
#19
Dec2208, 12:26 AM

P: 534

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



#20
Dec2208, 01:00 AM

P: 373





#21
Dec2208, 01:11 AM

P: 373





#22
Dec2208, 01:27 AM

P: 534

[tex]K = E  U = E + \frac{Gm^2}{2r}[/tex] 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. 



#23
Dec2208, 01:33 AM

P: 373

According to me 0.0000000000001 + 1 = 1 which is correct? 



#24
Dec2208, 01:40 AM

P: 534

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. 



#25
Dec2208, 01:48 AM

P: 373





#26
Dec2208, 01:52 AM

P: 534

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 = Gm^{2}/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.




#27
Dec2208, 02:04 AM

P: 373





#28
Dec2208, 05:58 AM

P: 114

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



#29
Dec2208, 08:53 AM

HW Helper
P: 1,932

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. 



#30
Dec2208, 09:16 AM

P: 373

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 E11 m/s^2 GPE (initial) = mg’h = G = 6.67 E11 Joules Total energy = PE (initial) = 6.67 E11 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 E11 Joules of KE which = 1/2 MV^2 So Max possible velocity = 8.167 E6 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. 



#31
Dec2208, 03:14 PM

Mentor
P: 40,878





#32
Dec2208, 05:18 PM

P: 373





#33
Dec2208, 05:49 PM

HW Helper
P: 5,004

Maybe putting some numbers into the equations will help you: For simplicity let's use a system of units where [itex]G=m_1=m_2=1[/itex] and so the potential energy is just [tex]U=\frac{1}{r}[/tex]....at [itex]r=1[/itex] let's say that the two masses are at rest so that the kinetic energy [itex]K=v^2[/itex] of each particle is zero. Then the total energy is initially [itex]E=K+U=0+1=1[/itex] and since energy is conserved you require that at all times [itex]t[/itex], you have [itex]U(t)+K(t)=1\implies K(t)=1U(t)[/itex] Now at some time after the objects are released, they will be much closer together...let's say at [itex]t=1[/itex], we have [itex]r=\frac{1}{1000000000}=10^{9}[/itex] (that is VERY close together). Clearly [itex]U(t=1)=10^9[/itex] and so [itex]K(t)=1(10^9)=10^9+1\approx 10^9[/itex]. 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). 



#34
Dec2208, 05:59 PM

P: 373

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"!




#35
Dec2208, 06:09 PM

HW Helper
P: 5,004

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. 



#36
Dec2208, 06:30 PM

Mentor
P: 40,878

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! 


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