How long for 2 particles to collide due to gravity?

[thecommexokid]

Homework Statement

Two particles (masses m₁, m₂) are released from rest a distance D apart in space. How long until they collide?

Homework Equations

The force between the particles is $F_G(t)=\frac{Gm_1m_2}{r(t)^2}$.

The center of mass is located a distance $r_1(t)=\frac{m_2}{m_1+m_2}r(t)$ from particle 1 and $r_2(t)=\frac{m_1}{m_1+m_2}r(t)$ from particle 2. Note that $r_1(t)+r_2(t)=r(t)$.

The Attempt at a Solution

The collision will take place at the center of mass. Particle 1 needs to traverse a total distance of $d_1=\frac{m_2}{m_1+m_2}D$ to reach the COM. Its acceleration over this distance is

$\ddot r_1(t)=\frac{m_2}{m_1+m_2}\ddot r(t)=\frac{Gm_2}{r(t)^2}$.​

OR! Another approach I thought of was to use potential and kinetic energy:

$-\frac{Gm_1m_2}{D}=-\frac{Gm_1m_2}{r(t)}+\frac{1}{2}m_1v_1(t)^2+\frac{1}{2}m_2v_2(t)^2$
$=-\frac{Gm_1m_2}{r(t)}+\frac{1}{2}m_1m_2\dot r(t)^2$.​

But either way, I wind up with a differential equation that's totally inhumane. I feel like I'm missing something here. Open to suggestion on how to further either of these two approaches, or other approaches entirely.

 

[spaderdabomb]

Using energy is definitely the right way to go about it. Start off with the fact that the center of mass is 0 potential energy for both objects. You should be able to find a useful relationship between $m_1, m_2, r_1$ and $r_2$. Next you will probably want to compare the kinetic energy of the two objects when the collide and the potential energy they begin with. We know kinetic energy has velocity in it. We also know we can relate distance, time and velocitz to each other. That is where you will get the amount of time it takes for them to collide.

Hope that helps.

 

[spaderdabomb]

Sorry double posted by accident

 

[rcgldr]

This problem has been posted at least twice before, but assuming you want to solve it yourself, start off by determining the rate of acceleration of each particle versus r(t), and subtract the accelerations (consider one of the accelerations to be negative, so you're really adding the magnitudes of accelerations) to determine the rate of acceleration of closure between the particles, and use that as the second derivative of r(t) (which will be negative since it's a rate of closure). This will end up with a tricky integral, but if you get that far, I or someone else can offer help.

 

[thecommexokid]

This problem has been posted at least twice before, but assuming you want to solve it yourself, start off by determining the rate of acceleration of each particle versus r(t), and subtract the accelerations (consider one of the accelerations to be negative, so you're really adding the magnitudes of accelerations) to determine the rate of acceleration of closure between the particles, and use that as the second derivative of r(t) (which will be negative since it's a rate of closure). This will end up with a tricky integral, but if you get that far, I or someone else can offer help.

Thanks for the reply, rcgldr. And sorry for the duplicate question but all the words in this problem are so common to physics problems that it was really hard to search for.

Anyhow, particle 1 feels a force of

$F_1(t)=\frac{Gm_1m_2}{r(t)^2}$​

and so it accelerates toward the COM at a rate of

$\ddot r_1(t)=\frac{F_G(t)}{m_1}=\frac{Gm_2}{r(t)^2}$.​

Having established the direction of particle 1's motion as positive, particle 2 then feels the equal and opposite force of

$F_2(t)=-\frac{Gm_1m_2}{r(t)^2}$​

and accelerates at

$\ddot r_2(t)=\frac{F_G(t)}{m_2}=-\frac{Gm_1}{r(t)^2}$.​

The distance between the particles is

$r(t)=r_1(t)+r_2(t)$,​

so the acceleration of closure is

$\ddot r(t)=\ddot r_1(t)+\ddot r_2(t)=\frac{G}{r(t)^2}(m_2-m_1)$.​

You say this acceleration is necessarily negative, but it looks to me like its sign would be based on which particle is more massive. So am I right so far? If so, I'm once again stuck with an intractable differential equation: I even tried feeding it to Wolfram|Alpha and I just got back a message saying “Standard computation time exceeded.”

 

[thecommexokid]

Ooh!

Let $v(t)=\frac{dr}{dt}$. Then $\ddot r(t)=\frac{dv}{dt}=\frac{dv}{dr}\cdot\frac{dr}{dt}=v\frac{dv}{dr}$.

So our equation becomes $v\frac{dv}{dr}=\frac{G(m_2-m_1)}{r}$, which is separable to $\int v dv = G(m_2-m_1)\int\frac{1}{r}dr$.

Got to go to work now but this seems pretty promising.

 

[rcgldr]

A minor change is needed. Assume m2 is to the right of m1, position of m1 is x1(t) (left of COM) and position of m2 is x2(t) (right of COM). For the integration constant, at t = 0, v = 0, r = r0.

$r(t)=x_2(t)-x_1(t)$

$\ddot r_1(t)=\frac{F_G(t)}{m_1}=\frac{Gm_2}{r(t)^2}$

$\ddot r_2(t)=\frac{F_G(t)}{m_2}=-\frac{Gm_1}{r(t)^2}$

$\ddot r(t)=\ddot r_2(t)-\ddot r_1(t)=-\frac{G}{r(t)^2}(m_1+m_2)$.

$v\frac{dv}{dr}=\frac{-G(m_1+m_2)}{r^2}$

$\int v dv = -G(m_1+m_2)\int\frac{1}{r^2}dr$.

​

 

[thecommexokid]

Thank you, thank you. Of COURSE r(t)=x₂(t)−x₁(t). It's always the little things that get me. Can't take the time to finish the problem right now because I'm at work, but you got me over the hurdle, I think.

 

[thecommexokid]

Alas, I spoke too soon. I finally returned to this problem to find myself still stuck. When we left off, I had this integral

$\displaystyle\int_0^{v(t)} v \,dv = -G (m_1 + m_2) \displaystyle\int_{r_0}^{r(t)} \dfrac {1} {r^2} \,dr$​

which I can evaluate to get

$\dfrac 1 2 v(t)^2 = G (m_1 + m_2) \bigg( \dfrac {1} {r(t)} - \dfrac {1} {r_0} \bigg)$.​

But where does that leave me, in terms of my original project of finding the time to collision? If I put back dr/dt in place of v, I have

$\dfrac {dr}{dt} = \bigg[ 2G (m_1 + m_2) \bigg( \dfrac {1} {r(t)} - \dfrac {1} {r_0} \bigg) \bigg]^{1/2}$,​

but I decidedly have no idea what to do when faced with the new integral that leaves me with:

$\displaystyle\int_0^{T_{coll}} dt = \displaystyle\int_{r_0}^0 \bigg[ 2G (m_1 + m_2) \bigg( \dfrac {1} {r} - \frac {1} {r_0} \bigg) \bigg]^{-1/2} dr$.​

Maybe there's just some trick to this integral that I'm missing, but it feels like I'm going in circles and like there must be some easier way to go about this.

 

[rcgldr]

You can rearrange the right side to:

$\displaystyle\int_0^{t_{coll}} dt = \displaystyle \sqrt{\frac{r_0}{2G (m_1 + m_2)}} \int_{r_0}^0 \sqrt{\frac{r}{r_0 - r}} \ dr$​

Then substitute

$\displaystyle u = \sqrt{\frac{r}{r_0 - r}}$​

so

$\displaystyle u^2 = {\frac{r}{r_0 - r}}$​

Then solve for r and dr in terms of u and du, and change the limits of the integral for u. You'll end up with an integral that can be solved more easily (integral table lookup, and integration by parts).

 

[BruceW]

or use a trigonometric substitution (sorry for butting in). I don't know if there is an 'easy way' to do this problem...

 

[rcgldr]

Assuming this isn't homework, perhaps this thread should be moved to general or classic physics section?

 

[rude man]

Has anyone come up with an answer? I only arrived at a solution for D = 1m, which was

T = (π/2)/k seconds where
k = (1+m₁/m₂)√(Gm₁m₂/M) and
M = (m₁/2)(1 + m₁/m₂).

So for m₁ = m₂ = m and D = 1m,
T = (π/2)/2√(mG).

(The units are consistent since "π/2" has dimensions of L^3/2 here.)

 

[rcgldr]

Has anyone come up with an answer? I only arrived at a solution for D = 1m

This problem has appeared at least twice before. For r_0 = 1 meter, m1 = m2 = 1kg, t_collision ~= 96136 seconds. Waiting for the OP to try and work out the final integral.

 

[rude man]

This problem had appeared at least twice before. For r_0 = 1 meter, t_collision ~= 96136 seconds. Waiting for the OP to try and work out the final integral.

It's not a function of m1 and m2? Not what I got, and doesn't appear that way from your post #10 either ...

I wound up with the same integral and, after sorting out my k coefficient, found it to be the same as yours.

But T has to be a function of m1 and m2 so your number 96136 s. must be for certain values of m1 and m2.

That second integral is not trivial - Wolfram alpha refused to evaluate it for me unless I paid - so I got T(D=1m) which was free.

 

[rcgldr]

It's not a function of m1 and m2?

It is, I missed an edit on my previous post where I set m1 = m2 = 1kg, it's fixed now.

That second integral is not trivial.

As far as I know, arildno was the first member here to have figured out the u = sqrt(...) substitution as a way to solve the second integral. What's left to do after my previous post is to figure out how to convert dr into a function of du, and change the limits of the integral based on u instead of r. Once this is done, the integral based on u and du will be solvable, including masses that are not point masses.

 

[voko]

or use a trigonometric substitution (sorry for butting in). I don't know if there is an 'easy way' to do this problem...

The trig sub was what I used in one of those earlier cases mentioned :)

In that same thread K^2 made this remark: "I'm going to cheat. This is still an orbital motion problem. By Keppler's Laws, period depends on semi-major axis only, and what we are looking for is half the orbital period." So much for the easy way.

 

[BruceW]

the difficult bit with the trig sub is getting the limits of the integral right.

 

[voko]

Frankly, I do not remember that was a problem. Guessing the sub and doing the requisite algebra was more of a bother.

 

[BruceW]

I think I have always been a bit unsure about converting limits of integrals, especially for sinusoid functions, which are not 1-to-1. I mean, how can you know what value of theta to use as the limit (because it does matter for the answer), but there is no way to tell which value you should use because the sinusoid is not a 1-to-1 function... you see what I mean? kinda tricky.

 

[voko]

I think I have always been a bit unsure about converting limits of integrals, especially for sinusoid functions, which are not 1-to-1. I mean, how can you know what value of theta to use as the limit (because it does matter for the answer), but there is no way to tell which value you should use because the sinusoid is not a 1-to-1 function... you see what I mean? kinda tricky.

Personally, I find it easiest just to take whatever interval where the trig function involved is monotonic, and then usually the one closest to $theta = 0$.

 

[BruceW]

yeah. that won't work in this case. (unless maybe I have just straight done it wrong for another reason). hmm. I might try to spend some time figuring this stuff out...

 

[voko]

yeah. that won't work in this case. (unless maybe I have just straight done it wrong for another reason).

It actually did :) If you want, I can PM you the link to the thread where it was done.

Frankly, I am having problem figuring out when this approach might legitimately not work.

 

[rcgldr]

You can rearrange the right side to ...

Since r decrease over time, I should have choosen the negative root for sqrt((dr/dt)^2) resulting in:

$\displaystyle\int_0^{t_{coll}} dt = \displaystyle - \sqrt{\frac{r_0}{2G (m_1 + m_2)}} \int_{r_0}^0 \sqrt{\frac{r}{r_0 - r}} \ dr$​

Then substitute

$\displaystyle u = \sqrt{\frac{r}{r_0 - r}}$​

so

$\displaystyle u^2 = {\frac{r}{r_0 - r}}$​

Then solve for r and dr in terms of u and du, and change the limits of the integral for u. You'll end up with an integral that can be solved more easily (integral table lookup, and integration by parts). In one of the previous threads on this an alternative method was used and produced the same results. I also created a spreadsheet to do numerical integration to confirm the results with a few test cases.

 

[BruceW]

@rcgldr - ah, nice spot. That is one of those things that you could so easily pass over without realizing. (I didn't realize until you said it, anyway).

@voko - I think I'll make another thread for my 'issues with limits'. Feel free to come along to help out.

 

[rude man]

Wolfram alpha would not let me get the definite integral from D to 0 without paying, but it did get me ∫dr/(1/r - 1)^1/2 from 0 to 1 = π/2. So if the masses are initially separated by D = 1m the answer is T = (π/2)(1/k). Post 10's expression for time to collide is equivalent to mine after all the m1's, m2's and D's are sorted out.

So that got me the answer for D = 1m.

We can then get the answer for any other D by rescaling the dimension L. E.g. if we want D = 4m we recompute G' = (4^3)G = 64G since G includes the dimension L^3 (and also replace D = 4m with D = 1 of course.)

As a footnote - I approached the problem from an energy conservation viewpoint by assuming small but finite-radii spheres. When it came time to evaluate the integral I let the radii = 0 and got away with it (wondered initially if the integral would converge for a lower limit of zero but of course it did).

 

[BruceW]

yeah, I kinda feel like we are 'lucky' that this integral does give a straightforward answer, even for particles of zero radius. Before attempting the question, I would not have been certain that the integral would converge. But it does, so yay.

Also, voko, I made this thread for my issues with limits after substitution: https://www.physicsforums.com/showthread.php?t=700689 and Curious3141 explained what I was doing wrong (which wasn't actually related to what was being put into the limits). Uh, so the take-away message from his answer is that you can use any value of theta which corresponds to the same 'r', and you still get the same answer for the integral. Which is nice. (It is what I was hoping for, really).

 

[voko]

I made this thread for my issues with limits after substitution

Very well, it is good to see your faith in calculus restored!

 

[BruceW]

haha yeah. you know it is a bad day when you start to doubt basic calculus!

 

[rcgldr]

Still waiting to see if the OP wants to continue with this before posting the rest of the answer using the original method.

 

[BruceW]

yeah... hopefully he will be back soon. I am curious about this 'original method'. I cannot guess how it would work from the short description you gave.

edit: I mean, that I can substitute r with u and dr with du. But then the integral it gives is not any nicer at first glance (maybe my brain works differently to whoever did the original method). But anyway, I can wait 'till the OP comes back.