A differential equation

[/0]

How do you solve:

where G is a constant

I'm not sure if I'm allowed to flip both sides...

lol thanks :p

[humph]

By inspection, try considering logarithm answers. And no, you can't flip both sides, you'd have to use techniques beyond Spesh. In fact, this ODE isn't linear, so you can't use most standard techniques. But it's pretty easy to find one family of solutions (though as to whether they'd work depends on your initial values).

[/0]

Hmm yeah somehow I can't seem to get them to work though.

The question came from trying to find how long it would take for two 1kg balls 10m apart to meet.

,

If we let

[QuantumJG]

Hmm yeah somehow I can't seem to get them to work though.

The question came from trying to find how long it would take for two 1kg balls 10m apart to meet.

,

If we let

What subject is this?

I'm guessing this is a question relating to how long it would take under the influence of their gravitation?

All I can think is that if:





(r: 10 -> 0 and t: 0 -> t)



My logic now is that if the system is held by G/10 J, then its Kinetic energy when r = 0 should be G/10 J (since they have no kinetic energy at r = 10)

Now,



therefore,



implying that,



Something tells me that my maths probably isn't right (never had to do a question like this before), its a difficult question since your accelleration is changing (depends on r). Interesting question though!

[/0]

Not really anything, I just made up the problem (well physics lol)

[QuantumJG]

It's great that you're thinking about things on your own, highly encourage it! Anyway, numerically solving if they're 10 cm apart then it takes just over 16 minutes and 1 second. For 1 meter apart around 8.44 hours. For 10 meters apart around 11.13 days. Hope I didn't make any mistakes

Where can we find those values? More importantly, how did they determine this?

[zzdfa]

make a simulation

[Ahmad]

It's great that you're thinking about things on your own, highly encourage it! Anyway, numerically solving if they're 10 cm apart then it takes just over 16 minutes and 1 second. For 1 meter apart around 8.44 hours. For 10 meters apart around 11.13 days. Hope I didn't make any mistakes

[/0]

LOL sorry I forgot to say what the radiuses of the balls were! I was thinking something like 1cm
(although hmm... I guess you could have point particles after all)
Thanks for trying to work it out Quantum but I don't really understand.. your second order DE seemed to turn into a first order DE

It's great that you're thinking about things on your own, highly encourage it! Anyway, numerically solving if they're 10 cm apart then it takes just over 16 minutes and 1 second. For 1 meter apart around 8.44 hours. For 10 meters apart around 11.13 days. Hope I didn't make any mistakes

Wow that's interesting, I never thought it would happen so fast. Did you use point particles?
(Oh and what program did you use? )

[Ahmad]

Firstly, I'd like to correct myself, the values I gave were for 10 kg point particles, just to keep things reasonable. How did I do it? I used mathematica. How could you do it if you only had a hand calculator? I've attached an excel file which shows how one might do something like this. I've used a very simple algorithm which is conceptually easy but not as accurate as something like mathematica would use. Even so, it seems to give the answer for 10 kg point particles separated by 10 cm correct to at least 1 decimal place. 

Back in year 12 I made a program which simulates gravity and shows how planets orbit each other (chaotically at times!) which I'm happy to share if anyone is interested.

[kamil9876]

I remember a similair, but more complicated DE comming from the same situation, except the balls both had a charge q

[/0]

I remember a similair, but more complicated DE comming from the same situation, except the balls both had a charge q

Do you remember how it was solved? Or was it solved by numerical methods?

[kamil9876]

actually, I made it up myself for the sake of revision, althuogh to be honest i had a slightly different situation: ball with mass m, charge q dropped from height h directly above another ball of same charge q fixed to the ground, (this is happening on Planet earth so i used constant acceleration of g). Heuristically one would imagine that initially it falls, then rises once the coloumb force gets big etc. then coloumb force gets weaker and eventually it turns back around. Using conservation of energy(remember; physics revision this was)(assuming no air resistance), you get that the equation for the values of y at which kinetic energy is zero(turning points) is a quadratic, hence two solutions only. Meaning that it is an osscilatary function with some constant amplitude!, however, trigonometric substitution didn't work for the actual differential equation. 

[/0]

lol nice
Coulomb vs. Gravity is like a really long spring in a changing gravitational field

[QuantumJG]

Hmm yeah somehow I can't seem to get them to work though.

The question came from trying to find how long it would take for two 1kg balls 10m apart to meet.

,

If we let

What subject is this?

I'm guessing this is a question relating to how long it would take under the influence of their gravitation?

All I can think is that if:





(r: 10 -> 0 and t: 0 -> t)



My logic now is that if the system is held by G/10 J, then its Kinetic energy when r = 0 should be G/10 J (since they have no kinetic energy at r = 10)

Now,



therefore,



implying that,



Something tells me that my maths probably isn't right (never had to do a question like this before), its a difficult question since your accelleration is changing (depends on r). Interesting question though!

Before doing this mathematical mess, I should have read what humph said:

In fact, this ODE isn't linear, so you can't use most standard techniques.

i.e. My first year uni maths is pretty useless.

So basically what I did was all wrong.