# Spaceballs

David113
Posers and Puzzles 20 May '08 10:11
1. 20 May '08 10:11
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?
2. 20 May '08 10:26
i don't know!
3. 20 May '08 10:26
too much effort!
4. 20 May '08 11:40
Who put them there...?
That's what I would like to know...
5. thyme
Undutchable
20 May '08 12:35
if they are motionless they will never touch.
6. 20 May '08 13:041 edit
Originally posted by thyme
if they are motionless they will never touch.
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)
7. 20 May '08 13:15
Originally posted by mtthw
What is t when x = r?
Rather, what is t when x = r minus twice the balls radius.
8. 20 May '08 13:171 edit
Originally posted by FabianFnas
Rather, what is t when x = r minus twice the balls radius.
r is the ball's radius 🙂 d is the initial distance between the centres.
9. 20 May '08 13:54
Originally posted by mtthw
r is the ball's radius 🙂 d is the initial distance between the centres.
Oh, sorry... 😞
10. 20 May '08 14:551 edit
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)
Based on that, I worked out:

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. 🙂
11. thyme
Undutchable
20 May '08 15:151 edit
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)
Ah I see. I hadn't though of there being no resistance in space ...
12. thyme
Undutchable
20 May '08 16:59
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)
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 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.
13. wolfgang59
20 May '08 18:51
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)
x= distance from sphere centre to mid-point

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!!

😕
14. PBE6
Bananarama
20 May '08 19:12
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.
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.
15. 20 May '08 19:172 edits
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?
People are trying integration 'n' stuff here while I consider the elementary school way much more simple, though longer.

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.