A question of physics

The Son of Him · Jul 3, 2004

Aeschylus said:
Galileo pre-dates Newton's law universal gravitation, so he would of had no way of knowing he was off by a tiny amount, that said his answer gives a good approximation as you're ever going to need.

I think you are not aware that Newton based himself in the fact that bodies fall at the same rate like GALILEO demonstrated.
Even Newton's gravitational law springs from that fact.

Aeschylus · Jul 3, 2004

Yes they do spring from the fcathat g is independnt of the mass of the object, but wha it also says is that all objects with exert the same (obviously quantitively different) attractive force that the Earth does, when you take this into consideration the time periods become different which is what this thread is about.

The Son of Him · Jul 3, 2004

Aeschylus said:
Yes they do spring from the fcathat g is independnt of the mass of the object, but wha it also says is that all objects with exert the same (obviously quantitively different) attractive force that the Earth does, when you take this into consideration the time periods become different which is what this thread is about.

There is a problem to that logic , and is:
If the time periods were different g will not be independent of the mass of the object becuase according to you it will seem as if heavier objects have a greater value of g because they would come in contact with earth faster than lighter objects and therefore g will lose its independence !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Aeschylus · Jul 3, 2004

The Son of Him said:
There is a problem to that logic , and is:
If the time periods were different g will not be independent of the mass of the object becuase according to you it will seem as if heavier objects have a greater value of g because they would come in contact with earth faster than lighter objects and therefore g will lose its independence !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

'g' is indpednt of mass but the accelartion of the Earth IS dependnt on the mass of the object, we can almsot ignore 'g' here (thoguh not quite), it's the acceleration of the Earth that's important here (well it's not important in real life, but it's crucial to the issue that's being discussed.

The Son of Him · Jul 3, 2004

Aeschylus said:
'g' is indpednt of mass but the accelartion of the Earth IS dependnt on the mass of the object, we can almsot ignore 'g' here (thoguh not quite), it's the acceleration of the Earth that's important here (well it's not important in real life, but it's crucial to the issue that's being discussed.

I give up my friend , according to you NASA is wrong, Galileo was wrong,
you use equations as you please . Good LUCK my friend .
But it was very enjoyable to discuss the matter with you sincerely. Take Care

Aeschylus · Jul 3, 2004

The Son of Him said:
I give up my friend , according to you NASA is wrong, Galileo was wrong,
you use equations as you please . Good LUCK my friend .
But it was very enjoyable to discuss the matter with you sincerely. Take Care

Just one last thing this isn't a crackpot thopery it's an oft over-looked aspect of Newton's law of unievrsal gravitation. Galileo and Nasa are right within certain limits.

Chi_Cygni · Jul 3, 2004

I just checked your answer and your method gives an approximation to the exact solution due to the fact you average the acceleration. In the limit of imagining the attraction of two point masses your solution gives the wrong result.

But on checking the error is only a factor of Pi/2 wrong. And your method does give the correct behaviour of the problem.

I don't know if I'll bother putting the full solution on a separate thread. I solved the problem starting off with the equation of motion for a central force orbit and that the bodies (Earth and mass) were static.

You can solve the equation of motion derived from the Lagrangian analytically from this to give the time for collision of the Earth and mass. You then difference this with the result for a different test mass.

As I said this is really the same thing you did since you can derive Lagranges equations from D'alemberts principle. The reason for the slight difference in the end result is that I carried through the integrations whereas you approximated by calculating an average acceleration.

Applying D'Alemberts principle really doesn't help or hinder in this problem.

Aeschylus · Jul 3, 2004

Chi_Cygni said:
I just checked your answer and your method gives an approximation to the exact solution due to the fact you average the acceleration. In the limit of imagining the attraction of two point masses your solution gives the wrong result.

But on checking the error is only a factor of Pi/2 wrong. And your method does give the correct behaviour of the problem.

I don't know if I'll bother putting the full solution on a separate thread. I solved the problem starting off with the equation of motion for a central force orbit and that the bodies (Earth and mass) were static.

You can solve the equation of motion derived from the Lagrangian analytically from this to give the time for collision of the Earth and mass. You then difference this with the result for a different test mass.

As I said this is really the same thing you did since you can derive Lagranges equations from D'alemberts principle. The reason for the slight difference in the end result is that I carried through the integrations whereas you approximated by calculating an average acceleration.

Applying D'Alemberts principle really doesn't help or hinder in this problem.

I'd like to see your solution, because the approach you describe should give indisputably the right answer, but you're incorrect about the average acceleration, as it's the mean average it should give the exact answer.

Chi_Cygni · Jul 3, 2004

Aeschylus do me a favour.

Give me the resultant time for collision for two point masses of M and m starting off at rest of separation 2R using your method.

It's basically what you have done and setting h = R.

Aeschylus · Jul 3, 2004

sqrt[8R^3/G(M+m)]

Chi_Cygni · Jul 3, 2004

For your method I get

t= 2*SQRT[R^3/G(M+m)]

The exact answer is

t= (Pi/2*SQRT(2)) * SQRT[R^3/G(M+m)]

That is you are off by a factor Pi/2 SQRT(2) according to me using my result for you or Pi/8 using your result.

You have to have a Pi in the answer because you are integrating an elliptical integral of the second kind.

Aeschylus · Jul 3, 2004

ah-ha, I might of seen a mistake in my orginal solution, I think I may of ignored an important factor in the nature of my equation for accelartion (which I know is definitely correct).

Chi_Cygni · Jul 3, 2004

Quickie version of full solution.

V(r) = -GMm/r (Potential energy)

T(r) = 1/2 * (Mm/M+m)*[rdot^2 + r^2 thetadot^2] (Kinetic energy - dot = differential with respect to time)

therefore

this leads to equation of motion

rdoubledot = r thetadot^2 - G(M+m)/r^2

Now if bodies in circulr orbits of radius R then Keplers 3rd Law gives

R= (G(M+m)T^2/4 Pi^2)^1/3 since thetadot = 2 Pi/T

Now for static masses then thetadot tends to zero therefore get

rdoubledot = -G(M+m)/r^2

Multiply both sides by integrating factor 2 *r*rdot and eventually we get

rdot = SQRT[(2G(M+m))]*SQRT((R-r/R*r))

Invert and solve for t(r)

get t = SQRT[(1/2G(M+m))]* Integral [ SQRT(R*r/(R-r)) from R to 0]

This leads to:

t= SQRT[R^3/2G(M+m)] *Pi/4

but this R is not the separation of the particles which is 2R

therefore for separation initially of R get

t= SQRT[R^3/G(M+m)] * Pi/(2*SQRT(2))

Chi_Cygni · Jul 3, 2004

The problem for a separation of just h the height above the Earth is that the integral is a ***** to do. But it is the correct answer.

This problem is a common on in grad astronomy courses where you calculate the limit of collision for point masses and you get a nice analytiical result as opposed to the result for just dropping the masses a distance h.

Chi_Cygni · Jul 4, 2004

For the following data :

height dropped from = 1000m

mass (1) = 20 Kg mass (2) = 10 Kg

I get the time difference is approx. 18.4 picoseconds. (1.84 X 10^-11 seconds).

This is a unbelievably small difference especially when you realise the fall was a 1km drop.

For a more realistic fall of 10m the difference is approx. 1.84 X 10^-12 seconds.

Aeschylus · Jul 4, 2004

Yep I thought about it more, my solution ignored a very impotant and screamingly obvious fact, that though:

a = -G(M + m)/r^2

Is the 'correct' equation to treat the problem pseudo-statically it's actually a differential equation which means that I cannot expect to intergrate with respect to r and get the correct solution.

Data · Jul 4, 2004

This thread is quite funny actually

Chi_Cygni · Jul 4, 2004

Aeschylus said:
Yep I thought about it more, my solution ignored a very impotant and screamingly obvious fact, that though:

a = -G(M + m)/r^2

Is the 'correct' equation to treat the problem pseudo-statically it's actually a differential equation which means that I cannot expect to intergrate with respect to r and get the correct solution.

Exactly. Thats what I mentioned in my post earlier.

This whole problem isn't particularly complicated but has subtleties that are easy to overlook.

Dragar · Jul 4, 2004

So, since I skimmed this entire thread and got very confused, is the answer basically that if we pretend the Earth does not move when we drop the small mass from a great height, objects of different masses will fall at the same rate (ignoring air resistance)?

But, if we take into account the fact that the tiny mass moves the Earth toward it by a tiny amount (normally ignored in such problems), we find that the object with a larger mass takes a tiny, tiny amount of less time to hit the Earth?

Dragar

Chi_Cygni · Jul 4, 2004

Correct.

A question of physics

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