Calculating Gravitational Attraction Between Two Masses in Empty Space

[Perillux]

I'm just curious how long it would take for two 1kg masses separated by 1m to attract each other gravitationally in empty space. The formula for gravitational force is:

F_{g} = G \frac{m_{1}m_{2}}{r^{2}}
where r is the distance between the two masses.

So if the midpoint for the two masses is centered at the origin then the two objects are located at x=-0.5 and x=0.5 respectively. So the value for r^2 in the equation above is always (2x)^{2} So we can say that the formula for force between the two objects based on either objects position is:

F_{g} = \frac{G}{(2x)^{2}}
I didn't include m1m2 because they are both 1.


and that is as far as I can go. If I had an equation for the force based on time then I could solve for how long it takes the two objects to meet.
So is it possible to find an equation of the force based on time from what I have? Or maybe something completely different? Any help is appreciated.

 

[Cspeed]

For distance, use 1m instead of 2x. Then, it will be F = G. The force on each mass will be equal to that at time = 0. Of course, the force will increase as they get closer. I'm not sure which equation you can use to solve this.

 

[Peeter]

You'll need two equations. One for the force on the first particle and one for the second. Example:

<br /> \text{Force on} \quad m_2 = - G m (x_2 - x_1)^{-2} = {x_2}&#039;&#039;<br />
<br /> \text{Force on} \quad m_1 = G m (x_2 - x_1)^{-2} = {x_1}&#039;&#039;<br />

subtraction of these gives you an integration problem for a difference of position variable. The first integration of this can give you the "velocity" that this difference changes by, and a second will give you the time.

 

[Perillux]

subtraction of these gives you an integration problem for a difference of position variable.

By subtraction of these do you mean:

-Gm(x_{2}-x_{1})^{-2} - ( Gm(x_{2}-x_{1})^{-2} )
-2Gm(x_{2}-x_{1})^{-2}

not sure how to integrate that, with the two different values of x.

 

[Peeter]

Subtract one equation from the other:

<br /> \begin{align*}<br /> (- G m_1 (x_2 - x_1)^{-2} = {x_2}&#039;&#039;) \\<br /> -(G m_2 (x_2 - x_1)^{-2} = {x_1}&#039;&#039;) \\<br /> = \\<br /> - G (x_2 - x_1)^{-2} (m_1 + m_2)= (x_2 - x_1)&#039;&#039; \\<br /> \end{align*}<br />

Now let u=x_2 - x_1, and use d/dt = (du/dt) d/du to get an equation for velocity (squared) from the second derivative term.

 

[Perillux]

I'm still not sure I follow. I haven't taken a differentials class yet (taking it next semester).
But I gave it a try:

so if u=x2 - x1 then du/dt = 0
and d/du of the entire function should be:
2G(m_{1}+m_{2}) u^{-3}
and since you said d/dt = (du/dt)d/du then I guess multiplying those two will give me d/dt? so...
d/dt =
2G(m_{1}+m_{2}) u^{-3} * 0

I know I'm probably way off... but there it is.

 

[Peeter]

I'm still not sure I follow. I haven't taken a differentials class yet (taking it next semester).

Here's a more detailed bash at the problem (too hard to do with inline tex in the forum) :

http://sites.google.com/site/peeterjoot/math/twobodies.pdf

however, if you haven't taken calculus, this may not be any better.

re: what you wrote. x_2 and x_1 are functions of t, so those derivatives aren't zero. Also, writing d/dt was using lazy notation for the chain rule (how to differentiate a function of a function), where I had left out the thing that was being differentiated.

Peeter

 

[schroder]

While I have nothing against integral calculus, it seems to me that this problem can be solved much more simply by use of conservation of energy.
The gravitational potential between two 1 kg spheres at a distance of 1 meter is: Gmm/r = G
So the potential energy of the system is G or 6.67E-11 Joules. (Also the acceleration is G)
Since the spheres are equal, they will share that PE equally as kinetic energy just before they crash together. So each sphere will have a KE of 3.34E-11 J at impact. From that we can get the velocity at impact since KE = 1/2 mV^2 so velocity at impact is 8.167E-6 m/s for each sphere or total velocity of 1.633E-5 m/s for the system.
So now we can use X = V*T -1/2 aT^2 to compute the time, where X = 1 meter, V = 1.633E-5 and a is equal to G = 6.67E-11 and we solve for T by quadratic equation which of course yields two answers: T = 71,750 seconds or 417,904 seconds
The second answer must be the correct one since they are starting at zero velocity and accelerating to the max at impact. So it will take 417,904 seconds or about 116 hours.
How does that compare with the answer you get by integrating?

 

[Perillux]

well I still haven't figured out how to do it through integration. However, I think your answer might be off. Because, if you just take the initial acceleration Gmm/r^2 = G
and use that as a constant acceleration I get 34 hours. It's obvious that the actual answer should be smaller than this value because as they get closer they acceleration will be stronger.

 

[schroder]

well I still haven't figured out how to do it through integration. However, I think your answer might be off. Because, if you just take the initial acceleration Gmm/r^2 = G
and use that as a constant acceleration I get 34 hours. It's obvious that the actual answer should be smaller than this value because as they get closer they acceleration will be stronger.

The quadratic did give two answers. I forget now why I chose the larger of these, probably I was tired. It should be the smaller at 71,750 seconds or about 20 hours. When I have more time I will recheck it. It would be interesting to do the integration also and compare, but I am fairly confident that the conservation of energy will give the right answer without the need for the more powerful mathematical treatment.

 

[turin]

...
The gravitational potential between two 1 kg spheres at a distance of 1 meter is: Gmm/r = G
So the potential energy of the system is G or 6.67E-11 Joules.
...
So each sphere will have a KE of 3.34E-11 J at impact. From that we can get the velocity at impact since KE = 1/2 mV^2 so velocity at impact is 8.167E-6 m/s for each sphere or total velocity of 1.633E-5 m/s for the system.
So now we can use X = V*T -1/2 aT^2 to compute the time, where X = 1 meter, V = 1.633E-5 and a is equal to G = 6.67E-11 and we solve for T by quadratic equation which of course yields two answers: T = 71,750 seconds or 417,904 seconds
...

This is flawed for another reason besides the use of a constant acceleration. The gravitational potential energy is NEGATIVE, and it only becomes more negative as the masses approach each other, without bound. So, the KE actually approaches infinity. How long does it take to accelerate a mass at constant acceleration to an infinite velocity?

BTW, I got about 27 hours using the integration method.

 

[schroder]

No. The kinetic energy is conserved and is no greater than the total PE that is calculated at the distance of one meter. The total energy does not increase as they get closer. As you said, the PE is negative, so even though the number gets higher, it gets higher in a negative direction, so the PE gets less and the KE increases, but the Total energy stays exactly the same at all times. The velocity does not go to infinity but is limited by the kinetic energy. The max velocity that is attained is what I calculated. I admit the time calculation I made is “flawed” as I did use constant acceleration. I was looking for a ball-park figure. If we assume constant acceleration, the max time it can take is 34 hours. If we assume constant max velocity, as computed, the min time it can take is 17 hours. The 20 hours does fall in that range but I know it is not correct. I am still having a bit of trouble evaluating the integral as I need to work it out long hand. The integrand keeps reducing to unity, leaving me only the external term and that works out to 24 hours, also in the required range. Your figure of 27 hours falls exactly in the center of the range and must be correct.
(Well, not exactly in the center, but I believe it)

EDiot: The kinetic energy is not conserved, the TOTAL energy is conserved, which limits the amount of kinetic energy. It does not approach infinity.

 

[adriank]

Time to throw differential equations at the problem, I think.

For simplicity, I'm saying the masses are equal; r is the distance from the centre to each mass. Initially, r = r₀, so you have

\frac{d^2r}{dt^2} = -\frac{Gm}{r^2}, r(0) = r_0, r&#039;(0) = 0

Say v = dr/dt, so d^2r/dt^2 = v \,dv/dr:

v \frac{dv}{dr} = -\frac{Gm}{4r^2}
v \,dv = -\frac{Gm}{4r^2} \,dr
\frac12 v^2 = \frac{Gm}{4r} + c&#039;_1

I'll figure out what that constant is now: at v = 0, r = r₀, so

0 = \frac{Gm}{4r_0} + c_1 \Longrightarrow c_1 = -\frac{Gm}{4r_0}
\frac12 v^2 = \frac{Gm}{4} \left( \frac{1}{r} - \frac{1}{r_0} \right)
v = \frac{dr}{dt} = -\sqrt{\frac{Gm}{2} \left( \frac{1}{r} - \frac{1}{r_0} \right) }

I took the negative square root because we want the masses moving toward each other.

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

Now the hard part: integrating that left side. If I integrate that from r₀ to 0, I should get the amount of time that takes:

\sqrt{\frac{2}{Gm}} \int_{0}^{r_0} \left( \frac{1}{r} - \frac{1}{r_0} \right)^{-\frac12} \,dr = T

I find that the integral there is r_0^{3/2} \pi/2, so the amount of time it takes in total is

T = \frac{\pi r_0^{3/2}}{\sqrt{2Gm}}

Putting in r₀ = 0.5 m, m = 1 kg, I get 96149 seconds, or 26.7 hours.The integral can be evaluated by using the trigonometric substitution r = r_0 \cos^2 \theta.

 

[Denton]

Although I am able to actually keep up with your working AdrianK unlike peeter's (sorry peeter, your math is beyond me heh) the solutions does not seem to be reasonable.

Only one day required for two one kilo objects a meter a part to come together? You would probably be able to see then very heavy objects hung on well lubricated rails slide together (albeit slowly).
Anyways always wanted to find a solution for this problem, and I've never realized it was so complex. A work of relativity in itself :)One last thing:

d^{}2r/dt^{}2 = -GM / r^{}2

Where does the d^{}2r/dt^{}2 come from and or how does that equal F?

 

[schroder]

Although I am able to actually keep up with your working AdrianK unlike peeter's (sorry peeter, your math is beyond me heh) the solutions does not seem to be reasonable.

Only one day required for two one kilo objects a meter a part to come together? You would probably be able to see then very heavy objects hung on well lubricated rails slide together (albeit slowly).



Anyways always wanted to find a solution for this problem, and I've never realized it was so complex. A work of relativity in itself :)


One last thing:

d^{}2r/dt^{}2 = -GM / r^{}2

Where does the d^{}2r/dt^{}2 come from and or how does that equal F?

The problem with that is, unlike most other forces, gravity cannot be shielded. The two masses being considered are assumed to be in deep space, far from any other sources of gravitational attraction and so can be analyzed as the only forces acting. Two masses hanging on rails here on Earth would be subjected to the gravitational attraction of earth, far outweighing the attraction between the two masses.

I also appreciate Adriank showing us the math and the very good explanation!

(That one last thing is the second derivative of distance, which is acceleration, not force.)

 

[adriank]

Well, I just thought of another solution that's quite a bit shorter, using a previously known result. You can consider the path of a mass as half of a (degenerate) elliptical orbit, with period 2T and semimajor axis r₀ (half the sum of the maximum and minimum distances between the two masses). Kepler's third law then gives

\left(\frac{2T}{2\pi}\right)^2 = \frac{r_0^3}{G(m + m)}
T = \frac{\pi r_0^{3/2}}{\sqrt{2Gm}}.

 

[Denton]

You are most wise.

 

[turin]

EDiot: The kinetic energy is not conserved, the TOTAL energy is conserved, which limits the amount of kinetic energy. It does not approach infinity.

If by "it" you mean TOTAL energy, then I agree. If by "it" you mean KINETIC energy, then you should just make a little table of kinetic energy vs. distance. I suggest r = 1 m, 0.1 m, 0.01 m, and 0.001 m, for example. Then, come back and tell us if you still believe that KE does not approach infinity.

 

[adriank]

Total energy does not limit kinetic energy. As the masses approach each other, the potential energy
U = -\frac{Gm^2}{2r}
approaches negative infinity. Since total energy is conserved, kinetic energy must approach infinity.

 

[schroder]

Total energy does not limit kinetic energy. As the masses approach each other, the potential energy
U = -\frac{Gm^2}{2r}
approaches negative infinity. Since total energy is conserved, kinetic energy must approach infinity.

That is absolutely wrong! Suppose the Total energy is 100 Joules, just for an example and we agree that Total energy is conserved and is the sum of kinetic enrgy and potential energy. Now let the potential energy approach negative infinity. The SUM still must be 100 Joules. Kinetic energy will approach the maximum allowed amount of 100 Joules, NOT infinity!

 

[schroder]

If by "it" you mean TOTAL energy, then I agree. If by "it" you mean KINETIC energy, then you should just make a little table of kinetic energy vs. distance. I suggest r = 1 m, 0.1 m, 0.01 m, and 0.001 m, for example. Then, come back and tell us if you still believe that KE does not approach infinity.

I suggest you make the table. Start with a Total energy of 100 Joules, which is conserved and consistes of the total of KE and PE. Now let PE vary with distance. You will find the PE, which is Negative, risess in magnitude, which means it is going more negative, towards negative infinity. But the Total energy remains at 100 Joules. So plot the KE, which is the difference between the Total and the PE. What limit does the KE approach? Please come back and tell us that it approaches 100 Joules, and not infinity! Thanks.

 

[adriank]

That is absolutely wrong! Suppose the Total energy is 100 Joules, just for an example and we agree that Total energy is conserved and is the sum of kinetic enrgy and potential energy. Now let the potential energy approach negative infinity. The SUM still must be 100 Joules. Kinetic energy will approach the maximum allowed amount of 100 Joules, NOT infinity!

I'm pretty sure you're the one who is wrong here. You have E = K + U, where E is total energy (a constant), K is kinetic energy, and U is potential energy. Then
K = E - U = E + \frac{Gm^2}{2r}
which goes to infinity as r goes to 0.

If kinetic energy approached 100 joules and potential energy went to negative infinity, then total energy would also go to negative infinity.

 

[schroder]

I'm pretty sure you're the one who is wrong here. You have E = K + U, where E is total energy (a constant), K is kinetic energy, and U is potential energy. Then
K = E - U = E + \frac{Gm^2}{2r}
which goes to infinity as r goes to 0.

If kinetic energy approached 100 joules and potential energy went to negative infinity, then total energy would also go to negative infinity.

Then, according to you -.000000000001 + 1 = -.0000000000001

According to me -0.0000000000001 + 1 = 1

which is correct?

 

[adriank]

Negative infinity. As in, a really really large negative amount. -0.0000000000001 is far from infinite.

Suppose total energy is E = 100 J. Potential energy is on its way to negative infinity, say, U = -100000000000 J. What's the kinetic energy?

You should get K = 100000000100 J.

 

[schroder]

Negative infinity. As in, a really really large negative amount. -0.0000000000001 is far from infinite.

Suppose total energy is E = 100 J. Potential energy is on its way to negative infinity, say, U = -100000000000 J. What's the kinetic energy?

You should get K = 100000000100 J.

Your concept of negative infinity is obviously not the same as my concept. I see it as a number smaller than any other number and when added to any positive number it cannot alter it. As PE approaches negative infinity, the KE approaches the Total energy value. This is the basis for the conservation of mechanical energy in a conservative system.

 

[adriank]

I think your concept of negative infinity is different from that of pretty much everyone else. The only number that when added to any positive number cannot alter it is zero. Calculate various values of U = -Gm²/2r, where r gets close to zero. It won't approach zero; it'll be a negative number that gets very large in magnitude. It's true that if potential energy approaches zero, then kinetic energy approaches total energy. However, that's not what's happening in this situation; as r decreases, potential energy attains a very large negative value, so kinetic energy must attain a very large positive value.

 

[schroder]

I think your concept of negative infinity is different from that of pretty much everyone else. The only number that when added to any positive number cannot alter it is zero. Calculate various values of U = -Gm²/2r, where r gets close to zero. It won't approach zero; it'll be a negative number that gets very large in magnitude. It's true that if potential energy approaches zero, then kinetic energy approaches total energy. However, that's not what's happening in this situation; as r decreases, potential energy attains a very large negative value, so kinetic energy must attain a very large positive value.

It seems we are thinking of the same thing in two different ways, but at least we agree that the Total energy is always conserved. Let me think some more on this and thanks for the integration lesson!

 

[Denton]

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

 

[epenguin]

You do get these infinite energies. You get, inside a finite time, infinite velocities, though only for an infinitesimal time so it all looks OK. What happens when the particles collide depends on physics not included in the formulation so far. If they bounce elastically they will just execute the same motion as before, backwards. Or you can imagine them passing through each other. If the particles are identical mass and indistinguishable both of these scenarios will look the same.

You can find integrals in books and do not need to wait till you have done them in class or someone shows you. Remember you do not need to know how an integral was arrived at to be able to check that it is true.

I worked this one out sometime this year, will give it if no one else does but may be delay due to present position of Earth in orbit.

 

[schroder]

Total energy does not limit kinetic energy. As the masses approach each other, the potential energy
U = -\frac{Gm^2}{2r}
approaches negative infinity. Since total energy is conserved, kinetic energy must approach infinity.

I will make one final attempt to explain about gravitational potential enrgy, even though I expect you will not agree. This may be a bit long, but I hope it is worth your while to read through it with an open mind. First, I believe it is useful to point out where our disagreement is; in other words, let us agree on what points we disagree on.

You are saying that gravitational potential is negative, and is expressed dimensionally as:
-GMm/R
Obviously, as R goes to zero, GPE will go to infinity. (I agree with this)

We also agree that Total energy is conserved. That is, the total energy remains unchanged in any process. We all agree this is true.

In a conservative system, that is, one in which only conservative forces are acting, the total energy is the sum of the PE and KE. When KE is zero, all of the energy is PE and vice versa. This can be seen in a simple pendulum. The gravitational system we are discussing here is a conservative system and we are starting out with zero KE and some fixed amount of PE. For the purpose of this discussion, let us say we are starting out with 100 Joules of PE, which must be equal to the total energy as KE is zero to start with. The number does not matter, as long as we agree it must remain constant throughout at 100 Joules, never more and never less. I think we are in agreement up to this point, but please correct me if I am mistaken.

Now the distance decreases from the starting distance of 1 meter. The PE must increase in Magnitude, according to the math, and is always Negative so it is decreasing. At the same time, the KE must increase, in order for the total energy to remain constant at 100 J.


Now we get into the area where we disagree: According to you, the PE increases to a Negative infinite value, so the KE must increase to a positive infinite value and thay add according to sign convention and the total remains constant.

I disagree right away, because what you are saying means that the KE is going to infinity, which means, according to KE = 1/2 MV^2 that velocity must also be going to infinity. What this clearly says is that the two spheres will come together at infinite velocity and infinite KE!
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!

But I will not base my disagreement on common sense alone. Here is my explanation, and I hope you will take the time to read and try to understand what I am saying. Yes, PE is always negative, and it does increase towards negative infinity as radius goes toward zero. The plot is a hyperbola. But this hyperbola is so not very useful at small values of radius since all PE are very close to negative infinity. It would not be useful at all for calculating the PE of gravity at the surface of the earth, for example. 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. The method of linearization is this:
PE2 – PE1 = - GM/(R + h) – ( - GM/R ) this is the Difference between two potentials where h is the difference in R.

Now we need to do a bit of algebraic simplification: = GM ( 1/R – 1/(R + h)
By LCD we get: = GMh/R(R + h) the final step is to drop the h as it is insignificant to the R.
And we have finally, GMh/R^2 which is gravitational acceleration multiplied by h and is
Gravitational potential and reduces to gh. For GPE just multiply by m and we have our familiar
Mgh. Which is the familiar linearized form of GPE. This is much more useful than using the more general form –GMm/R^2. And, because it is a difference in potential energy,, it is usually treated as a positive number. Although sign convention can be followed if the problem calls for it.

What this all boils down to for the case at hand of two 1 kg spheres at a distance of 1 meter:

g’ = Gmm/R^2 = G = 6.67 E-11 m/s^2
GPE (initial) = mg’h = G = 6.67 E-11 Joules
Total energy = PE (initial) = 6.67 E-11 Joules and is Constant
KE = (Total energy – PE ) at all times
At a distance of zero meters, when the spheres are touching, PE = mg’h and is also zero.
Thus KE is equal to 6.67 E -11 Joules and is shared equally by the two spheres.
Each sphere has 3.335 E-11 Joules of KE which = 1/2 MV^2
So Max possible velocity = 8.167 E-6 m/s which is considerably less than infinity!

The two spheres will pull together very slowly, by your own calculation, in a time of 28 hours.
At no time will velocity be infinite of KE be infinite. The KE cannot exceed the total energy that the system started with!
I sense that as a mathematical purist you may have trouble accepting this, but I assure you that such linearizations are done in physics all the time. I recommend Physics by K.R. Atkins for more such examples.

 

[Doc Al]

I will make one final attempt to explain about gravitational potential enrgy, even though I expect you will not agree. This may be a bit long, but I hope it is worth your while to read through it with an open mind. First, I believe it is useful to point out where our disagreement is; in other words, let us agree on what points we disagree on.

You are saying that gravitational potential is negative, and is expressed dimensionally as:
-GMm/R
Obviously, as R goes to zero, GPE will go to infinity. (I agree with this)

We also agree that Total energy is conserved. That is, the total energy remains unchanged in any process. We all agree this is true.

As r goes to 0, the GPE approaches negative infinity.

In a conservative system, that is, one in which only conservative forces are acting, the total energy is the sum of the PE and KE. When KE is zero, all of the energy is PE and vice versa. This can be seen in a simple pendulum. The gravitational system we are discussing here is a conservative system and we are starting out with zero KE and some fixed amount of PE. For the purpose of this discussion, let us say we are starting out with 100 Joules of PE, which must be equal to the total energy as KE is zero to start with. The number does not matter, as long as we agree it must remain constant throughout at 100 Joules, never more and never less. I think we are in agreement up to this point, but please correct me if I am mistaken.

You would be wise to use a negative number for the initial PE (using r = ∞ as your zero point for PE), and thus the total energy. Then you would be less tempted to associate KE with that total energy.

Now the distance decreases from the starting distance of 1 meter. The PE must increase in Magnitude, according to the math, and is always Negative so it is decreasing. At the same time, the KE must increase, in order for the total energy to remain constant at 100 J.


Now we get into the area where we disagree: According to you, the PE increases to a Negative infinite value, so the KE must increase to a positive infinite value and thay add according to sign convention and the total remains constant.

I disagree right away, because what you are saying means that the KE is going to infinity, which means, according to KE = 1/2 MV^2 that velocity must also be going to infinity. What this clearly says is that the two spheres will come together at infinite velocity and infinite KE!

But it's true! If two point masses under the influence of Newtonian gravity alone approach each other, they will end up going infinitely fast as they approach each other. Realize that this is just an exercise--not realizable in practice.

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!

Obviously, there are other factors that distinguish this case from the ideal case of two colliding point masses: You are not a point mass and other forces are involved. Two spheres coming together is quite different than two point masses colliding.

But I will not base my disagreement on common sense alone. Here is my explanation, and I hope you will take the time to read and try to understand what I am saying. Yes, PE is always negative, and it does increase towards negative infinity as radius goes toward zero. The plot is a hyperbola. But this hyperbola is so not very useful at small values of radius since all PE are very close to negative infinity. It would not be useful at all for calculating the PE of gravity at the surface of the earth, for example. 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. The method of linearization is this:
PE2 – PE1 = - GM/(R + h) – ( - GM/R ) this is the Difference between two potentials where h is the difference in R.

Now we need to do a bit of algebraic simplification: = GM ( 1/R – 1/(R + h)
By LCD we get: = GMh/R(R + h) the final step is to drop the h as it is insignificant to the R.
And we have finally, GMh/R^2 which is gravitational acceleration multiplied by h and is
Gravitational potential and reduces to gh. For GPE just multiply by m and we have our familiar
Mgh. Which is the familiar linearized form of GPE. This is much more useful than using the more general form –GMm/R^2. And, because it is a difference in potential energy,, it is usually treated as a positive number. Although sign convention can be followed if the problem calls for it.

What this all boils down to for the case at hand of two 1 kg spheres at a distance of 1 meter:

g’ = Gmm/R^2 = G = 6.67 E-11 m/s^2
GPE (initial) = mg’h = G = 6.67 E-11 Joules
Total energy = PE (initial) = 6.67 E-11 Joules and is Constant
KE = (Total energy – PE ) at all times
At a distance of zero meters, when the spheres are touching, PE = mg’h and is also zero.
Thus KE is equal to 6.67 E -11 Joules and is shared equally by the two spheres.
Each sphere has 3.335 E-11 Joules of KE which = 1/2 MV^2
So Max possible velocity = 8.167 E-6 m/s which is considerably less than infinity!

Sorry, but this "linearization" is nonsense as applied to the problem at hand. Of course, near the surface of the Earth we "linearize" gravity and use mgh for local changes in GPE.

The two spheres will pull together very slowly, by your own calculation, in a time of 28 hours.
At no time will velocity be infinite of KE be infinite. The KE cannot exceed the total energy that the system started with!

The system started out with negative energy. Are you saying that the KE energy must be negative?

I sense that as a mathematical purist you may have trouble accepting this, but I assure you that such linearizations are done in physics all the time. I recommend Physics by K.R. Atkins for more such examples.

Please scan in the pages where Atkins does this particular example.

 

[schroder]

As r goes to 0, the GPE approaches negative infinity.


You would be wise to use a negative number for the initial PE (using r = ∞ as your zero point for PE), and thus the total energy. Then you would be less tempted to associate KE with that total energy.
But it's true! If two point masses under the influence of Newtonian gravity alone approach each other, they will end up going infinitely fast as they approach each other. Realize that this is just an exercise--not realizable in practice.

Obviously, there are other factors that distinguish this case from the ideal case of two colliding point masses: You are not a point mass and other forces are involved. Two spheres coming together is quite different than two point masses colliding.


Sorry, but this "linearization" is nonsense as applied to the problem at hand. Of course, near the surface of the Earth we "linearize" gravity and use mgh for local changes in GPE.


The system started out with negative energy. Are you saying that the KE energy must be negative?

Please scan in the pages where Atkins does this particular example.

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. The PE and indeed the total energy they possesses is based on that distance and not on infinity. When you calculate the PE of a rock held at 1 meter above the ground, you do not reference it to infinity? The linearization I am using here is valid for a distance of 1 meter. 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!

 

[gabbagabbahey]

<incorrect mathematical and physical reasoning snipped> I am not the one talking nonsense here!

Actually you are.

Maybe putting some numbers into the equations will help you: For simplicity let's use a system of units where G=m_1=m_2=1 and so the potential energy is just U=\frac{-1}{r}...at r=1 let's say that the two masses are at rest so that the kinetic energy K=v^2 of each particle is zero. Then the total energy is initially E=K+U=0+1=1 and since energy is conserved you require that at all times t, you have U(t)+K(t)=1\implies K(t)=1-U(t)

Now at some time after the objects are released, they will be much closer together...let's say at t=1, we have r=\frac{1}{1000000000}=10^{-9} (that is VERY close together). Clearly U(t=1)=-10^9 and so K(t)=1-(-10^9)=10^9+1\approx 10^9. You see as the objects get closer, their kinetic energy gets larger and larger (of course in the real world there are other forces at work which will repel the objects once they get close enough to each other).

 

[schroder]

You are saying that they are accelerating from rest to approaching infinite velocity and infinite kinetic energy, over a distance of one meter. And yet, we have shown it takes 27 hours to cover that distance. I would call what you are saying "nonsense"!

 

[gabbagabbahey]

You are saying that they are accelerating from rest to approaching infinite velocity and infinite kinetic energy, over a distance of one meter. And yet, we have shown it takes 27 hours to cover that distance.

Yes.

I would call what you are saying "nonsense"!

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.

 

[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 <br /> 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:
<br /> \begin{align*}<br /> \int_0^{r_0} \left( \frac{1}{r} - \frac{1}{r_0} \right)^{-1/2} dr<br /> &amp;= \int_{\pi/2}^0 \left( \frac{1}{r_0 \cos^2 \theta} - \frac{1}{r_0} \right)^{-1/2}<br /> \frac{dr}{d\theta} \,d\theta \\<br /> &amp;= r_0^{1/2} \int_{\pi/2}^0 (\sec^2 \theta - 1)^{-1/2}<br /> (-2r_0 \sin \theta \cos \theta) \,d\theta \\<br /> &amp;= r_0^{3/2} \int_0^{\pi/2} 2 (\tan^2 \theta)^{-1/2} \sin<br /> \theta \cos \theta \,d\theta \\<br /> &amp;= r_0^{3/2} \int_0^{\pi/2} 2 \cos^2 \theta \,d\theta \\<br /> &amp;= \frac{\pi}{2} r_0^{3/2}.<br /> \end{align*}<br />

 

[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&#039;} - \frac{1}{r_0} \right)^{-1/2} dr&#039; = dt&#039;,

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&#039;} - \frac{1}{r_0} \right)^{-\frac12} \,dr&#039; = \int_0^t dt&#039; = t.

I make the substitution r&#039; = r_0 \cos^2 \theta, so \theta = \arccos \sqrt{r&#039;/r_0}:

\begin{align*}<br /> t &amp;= \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<br /> \\<br /> &amp;= \sqrt{\frac{2r_0^3}{Gm}} \int_{0}^{\arccos \sqrt{r/r_0}} 2 \cos^2 \theta \,d\theta<br /> \\<br /> &amp;= \sqrt{\frac{2r_0^3}{Gm}} \left[ \theta + \sin \theta \cos \theta \right]_{0}^{\arccos \sqrt{r/r_0}}<br /> \\<br /> &amp;= \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).<br /> \end{align*}

I can differentiate this to obtain the velocity:

\begin{align*}<br /> \frac{1}{v} &amp;= \frac{dt}{dr} = -\sqrt{\frac{2r}{Gm (1 - r/r_0)}}<br /> \\<br /> v &amp;= -\sqrt{\frac{Gm}{2r} \left( 1 - \frac{r}{r_0} \right)} = -\sqrt{\frac{Gm}{2r_0} \left( \frac{r_0 - r}{r} \right)}.<br /> \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*}<br /> E &amp;= U + K = -\frac{Gm^2}{2r} + 2 \left( \frac12 mv^2 \right)<br /> \\<br /> &amp;= -\frac{Gm^2}{2r} + m \frac{Gm}{2r_0} \left( \frac{r_0 - r}{r} \right)<br /> \\<br /> &amp;= -\frac{Gm^2}{2r} + \frac{Gm^2}{2r} - \frac{Gm^2}{2r_0}<br /> \\<br /> &amp;= -\frac{Gm^2}{2r_0}.<br /> \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!