Rate of Attraction

[Robittybob1]

 

Not weird at all. Look at Janus' equation - all that is varying is the mass. The ratio of the times is the ratio of the roots of the masses

 

sqrt(2)/sqrt(1.05) = 1.38013

 

I had checked it wasn't the sqrt(2) and left it at that. It is good to know that, thanks. Two objects will fall with a period "The ratio of the times is the ratio of the roots of the masses". This might be applicable situations of colliding galaxies or collapsing nebulae.

[StringJunky]

 

 

No it is not correct. The lighter ball accelerates at the same speed as a heavy ball would in the same position; but the light ball attracts its partner heavy ball proportionally less. The closing time is higher.

 

This is the weak equivalence principle

But by making ball A light you are changing the gravitational field that ball B experiences - it is no longer constant and you cannot use the above principle.

How is it then that a hammer and feather dropped on the Moon from the same height will strike the surface at the same time? Is it because they are part of the same system sharing the Moon's gravity?

Edited June 23, 2015 by StringJunky

[swansont]

How is it then that a hammer and feather dropped on the Moon from the same height will strike the surface at the same time? Is it because they are part of the same system sharing the Moon's gravity?

 

If you did the experiments separately, i.e. timed the fall to the point of contact, the answer would be different by an amount so small that it probably couldn't be measured. The assumption is that the moon remains at rest, which is a very good assumption (the mass of the moon being ~ 7 x 10²² kg), within the precision of the measurement.

 

Let's say you drop a 1 kg object for 1 sec, and it falls a distance of about 0.8 m. The moon's acceleration will be 1.4 x 10^-23 of the object's, so it will have moved around 10^-23 m. A 1 g object will cause the moon to move 10^-26 m.

[StringJunky]

 

If you did the experiments separately, i.e. timed the fall to the point of contact, the answer would be different by an amount so small that it probably couldn't be measured. The assumption is that the moon remains at rest, which is a very good assumption (the mass of the moon being ~ 7 x 10²² kg), within the precision of the measurement.

 

Let's say you drop a 1 kg object for 1 sec, and it falls a distance of about 0.8 m. The moon's acceleration will be 1.4 x 10^-23 of the object's, so it will have moved around 10^-23 m. A 1 g object will cause the moon to move 10^-26 m.

Right. The effect is there but it's too small

[md65536]

 

If you did the experiments separately, i.e. timed the fall to the point of contact, the answer would be different by an amount so small that it probably couldn't be measured. The assumption is that the moon remains at rest, which is a very good assumption (the mass of the moon being ~ 7 x 10²² kg), within the precision of the measurement.

 

There was a similar thread here a few years ago.

 

If you completely separate the two tests, so that you remove the hammer from the system when you're measuring the feather's drop, then the feather/moon collision will take negligibly more time than the hammer/moon collision would. However if you leave the hammer or feather on the moon when you drop the other, then the total mass of the system is the same in each drop and the closing time would be the same.

 

This MIGHT (I can't seem to decide) assume that you can still treat the moon+hammer and moon+feather as a uniform balls, and of equal size. The difference in mass is so slight that you couldn't assume that, and details such as the location of the hammer would become significant.

Edited June 23, 2015 by md65536

[Robittybob1]

When the hammer and the feather are dropped together the Moon is not attracted to each of them separately but to the combined dropped mass. Therefor the very small amount the Moon moves is one distance not two distances, so the hammer and the feather fall at the same speed, not at two speeds with an insignificant time difference because of distance.

Edited June 23, 2015 by Robittybob1

[imatfaal]

 

I had checked it wasn't the sqrt(2) and left it at that. It is good to know that, thanks. Two objects will fall with a period "The ratio of the times is the ratio of the roots of the masses". This might be applicable situations of colliding galaxies or collapsing nebulae.

 

Your summary is a bit misleading as it over simplifies. This is a model for two systems of two objects which only vary by some of the objects being a different mass. How this would apply to collapsing nebula is beyond me - it is a result of a highly constrained and artificial system; the ratio flows directly from the equation Janus provided.

 

Have another look at it; the ratio of the times t_a/t_b must equal the RHS of Janus' equation for System a over the RHS of Janus' equation for System b. If you do this in your head you will see that the bulk of the RHS does not vary with mass and is thus equal for System a and System b - ie algebraically you can just cancel - that leaves you with the sqrt[2G(M_b1+m_b2)] / sqrt[2G(M_a1+m_a2)], you can cancel out the 2G which leaves you with the ratio above.

[Robittybob1]

 

Your summary is a bit misleading as it over simplifies. This is a model for two systems of two objects which only vary by some of the objects being a different mass. How this would apply to collapsing nebula is beyond me - it is a result of a highly constrained and artificial system; the ratio flows directly from the equation Janus provided.

 

Have another look at it; the ratio of the times t_a/t_b must equal the RHS of Janus' equation for System a over the RHS of Janus' equation for System b. If you do this in your head you will see that the bulk of the RHS does not vary with mass and is thus equal for System a and System b - ie algebraically you can just cancel - that leaves you with the sqrt[2G(M_b1+m_b2)] / sqrt[2G(M_a1+m_a2)], you can cancel out the 2G which leaves you with the ratio above.

Could two nebulae fall toward each other? You could just give each one a specified spherical radius and compare the mass within that sphere. They would "fall" toward each other in proportional to the sqrt of their combined mass.

 

I don't know if it would be useful at all, for I was at a loss as to think what else this knowledge could be used for.

[DimaMazin]

 

If you did the experiments separately, i.e. timed the fall to the point of contact, the answer would be different by an amount so small that it probably couldn't be measured. The assumption is that the moon remains at rest, which is a very good assumption (the mass of the moon being ~ 7 x 10²² kg), within the precision of the measurement.

 

Let's say you drop a 1 kg object for 1 sec, and it falls a distance of about 0.8 m. The moon's acceleration will be 1.4 x 10^-23 of the object's, so it will have moved around 10^-23 m. A 1 g object will cause the moon to move 10^-26 m.

Big distance or big distinction of masses A and D creates big distinction of fall times. Take dark particle for example.

[Mordred]

Why would any particle species matter in a GR problem set of mass/ volume differential problem- set matter?

The problem covered by the OP and the formula provided by Janus is one of mass vs volume vs distance of travel/geometry.

 

All the mathematics has involved strictly the above.... RIGHTLY SO

 

Particle species outside don't matter except thermodynamic influences apply, (stress energy, pressure influence- via the Einstein field equations)

Edited June 24, 2015 by Mordred

[imatfaal]

Could two nebulae fall toward each other? You could just give each one a specified spherical radius and compare the mass within that sphere. They would "fall" toward each other in proportional to the sqrt of their combined mass.

 

NO!

 

A pair of nebulae would fall together in line with the equation that Janus gave or the earlier one provided from wikipedia. Look at either of those and you will note where the mass is! It is in the bottom of the fraction not the top.

[StringJunky]

Why would any particle species matter in a GR problem set of mass/ volume differential problem- set matter?

The problem covered by the OP and the formula provided by Janus is one of mass vs volume vs distance of travel/geometry.

 

All the mathematics has involved strictly the above.... RIGHTLY SO

 

Particle species outside don't matter except thermodynamic influences apply, (stress energy, pressure influence- via the Einstein field equations)

Yes, the whole idea of asking the question was to minimise the number of parameters.

[Robittybob1]

The "fall time" for two bodies such as in your scenario is found by

 

[math]t = \frac{ \arccos \left( \sqrt{ \frac{x}{r} }\right) + \sqrt {\frac{x}{r} \left ( 1 - \frac{x}{r} \right ) } }{ \sqrt{ 2 \mu } } r^{3/2}[/math]

 

Where r is the starting distance between the centers of the bodies, x is the ending distance and [math]\mu = G(M_1+M_2)[/math].

 

...

The mass component is the denominator when one looks to the period, but period is inverse to speed. So two objects will fall faster with higher combined mass, OK it then takes a shorter time for it to happen.

 

NO!

 

A pair of nebulae would fall together in line with the equation that Janus gave or the earlier one provided from wikipedia. Look at either of those and you will note where the mass is! It is in the bottom of the fraction not the top.

The two nebulae will combine faster if their combined mass is greater. I think this is correct. The period will be smaller with higher mass for they will move faster toward each other.

(This was not directly proportional but in proportion to the sqrt of the combined mass.)

Edited June 24, 2015 by Robittybob1