.s. I hope I haven't made any dumb mistakes because all the calculations I did where quickly scribbled without any thorough checking. I know I'm lazy.

EDIT: Ooops, the force isn't constant.