Gravity Revisited: Earth-Moon & Earth-Sun Comparisons

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The discussion revolves around the implications of Newton's Law of Gravity in hypothetical scenarios involving the Earth, Moon, and Sun. It explores how the gravitational forces between these bodies differ due to their varying masses, leading to different accelerations and velocities when they are not in orbit. The conversation highlights the confusion surrounding the universality of free fall, particularly in experiments where two objects of different masses fall simultaneously. Participants debate whether the differences in impact times between objects of different masses are significant or negligible, with some asserting that Newtonian physics may not fully account for these nuances. Ultimately, the thread emphasizes the complexities of gravitational interactions and the importance of frame of reference in understanding these phenomena.
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I am having some problems working this one out:
If the Earth and the moon were not in any orbit (and assume there are no other celestial bodies) then Newton's Law of Gravity would have them accelerating towards one another at some value X.
If the Earth and the sun were not in any orbit (and assume no other celestial bodies) then would they accelerate towards one another at the same value X?
 
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Newton's law of gravity says that the forces would be the same. Since they have different masses, the accelerations would be different since F=ma.
 
As part of my hypothetical, I forgot to include that the Earth and sun were separated by the same distance as the Earth and the moon.
What I am getting at is the following: according to Newton’s law of gravity, the gravitational force between two bodies is a function of their masses and their distance apart. The gravitational constant we have been told is a universal constant.
In my hypothetical, if the distance between the Earth and sun is the same between the Earth and the moon, then the gravitational force between the Earth and the sun would be greater than the gravitational force between the Earth and the moon (the product of the two masses).
If the above argument is correct, consider the forces and accelerations between say the Earth and the sun. For equilibrium to work, the force on each of the two bodies must be the same. If we now consider F = ma for the two bodies, the forces are the same, the masses are different, therefore the accelerations must be different so that their product is equal.
If the accelerations are different so are their velocities. So the Earth must travel faster towards the sun than the sun would travel towards the earth.
 
If all of the above makes sense, the accelerations of the two bodies are different. So are their velocities.

If this is the case, consider the Galileo experiment. The two masses say 1 kg and 2 kg, falling from the same height on Earth will NOT land at the same time because they will be traveling at different velocities (neglecting friction)!
 
My first comment is you know Newtonian physics at least at it's heart is wrong ... correct?
There are no forces they are fictional mathematical forces to make the mathematics work.

So let's see if we can make this all fly under Newtonian law.

Okay first there isn't 1 gravity force acting on each body there is 2 and they need to be vector summed.

Sun sees Earth and Moon
Moon sees Sun and Earth
Earth sees Sun and Moon

So you need to calculate each force record it's direction as if the force was acting at the mass centre. I think that will fix the Newtonian physics up to work.

I think what that will do is give you an average gravity force of the 3 bodies and the two lighter bodies will fall at the same rate .. let's hope so :-)
 
Apparently I have not made myself clear - apologies to those that have commented or have read my comments.
I will start at the end point this time: when two objects of different mass, say 1 kg and 2 kg, fall towards the Earth (neglecting air friction) they will NOT land at the same time. The 2 kg will land first. The difference will be so small that it cannot be observed. This can be proven if we use Newtons Laws.
 
Restructure said:
If all of the above makes sense, the accelerations of the two bodies are different. So are their velocities.

If this is the case, consider the Galileo experiment. The two masses say 1 kg and 2 kg, falling from the same height on Earth will NOT land at the same time because they will be traveling at different velocities (neglecting friction)!

Really? And you can prove this?
You would find that, absence of friction, a 10-ton mass will impact the Earth at exactly the same time as a penny from the same height.

Try it at home! Use a penny and some heavy object(bowling ball, etc...) at a 4-foot height.
Which hits the ground first?

They both do.
 
Difficult to observe the difference. Possibly less than a nanosecond
 
Ah, my favorite discussion again. Restructure is correct. The time to impact for the 2kg mass is less than the time to impact for the 1kg mass (even if it cannot be detected). But this has nothing to do with the universality of free fall. The accelerations of the two masses in the UFF is relative to the CoM frame of reference of the Earth and the falling body. Whereas the time to impact is determined by the relative accelerations, where the frame of reference is either the Earth or the falling body.

But DrGreg can explain it better than I can:
https://www.physicsforums.com/showpost.php?p=3276671&postcount=53
 
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  • #10
TurtleMeister said:
The time to impact for the 2kg mass is less than the time to impact for the 1kg mass (even if it cannot be detected)

What??
Please offer proof.

Furthermore, if, as you say, "it cannot be detected" what is a responsible person to make of this VERY APPARENT contradiction?

Mods; please chime in on this...
 
  • #11
And that answer is correct.

Newton had no basis to realize there were objects much heavier than Earth he never really thought about objects with significant mass greater than earth.

F = MA as you gave in you calculation is also an approximation you realize for similar reasons provided the accelerations are not extreme.

What are you trying to proove Newtonian physics is wrong .. we already know that!
 
  • #12
pallidin said:
TurtleMeister said:
The time to impact for the 2kg mass is less than the time to impact for the 1kg mass (even if it cannot be detected)
What??
Please offer proof.

Furthermore, if, as you say, "it cannot be detected" what is a responsible person to make of this VERY APPARENT contradiction?

Mods; please chime in on this...

Did you look at the link I provided? The UFF is about accelerations relative to the CoM of the two bodies. It is not about the time to impact. The time to impact is determined by the relative acceleration. Yes, the UFF is sometimes demonstrated as the time to impact. But even if it is unstated, it carries the caveat that the masses of the falling bodies are negligible to the mass of the earth. In your example of the 10-ton mass and the penny, try changing the mass of the 10-ton object to the mass of the moon, but keep its physical size the same. Will it's time to impact be the same as the penny?
 
  • #13
Thank you for the reference to DrGreg. It helps address most of my physics/mathematical problem. Not all of it however.

UFF exists because of emperical/observational data. From a mathematical/physics perspective it is incorrect.
 
  • #14
The whole situation you describe is ridiculous

In your example of the 10-ton mass and the penny, try changing the mass of the 10-ton object to the mass of the moon, but keep its physical size the same. Will it's time to impact be the same as the penny?

If you had a moon like mass held off the Earth you would have to consider it a totally different

1.) The mass of the moon added to Earth and therefore the Earth's gravity would be a hell of a lot bigger. Remember even under Newtonian law the total mass acts as gravity from the centre point.

2.) In that situation the two should should fall towards the larger body Earth the same as viewed by Newtonian law.

This idea you can have a mass not considered to be part of Earth's mass and yet drop it is where the falacy lies.
 
  • #15
I understand now the confusion. This is what I think is happening:

In the two-body problem (the Sun and the Earth), the force between them is the same. If we stand somewhere out in the rest frame of space, we see the Sun not accelerating as fast toward the Earth as the Earth is toward the Sun (this is of course neglecting orbit and treating them purely as "stationary" objects) in accordance with F=ma. However, standing on one of the objects, we would see the other body accelerating toward us at the exact same magnitude, no matter if we are standing on the Sun (!) or the Earth.

In Galileo's thought experiment, we are considering a problem involving relative accelerations and three objects. We are standing on the Earth and dropping two objects and measuring how fast they each accelerate toward the third (the Earth). If we use Newton's law of universal gravitation and the 2nd law, we get:

m1*a = G*m1*m_e/r^2
a = G*m_e/r^2

where m_e is the mass of the Earth and m1 is the mass of the 1st object. The results are the same with the 2nd object. That shows that the Earth accelerates everything toward itself with the same acceleration, thus proving the Law of Falling Bodies because the acceleration is independent of the mass of the object.

You could also see this result if you stand out in space somewhere or another suitable rest frame where you can consider the three bodies independently. The gravitational force between the Earth and object 1 would be F1=G*m_e*m1/r^2 and between the Earth and object 2 is would be F2 = G*m_e*m2/r^2. If the r's are the same and m1>m2, then F1>F2 since G, r and m_e are constants. This makes sense because if we stand on the Earth and see the objects accelerating towards us, more force is required to move m1 than m2 to make their accelerations equal.

I hope this explanation made sense.
 
  • #16
The above makes sense except that the acceleration of m1 and the corresponding acceleration of the Earth are not the same as the acceleration of say m2 and the corresponding acceleration of the earth. The mass ratios are different.
 
  • #17
Uglybb said:
1.) The mass of the moon added to Earth and therefore the Earth's gravity would be a hell of a lot bigger. Remember even under Newtonian law the total mass acts as gravity from the centre point.
If I'm understanding you correctly, Yes. I call that the center of mass (CoM).
Uglybb said:
2.) In that situation the two should should fall towards the larger body Earth the same as viewed by Newtonian law.
Are you are thinking about this in terms of the three body problem? That is difficult to do with simple math and thought experiments. It usually requires numerical analysis and computer simulation. Instead you should think in terms of the two body problem, which is what I was doing. Consider the gravitational interaction between the Earth and object 1, then consider the gravitational interaction between the Earth and object 2. Compare the accelerations and impact times.

It's amazing how this topic can bring out so many misunderstandings and misconceptions. And that includes myself. But I always enjoy it, even when I'm wrong. I've been through this before at PF, and I've put a lot of thought into it. I have a high degree of confidence that my understanding of this topic is correct. But I'm always willing to learn more. At this point I stand firm with my previous posts and DrGreg's post that I linked to in post #9.
 
  • #18
The whole argument gets convoluted one minute we are talking about three objects in space and the next minute we are talking about dropping two things on earth.

Newtonian physics, GR whichever your choose there are big differences between those two cases simply because you frames of reference change (one you are standing on Earth and the other you are standing in space). Neither frame is more correct than any other frame and observations may differ.

Thats all my comment is the two seem to get muddled up in this whole thread.

The one I really loved was the nuetron star being held off the Earth and it didn't fall at the same rate to Earth ... LMAO ... really no kidding ! Might try putting the frame of reference on the largest mass for Newtonian physics.
 
  • #19
Restructure is correct in that a 2kg mass will reach the Earth a tiny fracton of a moment earlier than a 1kg mass (assuming the experiments are done separately so that they do not interfere.)

Or more accurately, it is the Earth that will reach the 2kg mass soone than the Earth will reach the 1kg mass. :biggrin:



The 2kg mass will accelerate the Earth faster than the 1kg mass will, closing the distance sooner by an effectively immeasurable amount.

The point here is that Galileo's experiment assumes a neglible mass i.e. disparity in masses of the two objects (Earth and cannonball) is so large that the ball's mass can be considered zero (on the scale of 10^24, 10^0 is pretty much zero) so that Earth's own acceleration is effectively zero.

When the masses become more comparable, it makes mroe sense that you must consider both masses, and the disparity in closing times when using different 2ndary masses is more obvious.
 
  • #20
And I have problem with that dave.

Then they bring in a nuetron star and the moon we set our reference frame on Earth and we drop the objects and they act like the object is going to drop in a straight line to earth.

You know it isn't it's going to take a weird path that is something between the two gravities and they say ohh the Earth gets lifted towards the bigger mass a bit so it changes the time they don't discuss the other mass doesn't exactly drop in a straight line either it has attraction to the 3rd mass.

The references are all over the place and they selectively choose to move them around to support there argument.
 
  • #21
Uglybb said:
And I have problem with that dave.

Then they bring in a nuetron star and the moon we set our reference frame on Earth and we drop the objects and they act like the object is going to drop in a straight line to earth.

You know it isn't it's going to take a weird path that is something between the two gravities and they say ohh the Earth gets lifted towards the bigger mass a bit so it changes the time they don't discuss the other mass doesn't exactly drop in a straight line either it has attraction to the 3rd mass.

The references are all over the place and they selectively choose to move them around to support there argument.

Then simplify it.

Two experiments.
Earth and m1.
Earth and m2.

OK, three.

Earth and m2^20.

It's pretty straightforward.
 
  • #22
Exactly and you need to put positions on things.

one of the links we have a person holding a nuetron star in one hand and and a penny in the other and then this poster (restructure)has two masses equistant from a 3rd.

I laughed at the nuetron one with the Newtonian analysis because according to them the nuetron star fell to Earth while moving the Earth upwards a bit but the penny fell directly to earth, apparently it has no attraction to the nuetron star.

The flaw I am seeing over and over again in the analysis is they work m1 to earth, m2 to Earth they forget there is a relationship between m1 to m2. Now when we have a penny and cannon ball m1 to m2 reaction is incredibly tiny you can ignore it but you can't when you have a nuetron star or moon.
 
  • #23
Uglybb said:
I laughed at the nuetron one with the Newtonian analysis because according to them the nuetron star fell to Earth while moving the Earth upwards a bit but the penny fell directly to earth, apparently it has no attraction to the nuetron star.

The flaw I am seeing over and over again in the analysis is they work m1 to earth, m2 to Earth they forget there is a relationship between m1 to m2. Now when we have a penny and cannon ball m1 to m2 reaction is incredibly tiny you can ignore it but you can't when you have a nuetron star or moon.
Sorry, call me dense. Can you point me to any mention of a neutron star in this discussion?


[ EDIT] Oh. It's under the link. OK well, I'm only prepared to argue one thread at a time.
 
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  • #24
DaveC426913 said:
Restructure is correct in that a 2kg mass will reach the Earth a tiny fracton of a moment earlier than a 1kg mass (assuming the experiments are done separately so that they do not interfere.)

But that assumption contradicts the OP. He said "if you do the Galileo experiment," which implies two masses that are negligible compared with the earth, dropped from no more than arm's length apart, at the same time. Under those conditions, I'd guess that any effect of the Earth being slightly more attracted to the larger mass would be less than the effects of slight imperfections in the landing area, and the inability to release the objects at precisely the same instant.
 
  • #25
brocks said:
But that assumption contradicts the OP. He said "if you do the Galileo experiment," which implies two masses that are negligible compared with the earth, dropped from no more than arm's length apart, at the same time.

I would ask the OP for clarification on that point then. He may not have realized that, if the two experiments are done simultaneously, rather than discretely, it will have different results.
 
  • #26
Yes that's the key.

If they are done together they need to account for the m1<->m2 attraction as well and only if that force is really small will the fall actually look anything like a straight line. And I am guessing it will therefore change there results to what we expect.
 
  • #27
Uglybb said:
Yes that's the key.

If they are done together they need to account for the m1<->m2 attraction as well and only if that force is really small will the fall actually look anything like a straight line. And I am guessing it will therefore change there results to what we expect.

Still don't know what the straightness of the line has to do with anything.
 
  • #28
DaveC426913 said:
Still don't know what the straightness of the line has to do with anything.

I think he has a point. If only one mass is falling, it should go straight down, perpendicular to the ground. If two masses are falling side by side, they should attract each other, so their paths will be slightly off perpendicular, and the smaller mass will deviate more than the larger mass, and therefore have a slightly longer path to the ground, and therefore take slightly longer to hit it.

Without doing any calculations, I expect that this effect would be larger than the one discussed in previous posts.
 
  • #29
I think the issue is being over-complicated.

Consider 2 masses m1 and m2 such that m1 < m2

Also consider some arbitrary distance from earth, r.

If either m1 or m2 is placed at r, they will experience the same acceleration at that point, as explained by timthereaper's post.

However, they will hit Earth at different times, as m2 will accelerate Earth more than m1 will, and so Earth will come in contact with m2 more quickly.

I think that satisfies the OP's question, and we don't need to consider an interaction between m1 and m2.
 
  • #30
Agree with Morberticus.

Unless the OP specifically wants to discuss that 3-body scenario, it's needlessly complicated.
 
  • #31
Yes, I agree with Morberticus and Dave. You're basically saying the same thing I said in post #17. I don't even think the OP was even considering anything more than a two body problem. Using more than two bodies only serves to complicate the understanding of the universality of free fall and the equivalence principle. The EP has not been mentioned in this thread. But I think it is important to bring it up because the UFF and EP go hand in hand. I will use the following illustration:

uff_ep.gif


The dot located between m1 and m2 is the CoM. Notice that it is in the center an equal distance from m1 and m2. This tells us that the mass of m1 and m2 are equal. I cannot do animation here so you'll have to visualize the following.
Lets increase the mass of m2. Two things will happen. 1)The increase in inertial mass of m2 will cause the CoM to move to a position closer to m2. 2)The increase in gravitational mass of m2 will cause the relative acceleration of m1 and m2 toward each other to increase. Now, the equivalence principle tells us that gravitational mass is proportionally equal to inertial mass. So the shorter distance that m2 has to travel to reach the CoM (as a result of the increase in inertial mass of m2) exactly offsets the increase in acceleration (as a result of the increase in gravitational mass of m2). So the acceleration of m2 relative to the CoM remains constant, and the time that it takes to reach the CoM decreases. The only way to change the acceleration of m2 relative to CoM is to change the mass of m1.
 
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  • #32
Hello all.

I have a question.
Has any of the above discussion been experimentally verified?
Not saying it's wrong. Just want to know.
 
  • #33
Pallidin, I deleted part of my first paragraph in post #31 because I felt some parts of it may be incorrect. However, I have a high degree of confidence with the post as it is now. Please be more specific with your question. There have been lots of experimental tests for the UFF and EP.
 
  • #34
TurtleMeister said:
There have been lots of experimental tests for the UFF and EP.

Would you be willing to be more specific as to the sources?
 
  • #35
I think http://www.npl.washington.edu/eotwash/" has some of the most recent tests of the EP. Right off hand, I don't have a recent source for the UFF.
 
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  • #36
OK, good read.
But, I understand that this has not been endorsed by the NSF, DOE or NASA.
Am I missing something there?
 

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