If earths rotation stoped, the moon would be released.

[JaredJames]

You claim factual error but yet offer no evidence, just ridicule. This thread is to discuss what would happen to the moon if the Earth were to stop rotating, let's stick to the topic.

Your claims are supported by no evidence - you need to provide it. It is not up to us to disprove you. It is down to the claimant to support their statements with evidence.

 

[narrator]

You claim factual error but yet offer no evidence, just ridicule. This thread is to discuss what would happen to the moon if the Earth were to stop rotating, let's stick to the topic.

ttown okie, this is a science forum, and the people who have responded to you, who have attempted to correct you, are well versed in these topics (not laymen). Rather than complain, you would do better to learn from them. Even as a layman myself, I can see where you are mistaken. There is nothing wrong with mistakes, as long as you allow yourself to be corrected.

Back to my question in post #39, if the dissipation of energy by tidal friction averages about 3.75 terawatts, what's the total energy in the Earth Moon system?

 

[Borek]

Back to my question in post #39, if the dissipation of energy by tidal friction averages about 3.75 terawatts, what's the total energy in the Earth Moon system?

I doubt total energy can be calculated from this data. I would start from known masses, radii, distance and speeds. Simple high school physics.

The only problem is - I am not sure what to include, as it depends on the definition of "total". Does the motion around Sun counts, or not?

 

[sophiecentaur]

I think the Energy to go for would just be the energy that could be transferred by the inter-planetary effects. If the Solar orbit of the c.m. of the Moon-Earth didn't change then the energy of the orbit round the Sun wouldn't be relevant in this respect. You could see what a small change in this orbit would represent in the form of 'extra' energy but any tidal locking of the Moon-earth to the Sun would probably be very small, considering the geometry. Suck it and see?

 

[narrator]

Interesting problem. And back to what I asked earlier, about proportion. "What percentage or proportion of the total energy in the Earth-Moon system would be heat energy?"

This is purely a guess, so please excuse. Given the masses of Earth and Moon, and excluding other influences (Sun et al) I'm thinking that the total energy between the two would be some hundreds times more than that given over to friction.

Having been a mechanical engineer in my distant past, I'm looking at it as being something like the efficiency of two well oiled gears or cogs, with one difference being that gravity is like energy normally delivered through the shaft (which I guess is where the analogy breaks down), but in engineering terms, I would expect something like 98% efficiency. And with celestial bodies suffering no physical friction or atmospheric friction, I would expect something like 99.99% efficiency.

Am I looking at this right?

 

[Janus]

Interesting problem. And back to what I asked earlier, about proportion. "What percentage or proportion of the total energy in the Earth-Moon system would be heat energy?"

This is purely a guess, so please excuse. Given the masses of Earth and Moon, and excluding other influences (Sun et al) I'm thinking that the total energy between the two would be some hundreds times more than that given over to friction.

Having been a mechanical engineer in my distant past, I'm looking at it as being something like the efficiency of two well oiled gears or cogs, with one difference being that gravity is like energy normally delivered through the shaft (which I guess is where the analogy breaks down), but in engineering terms, I would expect something like 98% efficiency. And with celestial bodies suffering no physical friction or atmospheric friction, I would expect something like 99.99% efficiency.

Am I looking at this right?

It's pretty easy to figure out. The moon is receding by 3.8 cm per year.

It's orbital energy is found by:

E = -\frac{G M_{earth}M_{moon}}{2a}

with 'a' being the average Earth-Moon distance.

In the process, it lengthens the Earth's rotation period by 1.5 milliseconds/century.

The Rotational energy of the Earth is found by:

E= \frac{r^2 \omega^2M}{5}

Where r, \omega and M are the radius, angular velocity( in rads/sec) and mass of the Earth.

So, find the energy gained by the Moon when it climbs 3.8 meters further out and compare it to the amount of rotational energy the Earth loses by lengthening its rotation by 1.5 ms.

Compare the two and you have your answer. (you might be surprised.)

 

[shashankac655]

So , the earth-moon system is constantly losing it's energy in the form of heat energy.Will the system lose energy like this even after the tidal locking??...if yes that means the system will continuously lose energy, but for how long? it can't lose energy forever right? ,the system must come to an end some way,how?? (just assume that the sun is not going to be a red giant and interfere with the system)

 

[ttmark]

Your friend is correct. We do not know if mass creates gravity. As you are aware many things can make you experience a different gravity constant than mass itself such as changes in velocity, centrifugal force, magnetics, etc. Current physics does not understand what actually commutes gravity between points but we can observe that mass is proportional to gravity in some cases but since even this mass of planetary objects is "proven" by it's gravitational effects, one should be careful in accepting this circular reasoning.

 

[JaredJames]

As you are aware many things can make you experience a different gravity constant than mass itself such as changes in velocity, centrifugal force, magnetics, etc.

Don't confuse feeling a different force to changing gravity. There is a big difference between simulated gravity (accelerating at 1g in a spaceship) and real gravity (standing on Earth).

Whilst accelerating upwards, you experience a greater downwards force than gravity alone, but gravity itself is constant. The only way to alter the force of gravity on you is to increase either your own, or the objects mass.

There is no circular reasoning, it is observe and report. Your misunderstanding is not an excuse to attack scientific theory.

I'll leave the rest of your post to someone with more subject knowledge than myself.

 

[ttmark]

There is no attack here. A person on the moon certainly experiences different gravity than a person on Earth, or on any other planetary object. The effect of gravity is the summation of all the forces acting towards the systems center of mass. To describe a gravity constant you must pick a frame of reference somewhere, this is typically done at the center of mass of the system in question. To an object at rest within this system it is accepted to call that force a gravity constant and it is measured by the acceleration of that body if in free fall in a vacuum. We are able to measure a gravity constant for objects near Earth but Physics does not know what caused the gravity effect to begin with.

 

[DaveC426913]

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.

 

[ttmark]

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.

 

[Janus]

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.

 

[ttmark]

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.

 

[DaveC426913]

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.

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.

 

[sophiecentaur]

Do you mean g or G?

 

[DaveC426913]

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.

 

[ttmark]

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.

 

[DaveC426913]

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.

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.

 

[ttmark]

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 Langauge and Attitude:"

 

[ttmark]

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.

 

[DaveC426913]

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.

Your personal attacks are unneeded and against forum rules and will be ignored. Please read "Guidelines on Langauge 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.

...that mass is proportional to gravity in some cases...

In what cases is it not?

 

[narrator]

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. :)

 

[narrator]

that mass is proportional to gravity in some cases

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!

 

[Borek]

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.

 

[JaredJames]

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?

 

[narrator]

Check out this thread: Pioneer anomaly solved - no need for cosmology.

A very elegant solution to the puzzle :)

 

[ttmark]

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.

 

[narrator]

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.

 

[ttmark]

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.

 

[narrator]

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.

 

[DaveC426913]

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.