Comments on: A physics question. And no, you aren't doing my homework.

Question: A physics question. And no, you aren't doing my homework.

alby — Wed, 04 Jan 2006 13:14:06 -0800

Two objects, both mass 1kg are placed 1m apart in a total vacuum. How long would it take for the force of gravity to pull them together (assuming no other forces act)?

I've been trying to solve this iteratively by spreadsheet but I know there must be a proper integration/maths answer.

By: pwb503

pwb503 — Wed, 04 Jan 2006 13:26:28 -0800

Well, I don't know. But Two things come to mind. 1) The speed at which they would approach each other would increase exponentially I believe. Thus you might not get an exact answer as to when exactly they would touch. 2) I might be wrong about this but I would think that the answer would change depending on the shape of the objects. As I recall, gravity can be measured (strangely) from the absolute center of an object. Regardless however I believe that small dense balls would attract each other at different rates than tall thin cylinders.

By: I Love Tacos

I Love Tacos — Wed, 04 Jan 2006 13:31:31 -0800

The question is definitely answerable with a little calculus, but my first attempt to derive it made Haliday and Resnick cry.

By: Wild_Eep

Wild_Eep — Wed, 04 Jan 2006 13:38:35 -0800

distance = initial velocity * time + .5 * acceleration * time^2

If I'm remembering my High School physics correctly.

Solve for time.

By: teece

teece — Wed, 04 Jan 2006 13:38:48 -0800

From my (very quick, and possibly wrong) calculation, it should be around 122416 seconds after you "release" the masses that they touch.

F = ma, F = G (m1 m2)/r^2, a=x'(t), let the one on the left be initially at rest at position -0.5. Solve the differential equation: x(t) = 0.5(6.673 x 10^-11 t^2 -1). Solve that for when t = 0.

But again, I could very well be wrong, as I did this quickly and haven't used my physics brain in a while.

By: Wild_Eep

Wild_Eep — Wed, 04 Jan 2006 13:41:49 -0800

teece's answer is 1.41 days, more or less.

By: vacapinta

vacapinta — Wed, 04 Jan 2006 13:42:04 -0800

I'll take a bold leap here and notice that this is basically a small, binary star system. If they could go through each other they'd keep oscillating back and forth like any two stars in a degenerate elliptical orbit.

In that case, we just use Kepler's Harmonic Law and get:

T=sqrt (4*pi^2*R^3/GM)

where R is the radius of the orbit (0.5 meters) and M is the total mass (2 kg) The period is then twice the number you need.

By: teece

teece — Wed, 04 Jan 2006 13:43:14 -0800

Damn -- a = x double prime of t. It ate one of my apostrophes.

By: andrew cooke

andrew cooke — Wed, 04 Jan 2006 13:52:32 -0800

i can't solve it. it's certainly not exponential. sorry.

whoa - on preview, vacapinta is totally wrong, which surprises me somewhat.

you need to solve x' = -k/x^2 and there's no way that's sine/cosine. orbit is a completely different beast - the force is constant. here it tends to infinity as they close.

By: vacapinta

vacapinta — Wed, 04 Jan 2006 13:53:09 -0800

Btw, I pulled out the calculator and the answer I get is 1.1 days.

By: andrew cooke

andrew cooke — Wed, 04 Jan 2006 13:53:22 -0800

damn. one of my apostrophes was eaten too. x'' = -k/x^2

By: andrew cooke

andrew cooke — Wed, 04 Jan 2006 13:54:30 -0800

(and when i say sine/cosine i mean sine or cosine. it probably is somethig like a tangent substitution)

By: andrew cooke

andrew cooke — Wed, 04 Jan 2006 13:55:48 -0800

last comment - you can think of it as an orbit with infinite elipticity, i guess. that might help take vacapinta's insight further.

By: ewagoner

ewagoner — Wed, 04 Jan 2006 14:14:01 -0800

I get nearly exactly 24 hours, using the "back of the envelope" method found here (G = 6.67e-8, M=2000g, R=100cm).

By: gleuschk

gleuschk — Wed, 04 Jan 2006 14:19:10 -0800

Consider the two objects A and B. Let r(t) be the distance from A to the midpoint, so r(0)=-1/2 and we want to know when r(t)=0.

The force on A is (proportional to) 1/(2r(t))². Since force = mass x acceleration, we know that
1/(2r(t))² = r'(t).

This is a second-order differential equation. I never was very good at differential equations.

By: gleuschk

gleuschk — Wed, 04 Jan 2006 14:20:37 -0800

Huh. Why does it keep eating double apostrophes?

By: vacapinta

vacapinta — Wed, 04 Jan 2006 14:23:21 -0800

orbit is a completely different beast - the force is constant. here it tends to infinity as they close.

I genuinely dont understand, andrew. In this case they are both "orbiting" the constant center of mass of the system. In a highly elliptical orbit (say, a comet) the force increases too with approach as does the acceleration.

By: driveler

driveler — Wed, 04 Jan 2006 14:28:18 -0800

The differential equation suggested by teece, andrew cooke, and gleuschk is right.

I just put it into Mathematica, though, and it chokes on it. So, I can't give you a solution.

By: teece

teece — Wed, 04 Jan 2006 14:36:48 -0800

Checking my work, I'm pretty sure my answer is correct now. Solve the system x1"(t)=G, x2"(t)=-G, with initial conditions of x1(0)=-0.5, x2(0)=0.5, and x1'(0)=x2'(0)=0.

You can solve that system: the answer is x1(t)=0.5(Gt^2-1)
and x2(t)=0.5(1-Gt^2).

Pick either of those functions, and solve them for when x=0. You get 122416 seconds.

By: vacapinta

vacapinta — Wed, 04 Jan 2006 14:51:59 -0800

I rescind after reading ewagoner's link. I can see now why my "bold leap" only works as an approximation (although I will escape with some dignity by noting that I pulled it out on my own right now)

By: gleuschk

gleuschk — Wed, 04 Jan 2006 14:55:33 -0800

teece, I don't understand

Solve the system x1"(t)=G, x2"(t)=-G...

So this says that each particle has constant acceleration? That's not the case.

Maple claims to be able to solve the differential equation, but I don't understand what it's telling me for a solution.

By: paradroid

paradroid — Wed, 04 Jan 2006 14:56:57 -0800

each particle does indeed have constant acceleration, since they assumedly have constant mass.

By: mr_roboto

mr_roboto — Wed, 04 Jan 2006 14:59:58 -0800

Gravitational force (and therefore acceleration) is a function of distance: it's an inverse square law. Since the two objects are moving towards one another, the distance isn't constant.

By: Galvatron

Galvatron — Wed, 04 Jan 2006 15:11:35 -0800

If it helps anyone, the second-order diffeq can be reduced to first order. We start off with: d/dt(dx/dt) = - G m / (4 x^2)

We can obtain an equivalent first-order diffeq using this trick: d/dt(dx/dt) = dv/dt = (dv/dx)(dx/dt) = v (dv/dx)

So now we have: v dv/dx = - G m / (4 x^2)
Integrate once and use boundary condition v = 0 @ x = 0.5:
v^2 = (dx/dt)^2 = G m (1/(2 x) - 1)

I don't know how to solve that, and Mathematica chokes on it.

By: vacapinta

vacapinta — Wed, 04 Jan 2006 15:18:05 -0800

Galvatron: is this what you are trying to do?

Because, if so, that leads directly to the equation I stated above.

By: teece

teece — Wed, 04 Jan 2006 15:23:26 -0800

You are of course correct, gleuschk, I am a bonehead. The denominator is only 1 at the very beginning, it's a function of the two objects separation. Duh. My bad. I turned a function into a constant. That always makes things easier... and wrong.

F = ma, right? a=x"(t). m1=1, m2=1, d=1

Let particle 1 be on the left, and particle two be on the right.

The force on 1 due to 2 will be:
F12 = G (1*1)/(x2-x1)^2 = G/(x2-x1)^2 (this one is the left particle pulled to the right, hence a positive sign).

The force on 2 due to 1 will be:
F21 = - G (1*1)/(x2-x1)^2 = -G/(x2-x1)^2 (this one is the right particle pulled to the left, hence a negative sign).

Pick a coordinate system. I chose: the left particle is at x=-1/2. The right particle is at x=1/2. Initial velocity of both will be zero.

This is a system of 2 coupled, second order differential equations. I used Mathematica to solve it numerically. It is between 117569 and 117570 that they touch.

If you want to solve that system in Mathematica, copy and paste this monstrosity:
ans =
NDSolve[{x1'[t] ==6.673*10^-11/(x2[t] - x1[t]),
x1[0] == -1/2, x1'[0] = 0, x2'[t] == -6.673*10^-11/(x2[t] - x1[t]), x2[0] =1/2, x2'[0] = 0}, {x1, x2}, {t, 0,
117570}]

It will spit out a couple of interpolating functions, and they were saved in ans. Use ans to play with them, that is
{x1[117569],x2[117569]}/.ans
will give you two numbers close to zero. If you put 117570 in there, you will get two HUGE numbers out. It's because you've passed the point where x2-x1=0, and the system of differential equation we used are no longer a valid description of nature, and are thus wrong.

To reiterate: the answer I gave before was wrong.

You can also convert this to a problem with only one equation by letting x2-x1=x be your variable, but you throw information away if you do that. Since I have tools to easily solve systems of differential equations, I did not bother.

By: Decani

Decani — Wed, 04 Jan 2006 15:31:20 -0800

Now, is Askme really supposed to be used for doing folks' homework?

By: Galvatron

Galvatron — Wed, 04 Jan 2006 15:37:39 -0800

vacapinta: yeah, looks like that's probably equivalent.

By: teece

teece — Wed, 04 Jan 2006 15:38:16 -0800

Holy smokes, I can't do anything right today. I typo'ed the Mathematica input. So it gave me the correct answer for the wrong problem.

It should have been:
ans =
NDSolve[{x1'[t] ==6.673*10^-11/(x2[t] - x1[t])^2,
x1[0] == -1/2, x1'[0] = 0, x2'[t] == -6.673*10^-11/(x2[t] - x1[t])^2, x2[0] =1/2, x2'[0] = 0}, {x1, x2}, {t, 0,
117570}]

Which makes them touch between 96145 and 96146 seconds. (Yes, I've now given three different answers for this problem. *applause*). But this one is right.

And I actually did this very problem before in a differential equations class. I opened it to check my work, just now, and it's in agreement. I should have found that first.

Geez.

And Decani, we have albi's word that this is not homework. Good enough for me. If it is his/her homework? *shrug* They won't be able to do it on a test unless they understand it, so no skin off my teeth.

By: mr_roboto

mr_roboto — Wed, 04 Jan 2006 15:42:47 -0800

This is the two-body problem. I don't have time to read through the Mathworld explanation carefully, but maybe it'll help.

By: mbd1mbd1

mbd1mbd1 — Wed, 04 Jan 2006 15:53:23 -0800

I agree with teece's last answer. I used 6.6742*10^-11 for G and came up with about 96144 seconds, using a spreadsheet to do the double integration.

Of course we did all that work under the assumption that the two objects were point masses. Assuming they are spheres with radius r would mean they touch when x1 - x2 = 2*r.

By: folara

folara — Wed, 04 Jan 2006 16:53:41 -0800

This doesn't answer your question, but I'd like to point out that just because your two objects are in a vacuum doesn't mean that no other forces act on the objects. You still have to deal with the gravitational pull coming from every other object in the universe, which will affect how long it takes for the two objects in question to meet. (Is there a way to eliminate that? Obviously, putting distance between the objects in question and the "rest" helps... help me out here.) I'm not sure if that was your assumption, but I wanted to make sure there wasn't any confusion regarding the setup of the problem.

By: DevilsAdvocate

DevilsAdvocate — Wed, 04 Jan 2006 16:55:14 -0800

Picking up where Galvatron left off, i.e., (dx/dt)² = G(1/2x-1) (having substituted m=1):

dx/(1/2x-1)^1/2=±G^1/2dt

Substitute u=1/2x. du=-dx/2x², therefore dx=-du/2u².

du/[u²(u-1)^1/2] = ±2G^1/2dt

Integration of the left-hand side is given by formulas 137 and 136 in the CRC Handbook of Chemistry and Physics, 85th edition, p. A-27. Given the boundary condition x=0.5 (thus u=1) at t=0, it gives:

(u-1)^1/2/u + tan^-1[(u-1)^1/2] = ±2G^1/2t

And substituting back in for u gives:

t = ±{x(1/2x-1)^1/2 + (1/2)tan^-1[(1/2x-1)^1/2]}/G^1/2

(I'm not gonna try to solve for x in terms of t.)

According to this formula, t approaches pi/4G^1/2 as x approaches 0, or 96145.579 seconds, which corresponds with the answers others have given here.

By: teece

teece — Wed, 04 Jan 2006 17:05:16 -0800

For what it's worth, if you'd like to convert this problem to one differential equation, rather than two, as some seem to be doing, the equation you want to solve is:

x"(t)=-G (m1+m2)/x^2,

where in this problem x = x2-x1 (the separation between the two objects, which is the variable here), m1=m2=1. You get that by just subtracting the two differential equations I listed above, and simplifying.

That equation becomes x'(t)=v(t), v'(t)=-G 2/x^2. v(0)=0, x(0)=1.

That one has to be tackled numerically, too (well, there may be some analytic solution, but Mathematica does not know it, and I don't feel like hammering at it). It gives the same answer: just about 96145. It differs in accuracy, which would be an interesting fact to tackle in the filed of numerical approximation... But they're about the same: 96145.052154 for the former, 96145.2994996 for the latter.

On preview, I think that's maybe what DA just said. OK, I've killed enough time now ;-)

By: gleuschk

gleuschk — Wed, 04 Jan 2006 17:06:00 -0800

Rock on, DA.

By: treeshade

treeshade — Wed, 04 Jan 2006 17:25:10 -0800

Alby,

Are you testing us? From previous posts I can see that you claim to be a physicist and are studying teaching. The problem you are posing is not one you should find difficult. The question and the answer (at least my one) looks very much like it came from a textbook or an exam. What is up?

By: weapons-grade pandemonium

weapons-grade pandemonium — Wed, 04 Jan 2006 17:41:55 -0800

It seems to me that the density and/or shape of the objects would make a difference, since the gravitational attraction is calculated from the centers, whereas the diameters, or the outside configurations (leading edges) determine when they are "together". Even if you specified a shape, 1kg spheres of lead would come together at a different time than 1kg spheres of styrofoam, no?

By: weapons-grade pandemonium

weapons-grade pandemonium — Wed, 04 Jan 2006 17:48:54 -0800

Put another way: the one meter separation must be calculated from the leading edges, in which case the centers of gravity are farther apart in the styrofoam spheres than the lead spheres, and the time is slower. If the one meter separation is calculated from the centers of gravity, then I'll use a light foam, and have the spheres touching, so the answer is zero seconds.

By: teece

teece — Wed, 04 Jan 2006 17:50:34 -0800

weapons-grade pandemonium: You can approximate a spherical object as a point mass and get the correct answer. I can't remember if the sphere has to be uniform density right now, but I think no.

Indeed, it was this crucial step that led Newton to be the first one that published G (m1 m1)/r^2 for the force of gravity. Others had realized that that was probably the formula, but could not do the work without first making the (very crucial) substitution of the planets as point masses. Newton also delayed publishing his results, because he could not justify that step -- I think he eventually came up with the justification and published it. The now make you do the integral in college physics classes.

Non-spherical bodies are much more complicated, but will ultimately act a lot like point masses with their center of gravity as the point, BUT rotation may muck things up.

But to answer your query, no, lead and styrofoam would not come together at different times. Well, to get both to be 1kg, the styrofoam would be a much bigger sphere, so the correction mentioned by mbd1mbd1 would change things -- but not because the materials were different, but rather because the radii were different.

By: vacapinta

vacapinta — Wed, 04 Jan 2006 17:53:04 -0800

Shit. Looking at teece and DA's work I realized I had plugged in the wrong numbers: With R=1 , M=2, I get 96145,577 seconds from the Kepler formula (using 6.673 for G). Bleh.

This is because the Kepler derivation begins with d/dt(dx/dt)=-GM/x^2 for one body.

I'm still trying to understand the "totally wrong" comment from andrew. This is a valid model.

By: weapons-grade pandemonium

weapons-grade pandemonium — Wed, 04 Jan 2006 17:56:46 -0800

I understand the point mass point, teece, but I think you missed my point.
Read it again.

By: DevilsAdvocate

DevilsAdvocate — Wed, 04 Jan 2006 18:02:55 -0800

You can approximate a spherical object as a point mass and get the correct answer. I can't remember if the sphere has to be uniform density right now, but I think no.

It doesn't have to be uniformly dense, but only some non-uniform objects can be treated as a point mass.

To be more specific, you can treat a spherical object as a point mass as long as the density is uniform within any spherical shell centered on the center of the sphere. But different shells can have different densities. Or, to put it another way, as long as the density varies only as a function of the distance from the center, you're OK.

By: weapons-grade pandemonium

weapons-grade pandemonium — Wed, 04 Jan 2006 18:18:04 -0800

But this still doesn't address my point, DA. The problem needs to specify more information. If the one meter separation is measured from the point mass, then the time varies from zero to a day and a half, depending upon the materials used. If the separation is measured from the leading edges and the materials are wispy, the centers of mass could be light years apart, and the time could approach infinity.

By: chrismear

chrismear — Wed, 04 Jan 2006 18:21:05 -0800

As an alternative perspective on this sort of question, you can often get a surprisingly good ballpark solution by combining some basic physics with some bold approximations (and in the process save yourself a lot of calculus headaches). I did this:

I know the total potential energy at the start:

G m² / r

and I know that the total kinetic energy right at the point of collision must be

m v²

Conservation of energy tells me that these two must be equal: G m² / r = m v². A little rearranging gives

v² = G m / r

I remember that G is something like 6 x 10^-11 (in SI) -- let's round it up to 1 x 10^-10 to keep the numbers easy. Of course, m and r are just 1, so we get a final velocity of v = 10^-5 m/s.

Intuitively I know that they accelerated sort of smoothly from zero speed up to 10^-5 m/s, but instead of worrying about all that, let's just say that they travelled at a constant average speed halfway between 0 and 10^-5; in other words, 0.5x10^-5.

Then we can just use the familiar

time taken = distance / speed
= 0.5 m / 0.5x10^-5 m/s
= 100000 s

which is only about 5% out from the accurate answer.

By: weapons-grade pandemonium

weapons-grade pandemonium — Wed, 04 Jan 2006 18:23:01 -0800

Hello? Is anybody home?

By: gleuschk

gleuschk — Wed, 04 Jan 2006 18:31:45 -0800

knock it off, w-gd. We heard you, and we understand. Sure, as stated, there isn't enough information to solve the problem. You're right. So we make simplifying assumptions. This is standard stuff. Don't embarrass yourself.

By: teece

teece — Wed, 04 Jan 2006 18:35:47 -0800

WGP, I see what you are saying now.

Yes, the formula here is for point masses only. It is a very good approximation for r much less than d.

But the equations of motion are perfectly good regardless. So it's only a matter of using the solution you find, and looking for a different end point. So two point masses will touch at zero. Two 1 kg lead shot-puts won't, they will touch at 2*r meters apart. But it's the same equation. Now you just look for what time t gives you that separation, rather than zero.

Now, if you use styrofoam balls 1kg each and 1m apart, the balls are probably so big that you'll get significant error treating them like point masses. (because r is not much less than d). But you can still figure out exactly the time of impact, as long as you remember the spheres are not actually point masses and work accordingly towards the correct final separation that yields contact.

By: Jimbob

Jimbob — Thu, 05 Jan 2006 00:09:43 -0800

This post has reminded me why I fear and hate mathematics, and why I would so love to understand it like you people.

By: DevilsAdvocate

DevilsAdvocate — Thu, 05 Jan 2006 05:52:45 -0800

But this still doesn't address my point, DA.

I wasn't trying to address your point. I was addressing the part I quoted, which is why I quoted it, so it would be clear what I was addressing.

Your point is correct: I (and others) have assumed these are point masses. If that assumption is incorrect, then yes, we must consider whether the "1m" was between the centers of the masses, or their edges, and also that they touch when their edges come together, not their centers.

I didn't say, "yes, you are correct, wgp," because I'm not in the habit of making "me too!" comments.

I didn't make any attempt to calculate what would happen if the centers were 1m apart and the masses were spheres of 10cm diameter because a) without that information being given, that assumption is just as spurious as the assumption that they are point masses, and b) it's fairly easily calculated with the formula I gave above (plug in x=0.05).

I made the assumptions that a) the masses are point masses; b) there are no other masses in the universe; c) there are no light sources in the universe (light pressure, you know); d) space itself is not expanding or contracting to any significant extent; e) relativistic effects are negligible or non-existent; f) quantum effects are negligible or non-existent; g) this strange universe, clearly so different from our own in terms of contents, still operates under the same physical laws as ours. And probably a dozen others that I'm not thinking of off the top of my head. If any of those assumptions are invalid, the calculations will give an incorrect answer.

Happy now?

By: andrew cooke

andrew cooke — Thu, 05 Jan 2006 06:48:11 -0800

came to apologise. i suspect vacapinta is right, if there is an expression for the orbital period that depends only on the major axis, then it should be half that period (with the major axis being the initial separation). work's very busy, so i don't have time to check myself, but i don't why the limit to extreme elipticities shouldn't exist.

By: JackFlash

JackFlash — Thu, 05 Jan 2006 08:25:58 -0800

I found this to be a very interesting exercise. Given a first quick guess I would have expected a result in years instead of days given the 10^-11 magnitude of the gravitational constant. I particularly liked chrismear's solution because it takes only a minute to realize that your intuition is wrong -- that very tiny forces exerted for long periods lead to significant velocities. This is the principle behind interstellar ion engines and solar sails.

So why didn't I wake up this morning to find my stapler, mouse and coffee cup all in a clump in the center of my desk? Its because the gravitational forces between small objects is insignificant compared to frictional forces except in a weightless vacuum.

Chrismear's derivation also clearly shows that the velocity is proportional to the square roots of mass and distance. You have to have very large masses or very small distances to have noticeable effects.

By: alby

alby — Wed, 11 Jan 2006 12:39:03 -0800

A quick thanks to all, I'll be checking through and trying to understand this as soon as I get a minute. Once again, thanks!