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

January 4, 2006 1:14 PM Subscribe

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.

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

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

posted by I Love Tacos at 1:31 PM on January 4, 2006

posted by I Love Tacos at 1:31 PM on January 4, 2006

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

If I'm remembering my High School physics correctly.

Solve for time.

posted by Wild_Eep at 1:38 PM on January 4, 2006

If I'm remembering my High School physics correctly.

Solve for time.

posted by Wild_Eep at 1:38 PM on January 4, 2006

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.

posted by teece at 1:38 PM on January 4, 2006

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.

posted by teece at 1:38 PM on January 4, 2006

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.

posted by vacapinta at 1:42 PM on January 4, 2006

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.

posted by vacapinta at 1:42 PM on January 4, 2006

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

posted by teece at 1:43 PM on January 4, 2006

posted by teece at 1:43 PM on January 4, 2006

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.

posted by andrew cooke at 1:52 PM on January 4, 2006

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.

posted by andrew cooke at 1:52 PM on January 4, 2006

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

posted by vacapinta at 1:53 PM on January 4, 2006

posted by vacapinta at 1:53 PM on January 4, 2006

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

posted by andrew cooke at 1:53 PM on January 4, 2006

posted by andrew cooke at 1:53 PM on January 4, 2006

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

posted by andrew cooke at 1:54 PM on January 4, 2006

posted by andrew cooke at 1:54 PM on January 4, 2006

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

posted by andrew cooke at 1:55 PM on January 4, 2006

posted by andrew cooke at 1:55 PM on January 4, 2006

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

posted by ewagoner at 2:14 PM on January 4, 2006

posted by ewagoner at 2:14 PM on January 4, 2006

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))

1/(2r(t))

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

posted by gleuschk at 2:19 PM on January 4, 2006

The force on A is (proportional to) 1/(2r(t))

^{2}. Since force = mass x acceleration, we know that1/(2r(t))

^{2}= r'(t).This is a second-order differential equation. I never was very good at differential equations.

posted by gleuschk at 2:19 PM on January 4, 2006

*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.

posted by vacapinta at 2:23 PM on January 4, 2006

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.

posted by driveler at 2:28 PM on January 4, 2006

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

posted by driveler at 2:28 PM on January 4, 2006

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.

posted by teece at 2:36 PM on January 4, 2006

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.

posted by teece at 2:36 PM on January 4, 2006

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)

posted by vacapinta at 2:51 PM on January 4, 2006

posted by vacapinta at 2:51 PM on January 4, 2006

teece, I don't understand

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.

posted by gleuschk at 2:55 PM on January 4, 2006

*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.

posted by gleuschk at 2:55 PM on January 4, 2006

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

posted by paradroid at 2:56 PM on January 4, 2006

posted by paradroid at 2:56 PM on January 4, 2006

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.

posted by mr_roboto at 2:59 PM on January 4, 2006

posted by mr_roboto at 2:59 PM on January 4, 2006

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.

posted by Galvatron at 3:11 PM on January 4, 2006

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.

posted by Galvatron at 3:11 PM on January 4, 2006

Galvatron: is this what you are trying to do?

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

posted by vacapinta at 3:18 PM on January 4, 2006

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

posted by vacapinta at 3:18 PM on January 4, 2006

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

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.

posted by teece at 3:23 PM on January 4, 2006

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.

posted by teece at 3:23 PM on January 4, 2006

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

posted by Decani at 3:31 PM on January 4, 2006

posted by Decani at 3:31 PM on January 4, 2006

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

posted by Galvatron at 3:37 PM on January 4, 2006

posted by Galvatron at 3:37 PM on January 4, 2006

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.

posted by teece at 3:38 PM on January 4, 2006

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.

posted by teece at 3:38 PM on January 4, 2006

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

posted by mr_roboto at 3:42 PM on January 4, 2006

posted by mr_roboto at 3:42 PM on January 4, 2006

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.

posted by mbd1mbd1 at 3:53 PM on January 4, 2006

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.

posted by mbd1mbd1 at 3:53 PM on January 4, 2006

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.

posted by folara at 4:53 PM on January 4, 2006

posted by folara at 4:53 PM on January 4, 2006

Picking up where Galvatron left off, i.e., (dx/dt)

dx/(1/2x-1)

Substitute u=1/2x. du=-dx/2x

du/[u

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)

And substituting back in for u gives:

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

According to this formula, t approaches pi/4G

posted by DevilsAdvocate at 4:55 PM on January 4, 2006

^{2}= G(1/2x-1) (having substituted m=1):dx/(1/2x-1)

^{1/2}=±G^{1/2}dtSubstitute u=1/2x. du=-dx/2x

^{2}, therefore dx=-du/2u^{2}.du/[u

^{2}(u-1)^{1/2}] = ±2G^{1/2}dtIntegration 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^{1/2}tAnd 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.posted by DevilsAdvocate at 4:55 PM on January 4, 2006

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

posted by teece at 5:05 PM on January 4, 2006

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 ;-)posted by teece at 5:05 PM on January 4, 2006

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?

posted by treeshade at 5:25 PM on January 4, 2006

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?

posted by treeshade at 5:25 PM on January 4, 2006

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?

posted by weapons-grade pandemonium at 5:41 PM on January 4, 2006

posted by weapons-grade pandemonium at 5:41 PM on January 4, 2006

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.

posted by weapons-grade pandemonium at 5:48 PM on January 4, 2006

posted by weapons-grade pandemonium at 5:48 PM on January 4, 2006

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.

posted by teece at 5:50 PM on January 4, 2006

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.

posted by teece at 5:50 PM on January 4, 2006

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.

posted by vacapinta at 5:53 PM on January 4, 2006

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.

posted by vacapinta at 5:53 PM on January 4, 2006

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

Read it again.

posted by weapons-grade pandemonium at 5:56 PM on January 4, 2006

Read it again.

posted by weapons-grade pandemonium at 5:56 PM on January 4, 2006

*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.

posted by DevilsAdvocate at 6:02 PM on January 4, 2006

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.

posted by weapons-grade pandemonium at 6:18 PM on January 4, 2006

posted by weapons-grade pandemonium at 6:18 PM on January 4, 2006

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 m

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

m v

Conservation of energy tells me that these two must be equal: G m

v

I remember that G is something like 6 x 10

Intuitively I know that they accelerated sort of smoothly from zero speed up to 10

Then we can just use the familiar

time taken = distance / speed

= 0.5 m / 0.5x10

=

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

posted by chrismear at 6:21 PM on January 4, 2006

I know the total potential energy at the start:

G m

^{2}/ rand I know that the total kinetic energy right at the point of collision must be

m v

^{2}Conservation of energy tells me that these two must be equal: G m

^{2}/ r = m v^{2}. A little rearranging givesv

^{2}= G m / rI 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.

posted by chrismear at 6:21 PM on January 4, 2006

Hello? Is anybody home?

posted by weapons-grade pandemonium at 6:23 PM on January 4, 2006

posted by weapons-grade pandemonium at 6:23 PM on January 4, 2006

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

posted by gleuschk at 6:31 PM on January 4, 2006

**make simplifying assumptions**. This is standard stuff. Don't embarrass yourself.posted by gleuschk at 6:31 PM on January 4, 2006

WGP, I see what you are saying now.

Yes, the formula here is for point masses only. It is a very good approximation for

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*

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

posted by teece at 6:35 PM on January 4, 2006

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.posted by teece at 6:35 PM on January 4, 2006

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

posted by Jimbob at 12:09 AM on January 5, 2006

posted by Jimbob at 12:09 AM on January 5, 2006

*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?

posted by DevilsAdvocate at 5:52 AM on January 5, 2006

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.

posted by andrew cooke at 6:48 AM on January 5, 2006

posted by andrew cooke at 6:48 AM on January 5, 2006

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.

posted by JackFlash at 8:25 AM on January 5, 2006

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.

posted by JackFlash at 8:25 AM on January 5, 2006

Response by poster: 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!

posted by alby at 12:39 PM on January 11, 2006

posted by alby at 12:39 PM on January 11, 2006

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posted by pwb503 at 1:26 PM on January 4, 2006