If earths rotation stoped, the moon would be released.

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SUMMARY

The discussion clarifies that gravity is fundamentally a result of mass, not Earth's rotation. While Earth's rotation has a negligible effect on surface gravity, it does not influence the Moon's orbit significantly. The Moon is gradually receding from Earth at a rate of approximately 4 centimeters per year due to tidal interactions, but it will not break free from Earth's gravitational pull. Instead, the Earth and Moon will eventually reach a tidal locking state, where both bodies will rotate in sync.

PREREQUISITES
  • Understanding of Newton's Law of Universal Gravitation
  • Basic knowledge of tidal forces and their effects on celestial bodies
  • Familiarity with the concepts of angular momentum and energy transfer in gravitational systems
  • Awareness of the Earth-Moon system dynamics and orbital mechanics
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  • Research the effects of tidal locking on celestial bodies
  • Explore the concept of gravitational potential and its implications in astrophysics
  • Learn about the dynamics of the Earth-Moon system and its long-term evolution
  • Investigate the role of tidal friction in the Earth-Moon distance changes
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Astronomy enthusiasts, physics students, and anyone interested in understanding gravitational interactions and the dynamics of the Earth-Moon system.

  • #61
ttmark said:
A person on the moon certainly experiences different gravity than a person on Earth,

A person on the Moon experiences gravity exactly the same as they do on Earth - namely, that it will be proportional to the mass multiplied by the square of the distance to that mass.

Experiencing a different weight on the Moon is not the same as experiencing different gravity.

Your statement is quite misleading.
 
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  • #62
DaveC426913 said:
A person on the Moon experiences gravity exactly the same as they do on Earth - namely, that it will be proportional to the mass multiplied by the square of the distance to that mass.

Experiencing a different weight on the Moon is not the same as experiencing different gravity.

Your statement is quite misleading.

The Gravity constant is defined by the freefall acceleration in a vacuum of a body near the surface of the Earth. (m/s^2) I did not say anything at all about weight. This gravity constant is different on the Moon than on the Earth and as such the person on the moon certainly does experience different gravity than the person on Earth. The force is stronger on the Earth than it is on the moon. The person on the moon experiences gravity towards the Moon's center of mass, the person on Earth experience gravity towards Earth's center of mass. If this wasn't the case the person on the Moon would fall off and fall back to Earth.
 
  • #63
ttmark said:
The Gravity constant is defined by the freefall acceleration in a vacuum of a body near the surface of the Earth. (m/s^2)
That is not the gravitational constant, that is the acceleration due to gravity at the surface of the Earth (signified by "g").

The gravitational constant is 6.673e_11 m^3/kgs^2(signified by G) and this is the same for all masses. You use the gravitational constant, the mass and radius of the body to find the acceleration due to gravity at that body's surface.
 
  • #64
Janus said:
That is not the gravitational constant, that is the acceleration due to gravity at the surface of the Earth (signified by "g").

It is the gravity constant for Earth. I see you agree that the gravity constant for the moon is something else entirely.
 
  • #65
ttmark said:
It is the gravity constant for Earth. I see you agree that the gravity constant for the moon is something else entirely.

It's not a constant. You can tell this because it's not constant. :wink:

It is simply the value of a force at a given location, and it has a different value at a location three feet to your left or three feet above your head.

The gravity constant G is the same everywhere on the Earth, on the Moon, and, if we understand correctly, everywhere else in the universe.
 
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  • #66
Do you mean g or G?
 
  • #67
sophiecentaur said:
Do you mean g or G?

That is precisely ttmark's confusion.


ttmark, this would not be under discssuon at all if you would care to read up a little about G versus g. The first paragraph of Wiki will do for now:

The gravity of Earth, denoted g, refers to the acceleration that the Earth imparts to objects on or near its surface. ... It has an approximate value of 9.81 m/s2, which means that, ignoring air resistance, the speed of an object falling freely near the Earth's surface increases by about 9.81 metres per second every second. This quantity is informally known as little g (contrasted with G, the gravitational constant, known as big G).
http://en.wikipedia.org/wiki/Gravity_of_Earth


g on Earth is no more a constant than is v in a moving car.
 
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  • #68
DaveC426913 said:
That is precisely ttmark's confusion.

We all know this, there is no confusion on my part. You can use the universal gravity constant if you want but it doesn't change the fact that a person on the moon experiences gravity different than a person on Earth. There is more gravity on Earth than the Moon and each is towards their own center of mass.
 
  • #69
ttmark said:
We all know this, there is no confusion on my part.
You have just figured it out now - after we practically beat it into you. Don't make me quote the number of times you mixed them up.

ttmark said:
You can use the universal gravity constant if you want but it doesn't change the fact that a person on the moon experiences gravity different than a person on Earth. There is more gravity on Earth than the Moon and each is towards their own center of mass.
Of what relevance is that?
Here:
https://www.physicsforums.com/showpost.php?p=3336689&postcount=58
you claim it changes all the time and then call it circular reasoning.

Our measurement of G did not depend on g, so why have you been trying so hard to claim that there's circular reasoning?

That's a rhetorical queation. We know the answer. You were confused. Again, don't make me quote you.
 
  • #70
DaveC426913 said:
You have just figured it out now - after we practically beat it into you. Don't make me quote the number of times you mixed them up.
That's a rhetorical queation. We know the answer. You were confused. Again, don't make me quote you.

As you can see in my post, "As you are aware many things can make you experience a different gravity constant". Notice gravity is lower case and referencing an experience felt by a person. This is clearly written referring to a point of reference of someone on Earth.

Your personal attacks are unneeded and against forum rules and will be ignored. Please read "Guidelines on language and Attitude:"
 
  • #71
DaveC426913 said:
Y
you call it circular reasoning..

The circular reasoning is because we are using the gravity constant between two equal in mass objects which was measured on Earth to then claim the mass of planetary objects outside of Earth. One must be very careful in doing this because the motion of the bodies within gravitational field must be considered about it's own inertial frame for Newtons laws of motion to work.
 
  • #72
ttmark said:
As you can see in my post, "As you are aware many things can make you experience a different gravity constant". Notice gravity is lower case and referencing an experience felt by a person. This is clearly written referring to a point of reference of someone on Earth.
You have trieds to define the phrase "gravitational constant" several times, incorrectly. Backpedal all you want.

ttmark said:
Your personal attacks are unneeded and against forum rules and will be ignored. Please read "Guidelines on language and Attitude:"
I have made no personal attacks; I have pointed out errors. Forum rules do not cover being wrong.


All this aside, your original comment still makes no sense.
ttmark said:
...that mass is proportional to gravity in some cases...
In what cases is it not?
 
  • #73
Janus said:
The gravitational constant is 6.673e_11 m^3/kgs^2(signified by G) and this is the same for all masses.

I had always wondered what that value is and how constant it is. Many thanks. :)
 
  • #74
ttmark said:
that mass is proportional to gravity in some cases

DaveC426913 said:
In what cases is it not?

Even as a layman I can see this. As far as I know, mass is always proportional to gravity. Other factors can change the acceleration, but that doesn't affect the proportionality of mass to gravity. We experience a different gravity on the moon because the moon is a different volume of mass and follows the proportion exactly.

As an interesting contradiction though, I was surprised to read about the http://en.wikipedia.org/wiki/Pioneer_anomaly" . Freakish, especially since even cosmology was proven not to be the missing link!
 
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  • #75
narrator said:
As an interesting contradiction though, I was surprised to read about the http://en.wikipedia.org/wiki/Pioneer_anomaly" . Freakish, especially since even cosmology was proven not to be the missing link!

Check out this thread: Pioneer anomaly solved - no need for cosmology.
 
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  • #76
ttmark seems to be confusing the force due to gravity on its own and the net force with all other factors involved.

Aside from that note, I don't see what the point of his posts are. Perhaps an explanation ttmark?
 
  • #77
Borek said:
Check out this thread: Pioneer anomaly solved - no need for cosmology.

A very elegant solution to the puzzle :)
 
  • #78
JaredJames said:
ttmark seems to be confusing the force due to gravity on its own and the net force with all other factors involved.

Aside from that note, I don't see what the point of his posts are. Perhaps an explanation ttmark?

I have been away harvesting wheat all weekend, sorry guys... I do not remember if there is any point, but I do not have time to argue back and forth with the poster who claims I am confused with G or g. If there is any point it is that just because gravity is proportional to mass within our inertial frame of reference does not mean that we know that gravity is caused by mass. So when one is considering planetary objects outside our inertial frame of reference you have to be careful applying the gravity constant because that constant was measured with two objects of equal mass on Earth. So then we go out and assign a mass to every object we see in the sky with this constant, maybe right, never understanding what creates gravity to begin with.
 
  • #79
ttmark said:
So then we go out and assign a mass to every object we see in the sky with this constant, maybe right, never understanding what creates gravity to begin with.

It's not that simple ttmark. Science isn't science without the process of falsification. You don't just apply a value to something then move on. You test to make sure that applying that value actually matches with other evidence. If the constant isn't constant, then observations would have shown that long ago. Orbits alone would disprove it.

The proportionality of gravity to mass has been proven over and over. Other contributions to inertia, force or whatever are also proven. And all of it through science's most rigorous test - falsification.
 
  • #80
narrator said:
It's not that simple ttmark. Science isn't science without the process of falsification. You don't just apply a value to something then move on. You test to make sure that applying that value actually matches with other evidence. If the constant isn't constant, then observations would have shown that long ago. Orbits alone would disprove it.

The proportionality of gravity to mass has been proven over and over. Other contributions to inertia, force or whatever are also proven. And all of it through science's most rigorous test - falsification.

My understanding is we took this constant we measured on Earth and used it to establish the mass of the sun based upon Earths orbital period and distance. From there we used this mass of the sun to branch out to the other planets based upon their orbits. Changing the constant would not affect any orbits at all, merely the mass that we claim each is. Is there some over method that can prove this? As I see it right now we are just solving for whatever mass is needed to fit the orbit of the object in question.
 
  • #81
As I said, it's not that simple. It you were dealing with just one variable, and only ever used that one equation for one thing, then your argument could have merit. But the geometry of space has many equations (including complex calculus) with so many variables that a mountain of anomalies would have proven your case a long time ago. Except perhaps in exotic conditions (black holes, pre-plank time, and quantum physics), the relationship between mass and gravity is proven beyond doubt.
 
  • #82
ttmark said:
My understanding is we took this constant we measured on Earth and used it to establish the mass of the sun based upon Earths orbital period and distance. From there we used this mass of the sun to branch out to the other planets based upon their orbits. Changing the constant would not affect any orbits at all, merely the mass that we claim each is. Is there some over method that can prove this? As I see it right now we are just solving for whatever mass is needed to fit the orbit of the object in question.

There is what is known as a preponderance of evidence. We do not simply take maeasurments in one place and assume it holds everywhere. Our understanding of the cosmos is a collection of many, many interdependent pieces. If our numbers are wrong, it will show up in many places, both in our measurements of the cosmos and in our formulae and theories.
 

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