Calculating Gravitational Attraction Between Two Masses in Empty Space

[Doc Al]

I think it would be even wiser to stick with the problem as stated : two 1 kg masses starting out (stationary) at a distance of 1 meter.

I did. Of course you must realize for the exercise that these are point masses, not macroscopic objects.

The PE and indeed the total energy they possesses is based on that distance and not on infinity.

All measures of PE depend upon a reference point. When talking about the gravitational PE between two point masses (or astronomical bodies), the natural reference point is to take PE = 0 when they are infinitely far apart. You earlier agreed that the gravitational PE between two bodies is given by -GMm/R; that relationship assumes a reference point at infinity.

When you calculate the PE of a rock held at 1 meter above the ground, you do not reference it to infinity?

Not generally, because you are talking about Earth gravity and you are only concerned about small changes in position compared to the distance to the Earth's center. But if you were to move that rock to a height equal to the Earth's radius, you'd certainly use the more general expression. Or perhaps you want to calculate the escape velocity of that rock, again you'd use the more general expression.

In raising the rock a mere 1 m above the Earth's surface, you are only changing the distance to the Earth's center by 1 part in about 6,400,000. That's a realm where you can treat the force of gravity as being constant. Not so when the bodies are going from 1m to "contact". Just to get to within a mm of each other changes the distance between them by a factor of 1000!

The linearization I am using here is valid for a distance of 1 meter.

Nope. As explained above, over this range the force is wildly nonlinear.

It takes about 27 hours for them to cover 1 meter. Do you really think their velocity is approaching infinity? When that tool bag floated away from the astronaut in space, would it have come back to her at infinite velocity and infinite energy? I am not the one talking nonsense here!

Again, you are confusing your common sense experience with ordinary objects with this exercise in Newtonian gravity involving point masses.

 

[adriank]

In other words, under the weak force of gravity, we would have for all intents and purposes a nuclear explosion caused by two spheres coming together in such a way. In fact, if I jump in the air, and return to earth, I should land with infinite energy at an infinite velocity using the exact same mathematics. Obviously, by common sense alone, this is wrong!

Without involving general relativity, you know from special relativity that as kinetic energy goes to infinity, the speed approaches c (the speed of light), not infinity. Even so, we are also assuming that these are point masses, which, from our experience, also do not exist in reality. They would really collide at some point before r = 0.

Anyway, let's do some specific calculations. For simplicity I'll pick my units such that Gm² = 1, so U = -1/2r.

Initially, the masses are not moving, so the kinetic energy is K = 0. Since the distance between the masses is 2r = 1, the potential energy is U = -1. Thus the total energy is E = -1.

Now remember, total energy is conserved, so E = -1 always. Now let's say the distance between the masses is now 0.1. The potential energy is U = -1/0.1 = -10, so kinetic energy is K = E - U = (-1) - (-10) = 9. If the distance is 0.001, U = -1000, so K = E - U = 999. It should be clear that K increases without bound, as long as r is allowed to go to zero.

The value of E does not in any way limit the value of K. In this particular example this is most apparent, since E is negative, but K must always be nonnegative.

Remember that potential energy is always measured relative to some reference point, so we can make, in our calculations, total energy to be any value we like. Kinetic energy, however, is absolute.

If you're still having trouble understanding what's going on, I suggest that it would be more productive to talk to a physicist in person.

Adrian, if you have the time could you explain a little better how you integrated the (1/r - 1/r0). Namely how you obtained $$r = r_0 \cos^2 \theta$$

It's one of the usual trigonometric substitutions, although in a slightly unusual form. I wanted to have $$1/r = (1/r_0) \sec^2 \theta$$. Once you have that, the rest is straightforward:
$$\begin{align*} \int_0^{r_0} \left( \frac{1}{r} - \frac{1}{r_0} \right)^{-1/2} dr &= \int_{\pi/2}^0 \left( \frac{1}{r_0 \cos^2 \theta} - \frac{1}{r_0} \right)^{-1/2} \frac{dr}{d\theta} \,d\theta \\ &= r_0^{1/2} \int_{\pi/2}^0 (\sec^2 \theta - 1)^{-1/2} (-2r_0 \sin \theta \cos \theta) \,d\theta \\ &= r_0^{3/2} \int_0^{\pi/2} 2 (\tan^2 \theta)^{-1/2} \sin \theta \cos \theta \,d\theta \\ &= r_0^{3/2} \int_0^{\pi/2} 2 \cos^2 \theta \,d\theta \\ &= \frac{\pi}{2} r_0^{3/2}. \end{align*}$$

 

[schroder]

Yes.


Then you would be wrong again.

Think of it this way...at r=1/2, U=-2 and so E=1 and hence v=1...that means that over the first half of their journey (distance wise), the particles have been slowly accelerating up to a speed of 1. In that first half, most of the time necessary for the entire journey has passed...it isn't until they get very very close together that they start to accelerate to enormous speeds and by that time most of the 27 hours has already passed.

It's sort of like driving most of the way from LA to New York at 1km/h and then switching on a jet pack when you reach Washington...you'll get from Washington to New York very quickly, but your entire journey will take considerably longer.

What you, as well as others here, are doing, is taking that hyperbolic acceleration curve, which approaches asymptotes at a radious of infinity and PE of zero, and other asymptote of radius zero and PE infinity, and you are applying that entire curve to a distance of one meter! It has long ago been established that you cannot do that. On the earth’s surface, for near Earth objects we linearize that curve to get mgh. And for two objects ( I don’t give a damn whether they are “point sources” or spheres as it makes no difference in the analysis ) at a distance of one meter you also must linearize that curve. It is ridiculous in the extreme to have two objects starting from rest at a distance of one meter accelerating to infinite velocity under the extremely weak force of gravity! If gravity were that powerful, nothing would ever float away in space, like that astronaut’s tool kit, but it would be stuck to her like the most powerful magnet imaginable. For once, think about what you are saying instead of blindly applying the math! LINEARIZE over short distances. Get a good physics book and READ IT!

 

[adriank]

Don't you find it odd that everyone else is disagreeing with you?

It is ridiculous in the extreme to have two objects starting from rest at a distance of one meter accelerating to infinite velocity under the extremely weak force of gravity!

This doesn't ever happen in reality because point masses aren't real.

Why are you talking about linearization? Why use an approximation? The exact solution has already been given.

Alright, so let's say kinetic energy doesn't increase without bound. What value will it approach as r goes to zero? (It won't be the total energy, since we've already established that total energy is negative; using our gauge choice for the potential energy, U = -Gm²/2r, so that E = -Gm²/2r₀.) Where is your math?

 

[schroder]

Don't you find it odd that everyone else is disagreeing with you?


This doesn't ever happen in reality because point masses aren't real.

Why are you talking about linearization? Why use an approximation? The exact solution has already been given.

Alright, so let's say kinetic energy doesn't increase without bound. What value will it approach as r goes to zero? (It won't be the total energy, since we've already established that total energy is negative; using our gauge choice for the potential energy, U = -Gm²/2r, so that E = -Gm²/2r₀.) Where is your math?

I don't find that at all odd, on THIS forum, where at least one of the so-called "mentors" firmly believes in DDWFTTFW! So why should I be surprised at this?

I have already, several times, calculated the MAX velocity that is reached. You can find that a couple of posts on up. IF that max velocity was constant (it isn't) but IF it were, it would take 17 hours to pull them together. IF the acceleration was constant (it isn't) it would take 34 hours. By integration, we all agree total time is around 27 hours. Each sphere or "point mass" take you pick covers a half meter in 27 hours starting from rest. And somehow you are saying and actually believing, that they approach infinite velocity! That is one HELL of an acceleration curve over one meter and all under the force of gravity! THat makes sense to you? The problem comes from applying that acceleration curver over the distance of one meter. THAT is why it must be linearized.
They may as well draw that staright line through my name right now, as this physics forum never admits they are wrong they just delete posts and ban people. I have had enough of this. Incidently, I am a degreed engineer in both civils engineering and EE and have been a team leader on advanced R+D for satellite communications as well as deep space probes. I know what I am talking about.

 

[Crazy Tosser]

Gravity is inversely proportional to r^2. With pointlike masses, as r approaches zero, G approaches infinity. From the classical mechanics viewpoint, anyway.
That's why I hate equations, they never make any sense. Just take it easy, schroder ;)

 

[adriank]

schroder: Clearly you don't know what your talking about, and throwing credentials around will get you nowhere if you can't make sense.

I'll now respond to your post where you gave the maximum speed, which I really didn't read fully before.

So physicists have linearized the gravitational potential and potential energy to make it more useful for everyday experience, which includes our consideration of two bodies separated by only one meter.

Do you understand that linearization is an approximation? It's just nonsense to use an approximation to claim that an exact solution is wrong! You won't get the correct time to collision that way, either.

By LCD we get: = GMh/R(R + h) the final step is to drop the h as it is insignificant to the R.

Assuming I'm interpreting what you mean by h, that is far from insignificant compared to R! That isn't valid. The amount the masses move is comparable (actually equal!) to the distance between them.

At a distance of zero meters, when the spheres are touching, PE = mg’h and is also zero.

That only works for variations in height that are very small compared to the distance between the masses, so that g' is effectively constant. That's only a first-order approximation of the actual potential energy, U = U₀ - GMm/r². You really should know this. Your problem there is that the gravitational acceleration is not constant. You even admitted earlier that potential energy does indeed go to negative infinity.

Prove to me that, given your solution, energy is conserved. If you can't do that, then there's a problem with your argument, isn't there?

The problem comes from applying that acceleration curve over the distance of one meter.

There is no problem.
If you continue to insist that you are correct, I will consider you to be trolling and ignore you.

 

[adriank]

One more thing: I will find the velocity in terms of r directly. Recall from my initial post that

$$-\frac{dr}{\sqrt{\frac{Gm}{2} \left( \frac{1}{r} - \frac{1}{r_0} \right) }} = dt,$$

or equivalently,

$$-\sqrt{\frac{2}{Gm}} \left( \frac{1}{r'} - \frac{1}{r_0} \right)^{-1/2} dr' = dt',$$

where I changed r to r' and t to t' so I could use r and t inside the limits of integrals.

I integrate this to find the time required to reach any particular position. At t' = 0, r' = r₀, and at t' = t, r' = r, so

$$-\sqrt{\frac{2}{Gm}} \int_{r_0}^{r} \left( \frac{1}{r'} - \frac{1}{r_0} \right)^{-\frac12} \,dr' = \int_0^t dt' = t.$$

I make the substitution $$r' = r_0 \cos^2 \theta$$, so $$\theta = \arccos \sqrt{r'/r_0}$$:

$$\begin{align*} t &= \sqrt{\frac{2}{Gm}} \int_{0}^{\arccos \sqrt{r/r_0}} \left( \frac{1}{r_0 \cos^2 \theta} - \frac{1}{r_0} \right)^{-\frac12} 2r_0 \sin \theta \cos \theta \,d\theta \\ &= \sqrt{\frac{2r_0^3}{Gm}} \int_{0}^{\arccos \sqrt{r/r_0}} 2 \cos^2 \theta \,d\theta \\ &= \sqrt{\frac{2r_0^3}{Gm}} \left[ \theta + \sin \theta \cos \theta \right]_{0}^{\arccos \sqrt{r/r_0}} \\ &= \sqrt{\frac{2r_0^3}{Gm}} \left( \arccos \sqrt{\frac{r}{r_0}} + \sqrt{1 - \frac{r}{r_0}}\sqrt{\frac{r}{r_0}} \right). \end{align*}$$

I can differentiate this to obtain the velocity:

$$\begin{align*} \frac{1}{v} &= \frac{dt}{dr} = -\sqrt{\frac{2r}{Gm (1 - r/r_0)}} \\ v &= -\sqrt{\frac{Gm}{2r} \left( 1 - \frac{r}{r_0} \right)} = -\sqrt{\frac{Gm}{2r_0} \left( \frac{r_0 - r}{r} \right)}. \end{align*}$$

I plot that below.

http://img56.imageshack.us/img56/1972/graveo8.png

It's pretty apparent from both the equation and the graph that as r goes to 0, v goes to negative infinity.

Remember that I've been using purely classical mechanics here, so the kinetic energy of one of the masses is $$mv^2/2$$. Let's see if this expression for velocity conserves energy:

$$\begin{align*} E &= U + K = -\frac{Gm^2}{2r} + 2 \left( \frac12 mv^2 \right) \\ &= -\frac{Gm^2}{2r} + m \frac{Gm}{2r_0} \left( \frac{r_0 - r}{r} \right) \\ &= -\frac{Gm^2}{2r} + \frac{Gm^2}{2r} - \frac{Gm^2}{2r_0} \\ &= -\frac{Gm^2}{2r_0}. \end{align*}$$

This is what was desired. Note that I have made no approximations or unjustified steps anywhere. This is exact.

 

[Doc Al]

What you, as well as others here, are doing, is taking that hyperbolic acceleration curve, which approaches asymptotes at a radious of infinity and PE of zero, and other asymptote of radius zero and PE infinity, and you are applying that entire curve to a distance of one meter! It has long ago been established that you cannot do that.

Kind of silly for you to keep insisting this, when not only can you do it, it's been done for you right in this thread.

On the earth’s surface, for near Earth objects we linearize that curve to get mgh.

Yes, because h << radius of the earth!

And for two objects ( I don’t give a damn whether they are “point sources” or spheres as it makes no difference in the analysis ) at a distance of one meter you also must linearize that curve.

Still comically wrong! And why even attempt a linear approximation when you can solve it exactly?

It is ridiculous in the extreme to have two objects starting from rest at a distance of one meter accelerating to infinite velocity under the extremely weak force of gravity!

Please give me an estimate of the gravitational force between two 1 kg point masses that are 10^-12 m apart?

If gravity were that powerful, nothing would ever float away in space, like that astronaut’s tool kit, but it would be stuck to her like the most powerful magnet imaginable.

Still not relevant. Again, such examples have nothing to do with idealized point masses.

For once, think about what you are saying instead of blindly applying the math! LINEARIZE over short distances.

This is getting comical. You linearize as an approximation, and only when Δx << x. That is far from applicable to this elementary exercise.

Get a good physics book and READ IT!

Again let me remind you that this is just an exercise in applying Newtonian gravity to idealized point masses in Newtonian mechanics: There are no real point masses that are only subject to Newtonian gravity, and Newtonian mechanics breaks down at small distances and high speeds.

 

[gabbagabbahey]

What you, as well as others here, are doing, is taking that hyperbolic acceleration curve, which approaches asymptotes at a radious of infinity and PE of zero, and other asymptote of radius zero and PE infinity, and you are applying that entire curve to a distance of one meter!

Yes. And the reason we are doing that is because (in the context of Newtonian gravity) it is the correct approach.

It has long ago been established that you cannot do that.

Your definition of established clearly differs greatly from the definition that reasonable people use.

On the earth’s surface, for near Earth objects we linearize that curve to get mgh. And for two objects ( I don’t give a damn whether they are “point sources” or spheres as it makes no difference in the analysis ) at a distance of one meter you also must linearize that curve. It is ridiculous in the extreme to have two objects starting from rest at a distance of one meter accelerating to infinite velocity under the extremely weak force of gravity! If gravity were that powerful, nothing would ever float away in space, like that astronaut’s tool kit, but it would be stuck to her like the most powerful magnet imaginable. For once, think about what you are saying instead of blindly applying the math! LINEARIZE over short distances.

This is complete and utter nonsense. The reason you linearize gravity near the surface of the Earth is because your PE and hence acceleration due to gravity are vary slowly over reasonable height distances. This (as Doc Al has already pointed out to you) is because the radius of the Earth is MUCH larger than normal height distances and so $R+h \approx R$. In the case of these two particles which are subject only to each others gravity, there is no large Earth that they sit on. 'R' is zero and so a change of one meter in h changes the PE and the acceleration GREATLY.

Get a good physics book and READ IT!

Get a less defective brain and use it.

 

[schroder]

Kind of silly for you to keep insisting this, when not only can you do it, it's been done for you right in this thread.

Yes, because h << radius of the earth!

Still comically wrong! And why even attempt a linear approximation when you can solve it exactly?

Please give me an estimate of the gravitational force between two 1 kg point masses that are 10^-12 m apart?

Still not relevant. Again, such examples have nothing to do with idealized point masses.

This is getting comical. You linearize as an approximation, and only when Δx << x. That is far from applicable to this elementary exercise.



Again let me remind you that this is just an exercise in applying Newtonian gravity to idealized point masses in Newtonian mechanics: There are no real point masses that are only subject to Newtonian gravity, and Newtonian mechanics breaks down at small distances and high speeds.

Let me remind you, that until you entered this thread, in post number 31, there was NO mention of idealized “point masses” The only thing mentioned was two 1 kg masses and “time for two bodies to converge in space”. Read the thread and tell me where and when “point masses” entered the discussion. Until you made it an “exercise” in abstract mathematical concepts that have absolutely NO bearing on the physical world, this was a PHYSICS question. And in PHYSICS, there is such a thing as the conservation of mechanical energy in a conservative system. THAT is what this system is, or was, until you decided to change the subject entirely.
In the context of physics, in the real world, the TOTAL amount of energy in such a conservative SUSTEM, with two stationary masses, is determined by the gravitational PE BETWEEN the two masses over the distance between them. ALL the energy is PE at the start and during the transit over the distance the energy is converted to KE, until at the point of contact ALL of the energy is KE and PE is zero. Tell me how much PE there is between two masses that are in contact with each other, when the distance is zero. I will give you the answer without the need for any integration; the PE is zero, NOT infinity. The only thing that is comical about this is the way you are blindly applying abstract mathematical concepts to what is a practical real world physics problem and coming up with comical answers. There are NO “point particles” there is NO infinite energy and there is NO infinite velocity except in your abstract math which has no application to the real physical universe. I am done with this joke of a physics forum. This is a sad reflection on the state of physics education in the USA today.

 

[Hootenanny]

I am done with this joke of a physics forum. This is a sad reflection on the state of physics education in the USA today.

Your woefully inadequate ability to comprehend simple physical reasoning is the only sad reflection on the state of physics education.

This thread is done. If the OP would like to discuss this further they are more than welcome to PM me and I'll re-open the thread.

 

[Doc Al]

Let me remind you, that until you entered this thread, in post number 31, there was NO mention of idealized “point masses”

Right. I was the one who made explicit the assumption that everyone--including you--was making. If you weren't making such an assumption, tell me how big those two objects were and how far apart they were when they touched?

The only thing mentioned was two 1 kg masses and “time for two bodies to converge in space”. Read the thread and tell me where and when “point masses” entered the discussion. Until you made it an “exercise” in abstract mathematical concepts that have absolutely NO bearing on the physical world, this was a PHYSICS question. And in PHYSICS, there is such a thing as the conservation of mechanical energy in a conservative system. THAT is what this system is, or was, until you decided to change the subject entirely.

Get a clue: I and everyone else agree that mechanical energy is conserved in this system.

In the context of physics, in the real world, the TOTAL amount of energy in such a conservative SUSTEM, with two stationary masses, is determined by the gravitational PE BETWEEN the two masses over the distance between them. ALL the energy is PE at the start and during the transit over the distance the energy is converted to KE, until at the point of contact ALL of the energy is KE and PE is zero.

Still comically wrong. You might have squeaked by in your civil engineering classes with such a shoddy understanding, but clearly you never took a serious course in classical mechanics.

Just for review: Conservation of mechanical energy means that PEi + KEi = PEf + KEf, not that KEf = PEi. (That latter is only true in certain simple cases.)

Tell me how much PE there is between two masses that are in contact with each other, when the distance is zero.

Two macroscopic masses? Easy: How big are they? How far apart are their centers?

I will give you the answer without the need for any integration; the PE is zero, NOT infinity.

It's easy to get the wrong answer with no calculation at all!
Don't confuse macroscopic objects with point masses. (Of course, that's just one of your confusions.) In any case, the PE is not zero!

The only thing that is comical about this is the way you are blindly applying abstract mathematical concepts to what is a practical real world physics problem and coming up with comical answers. There are NO “point particles” there is NO infinite energy and there is NO infinite velocity except in your abstract math which has no application to the real physical universe.

Note that the "real world" problem of two macroscopic objects (let's say two perfect spheres, for simplicity) being attracted by gravity alone is solved in exactly the same way as for two point masses! The only difference is that you don't go from r = 1 m to r= 0, you go from r = 1 m to r = D (the distance between the centers on contact). In both problems the PE is given by -GMm/r; it is most certainly not zero.

I am done with this joke of a physics forum. This is a sad reflection on the state of physics education in the USA today.

My thoughts exactly!