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Copernicus

2018-Jan-28, 11:28 PM

There must be a point between two gravitational objects where they pull equally on each other. Lets say there is the point between the earth and the sun. Is this point affected dependent on where other planets are in their

orbit relative to the sun and earth?

DaveC426913

2018-Jan-28, 11:48 PM

Indeed. It is called Lagrange One.

All two body systems have such a point.

Copernicus

2018-Jan-28, 11:54 PM

Indeed. It is called Lagrange One.

All two body systems have such a point.

Do they have more than one Lagrange?

WaxRubiks

2018-Jan-29, 12:08 AM

Do they have more than one Lagrange?

five

22917

Hornblower

2018-Jan-29, 12:17 AM

At any given point, the gravitational components from any two bodies will be independent of other bodies. Of course those other bodies will perturb something at that point.

DaveC426913

2018-Jan-29, 12:47 AM

Do they have more than one Lagrange?

Yes. As WaxRubiks pints out, there are actually five in any given two-body system.

But only L1 is relevant to your question. It's the only one that is balanced between the two systems.

That's why I manually edited the other 4 out of my diagram for clarity in post 2.

DaCaptain

2018-Jan-29, 12:58 AM

five

22917

So being a newbie, can you explain what is happening at each point. L1 is pretty obvious but the rest??? Thanks :p

DaveC426913

2018-Jan-29, 01:04 AM

So being a newbie, can you explain what is happening at each point. L1 is pretty obvious but the rest??? Thanks :p

Have a read here (https://en.wikipedia.org/wiki/Lagrangian_point). And see if you have more questions.

It explains, fairly concisely, what is happening at each point.

WaxRubiks

2018-Jan-29, 01:36 AM

So being a newbie, can you explain what is happening at each point. L1 is pretty obvious but the rest??? Thanks :p

if there were a boulder at L4, it would be orbiting the Sun, but in one orbit of the Sun, it will have also made one orbit of Earth. So it is in a position to keep orbiting both Earth and Sun.

Hornblower

2018-Jan-29, 02:37 AM

if there were a boulder at L4, it would be orbiting the Sun, but in one orbit of the Sun, it will have also made one orbit of Earth. So it is in a position to keep orbiting both Earth and Sun.

That's a clever way of analyzing it.

Copernicus

2018-Jan-29, 02:13 PM

Lets look at the two body problem from a different angle. Lets say we have two solid uniform spheres touching each other. Sphere A and Sphere BThe gravity at the contact point is one. We want to look at the point in Sphere A where the the gravity decreases linearly to the center, at which point it is zero. Sphere B, the gravity from its surface decreases inversely to the square of the distance from Sphere B's center. At which point is the gravity from A and B equal. I set up the following equation, which I think is correct.

x=(\frac{1}{2-x})^2

x has the value \frac{1}{\phi^2}, 1, and \phi^2 which is one over the golden ratio squared, one, and the golden ratio squared. Is my equation correct for this problem.

Reality Check

2018-Jan-29, 11:46 PM

Lets look at the two body problem from a different angle. Lets say we have two solid uniform spheres touching each other. ...

This is not the two body problem (https://en.wikipedia.org/wiki/Two-body_problem)

In classical mechanics, the two-body problem is to determine the motion of two point particles that interact only with each other.

This is about the gravity outside of the point particles and thus outside of any equivalent solid uniform spheres. You are trying to look at the gravitation inside of the spheres.

grant hutchison

2018-Jan-30, 12:30 AM

Lets look at the two body problem from a different angle. Lets say we have two solid uniform spheres touching each other. Sphere A and Sphere BThe gravity at the contact point is one. We want to look at the point in Sphere A where the the gravity decreases linearly to the center, at which point it is zero. Sphere B, the gravity from its surface decreases inversely to the square of the distance from Sphere B's center. At which point is the gravity from A and B equal. I set up the following equation, which I think is correct.

x=(\frac{1}{2-x})^2

x has the value \frac{1}{\phi^2}, 1, and \phi^2 which is one over the golden ratio squared, one, and the golden ratio squared. Is my equation correct for this problem.Yes. The absolute values are the same, but your solution gives you two points where the gravity cancels, and one where it adds.

And there doesn't seem to be any relevance to the OP.

Grant Hutchison

grant hutchison

2018-Jan-30, 01:13 PM

Yes. The absolute values are the same, but your solution gives you two points where the gravity cancels, and one where it adds.

And there doesn't seem to be any relevance to the OP.Actually, the addition point is non-physical, since it extends the gravity rules for inside to the outside, and for outside to the inside.

You can find the third balance point by recasting the equation, or just from symmetry. It's at 2 - 1/φ² = φ.

Grant Hutchison

Copernicus

2018-Jan-30, 02:30 PM

Actually, the addition point is non-physical, since it extends the gravity rules for inside to the outside, and for outside to the inside.

You can find the third balance point by recasting the equation, or just from symmetry. It's at 2 - 1/φ² = φ.

Grant Hutchison

That is a very good point! Balance point!

Ken G

2018-Jan-30, 03:44 PM

Also, it is a common misconception that the Langrange point is where the gravity of the two objects cancel. That's not the case, to get the Lagrange point you go into a reference frame that is corotating with the orbit of the two bodies (which are assumed to be in circular orbits). Then the Lagrange points are where the gravity of the two objects add up to cancel the centrifugal force in that rotating frame, so that's effectively three forces that add up to zero, not two.

Also, it should probably be noted that both the center of the Earth, and the center of the Sun, would be Lagrange points in the Earth/Sun system (neglecting all other bodies), but those two Lagrange points are rarely counted, I presume because they are regarded as inaccessible.

Grey

2018-Jan-30, 05:58 PM

There must be a point between two gravitational objects where they pull equally on each other. Lets say there is the point between the earth and the sun. Is this point affected dependent on where other planets are in their

orbit relative to the sun and earth?I want to point out something pedantic. It definitely seems like the question you want to ask is about the Lagrange points, and other people have answered that reasonably well. But that's when two objects pull equally on a third object (and there's also the effective centrifugal force to balance, as Ken G brings up; thinking about that makes it easier to understand what's going on at the Lagrange points other than L1).

The question you actually asked is when two objects pull "equally on each other". And the answer to that is that they always do. The gravitational force on the Earth from the Sun is always exactly equal (but in the opposite direction) to the force on the Sun from the Earth. That's just Newton's Third Law.

Copernicus

2018-Jan-30, 09:04 PM

I want to point out something pedantic. It definitely seems like the question you want to ask is about the Lagrange points, and other people have answered that reasonably well. But that's when two objects pull equally on a third object (and there's also the effective centrifugal force to balance, as Ken G brings up; thinking about that makes it easier to understand what's going on at the Lagrange points other than L1).

The question you actually asked is when two objects pull "equally on each other". And the answer to that is that they always do. The gravitational force on the Earth from the Sun is always exactly equal (but in the opposite direction) to the force on the Sun from the Earth. That's just Newton's Third Law.

Makes sense. What I was thinking, was not then a Lagrange point.

DaveC426913

2018-Jan-31, 12:47 AM

You might be thinking of the barycentre.

Every pair of bodies has a common point about which they orbit as a system.

The Earth-Moon's barycentre is actually inside the Earth, about 1700 km below the surface.

Two bodies that are of similar mass will have their baryentre proportionately near the physical midpoint of the two of them.

https://spaceplace.nasa.gov/barycenter/en/jupiter-sun.en.png

Copernicus

2018-Jan-31, 03:40 AM

You might be thinking of the barycentre.

Every pair of bodies has a common point about which they orbit as a system.

The Earth-Moon's barycentre is actually inside the Earth, about 1700 km below the surface.

Two bodies that are of similar mass will have their baryentre proportionately near the physical midpoint of the two of them.

https://spaceplace.nasa.gov/barycenter/en/jupiter-sun.en.png

I'm not really sure what would be going. I was really thinking about David's Static Universe. It got me thinking about what would dark energy be. Which I think dark energy could be consistent with the observations. Then I was trying to figure out something for red shift and as I was playing around with equations trying to figure out how it would be consistent with my own theories of the universe being a sphere made of spheres. I came up with the equation which I brought up.

x=(\frac{1}{2-x})^2

I had heard of lagrange points before, but I am not always familiar with the language. I also have heard about the rotation around a common center of gravity. I thought it was interesting that my equation had come up with relation to the golden ratio or as like grant hutchison had noted 2 - 1/\phi^² =\phi.

In my own theory the spheres made of spheres have many built in packing defects, so I think it would probably be a center of gravity of a cloud of defects traveling around some center of gravity of a cloud of defects. But I don't know that this is happening, but if it is, how it would be characterized, and what it would be called. I don't know that anyone ever examined this particular relationship, x=(\frac{1}{2-x})^2, before either.

tusenfem

2018-Jan-31, 07:49 AM

I'm not really sure what would be going. I was really thinking about David's Static Universe. It got me thinking about what would dark energy be. Which I think dark energy could be consistent with the observations. Then I was trying to figure out something for red shift and as I was playing around with equations trying to figure out how it would be consistent with my own theories of the universe being a sphere made of spheres. I came up with the equation which I brought up.

x=(\frac{1}{2-x})^2

I had heard of lagrange points before, but I am not always familiar with the language. I also have heard about the rotation around a common center of gravity. I thought it was interesting that my equation had come up with relation to the golden ratio or as like grant hutchison had noted 2 - 1/\phi^² =\phi.

In my own theory the spheres made of spheres have many built in packing defects, so I think it would probably be a center of gravity of a cloud of defects traveling around some center of gravity of a cloud of defects. But I don't know that this is happening, but if it is, how it would be characterized, and what it would be called. I don't know that anyone ever examined this particular relationship, x=(\frac{1}{2-x})^2, before either.

Yeah, and let's not go there any further.

grant hutchison

2018-Jan-31, 01:32 PM

You might be thinking of the barycentre.

Every pair of bodies has a common point about which they orbit as a system.

The Earth-Moon's barycentre is actually inside the Earth, about 1700 km below the surface.

Two bodies that are of similar mass will have their baryentre proportionately near the physical midpoint of the two of them.The barycentre doesn't match what Copernicus seems to be looking for, except in the case where the two bodies are of equal mass. The barycentre is always closer to the more massive of the two bodies, so any test mass placed at the barycentre will fall towards the more massive body.

Grant Hutchison

Ken G

2018-Jan-31, 02:14 PM

It definitely seems like the question you want to ask is about the Lagrange points, and other people have answered that reasonably well. But that's when two objects pull equally on a third object (and there's also the effective centrifugal force to balance, as Ken G brings up; thinking about that makes it easier to understand what's going on at the Lagrange points other than L1).It is essential to include the centrifugal force, even at L1, or it's not L1.

Grey

2018-Jan-31, 05:20 PM

It is essential to include the centrifugal force, even at L1, or it's not L1.Absolutely. I think it's just that L1 looks vaguely right (qualitatively, at least; it wouldn't work out to the correct distance) without factoring that in, so someone might think they understand how that one works, but then end up confused about how the other points can end up being balanced, as DaCaptain was in post 7.

Copernicus

2018-Jan-31, 05:48 PM

The barycentre doesn't match what Copernicus seems to be looking for, except in the case where the two bodies are of equal mass. The barycentre is always closer to the more massive of the two bodies, so any test mass placed at the barycentre will fall towards the more massive body.

Grant Hutchison

This would be for exactly equal masses. The equations could be rewritten as follows.

1+1-y=\frac{1}{y^{2}}

The solutions are

y=1

y=\frac{1}{-\psi}=-0.618033988749895

And

y=\psi=1.61803398874989

Geo Kaplan

2018-Jan-31, 06:45 PM

This would be for exactly equal masses. The equations could be rewritten as follows.

1+1-y=\frac{1}{y^{2}}

The solutions are

y=1

y=\frac{1}{-\psi}=-0.618033988749895

And

y=\psi=1.61803398874989

And thus phi, the golden ratio, appears again.

Copernicus

2018-Jan-31, 07:10 PM

And thus phi, the golden ratio, appears again.

1+1-y=\frac{1}{y^{2}}

The solutions are

y=1

y=\frac{1}{-\psi}=-0.618033988749895

And

y=\psi=1.61803398874989

The equation would represent the point of gravity from one sphere(sphere A) decreasing linearly as one goes from the outer edge of the sphere towards the center and the gravity of an adjacent sphere (Sphere B) decreasing inversely to the square from the center of (Sphere B). Assuming each sphere is uniform and exactly equal in mass.

What I can't figure out is if this relationship can be converted to an equation for the spiral of the golden ratio.

grapes

2018-Jan-31, 08:16 PM

1+1-y=\frac{1}{y^{2}}

The solutions are

y=1

y=\frac{1}{-\psi}=-0.618033988749895

And

y=\psi=1.61803398874989

The equation would represent the point of gravity from one sphere(sphere A) decreasing linearly as one goes from the outer edge of the sphere towards the center and the gravity of an adjacent sphere (Sphere B) decreasing inversely to the square from the center of (Sphere B). Assuming each sphere is uniform and exactly equal in mass.

What I can't figure out is if this relationship can be converted to an equation for the spiral of the golden ratio.

You mean, using the definition of golden ratio? That a golden-ratio rectangle is such that removing a square from it leaves a golden-ratio rectangle? That (x)/1 = gr = (x-1)/1 ?

Is that what you mean?

Copernicus

2018-Jan-31, 09:25 PM

You mean, using the definition of golden ratio? That a golden-ratio rectangle is such that removing a square from it leaves a golden-ratio rectangle? That (x)/1 = gr = (x-1)/1 ?

Is that what you mean?

What I am thinking is that instead of going directly from the edge of the sphere to the center of the sphere there must be points of a triangle where the base of the triangle is from the edge of the sphere to the center of the sphere. The angles of the triangle would be a, b and 180-(a+b). The triangle could be made into to right triangles. Like the following link to a triangle. http://4.bp.blogspot.com/-LOFOX0eqT4I/UY3lZBgTXMI/AAAAAAAAATs/0S1hCkRjBmw/s1600/coslaw3.png

The equation would then change to, I believe,

2-a=1/(1+c)^2

where c^2=(a sin C)^2+(b-a cos C)^2 and some how this equation would eventually equal

r=ae^{b\theta} where a and b are different then the triangle a and b and c. Which is the equation for the spiral of the golden ratio.

DaveC426913

2018-Feb-01, 03:41 AM

The barycentre doesn't match what Copernicus seems to be looking for

I know, but he was reaching for something.

tusenfem

2018-Feb-01, 06:03 AM

Let's keep this thread on topic, based on the OP please.

No golden ratio discussions.

George

2018-Feb-01, 09:52 PM

Also, it is a common misconception that the Langrange point is where the gravity of the two objects cancel. That's not the case, to get the Lagrange point you go into a reference frame that is corotating with the orbit of the two bodies (which are assumed to be in circular orbits). Then the Lagrange points are where the gravity of the two objects add up to cancel the centrifugal force in that rotating frame, so that's effectively three forces that add up to zero, not two.

.That makes sense and I was curious what size vector the c.f. is relative to the other two. Solving for a 1 kg mass where the two grav. forces act equally upon it, it is just shy of the actual L1 by a factor of...5.79! Wow. If my napkin work is correct, that's surprising, huh?

iphone

Hornblower

2018-Feb-01, 11:14 PM

That makes sense and I was curious what size vector the c.f. is relative to the other two. Solving for a 1 kg mass where the two grav. forces act equally upon it, it is just shy of the actual L1 by a factor of...5.79! Wow. If my napkin work is correct, that's surprising, huh?

iphone

My bold. Not at all surprising. The booby trap here is thinking in terms of a static solution, when in fact we have orbital motion to consider. The Sun's force at L1 is vastly stronger, but the Earth's component creates just enough of an offset that a spacecraft there can orbit the Sun at that shorter radius in one year. As Ken G has shown above, this is a situation where the rotating frame of reference with its fictitious centrifugal force is a handy device for calculation.

Ken G

2018-Feb-02, 12:52 AM

That makes sense and I was curious what size vector the c.f. is relative to the other two. Solving for a 1 kg mass where the two grav. forces act equally upon it, it is just shy of the actual L1 by a factor of...5.79! Wow. If my napkin work is correct, that's surprising, huh?

What did you use for the mass ratio between the two bodies? If the bodies are equal mass, the L1 is at the barycenter and there's no centrifugal contribution. If one body is way more massive than the other, the balance is all between the big gravity and the centrifugal force. (The mass of the test particle is of no consequence.)

George

2018-Feb-02, 12:54 AM

I suppose the fact that the c.f. wasn't addressed until Ken's post led me to assume it was more of a tweak than the dominant vector. So much so that erroneous math won't be too surprising, huh? ;) [I stumbled in obtaining an exact solution with these simple three vectors; the algebra got tricky for me, though I'm on the run using iPhone & iPad.]

Ken G

2018-Feb-02, 01:48 AM

Yeah, it's no tweak, it's key.

George

2018-Feb-02, 03:30 AM

What did you use for the mass ratio between the two bodies? Sun & Earth, thus I can steal the actual L1 and compare it to my calculated point where the grav force balance for any given mass, giving the difference factor of 5.79 (using 1.5M km for L1).

If the bodies are equal mass, the L1 is at the barycenter and there's no centrifugal contribution.. Good point, pun unintended.

(The mass of the test particle is of no consequence.)Right since this mass cancels in the equations, but I mentioned it to separate it from the earlier point about gravitational forces being equal, which is true but different for our test mass.

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