Gravity!

[andrewcellini]

 

 

 

 

 

They still repel if i let go until they reach their very own event horizon, an equilibrium of force. But they are still repelling.

mass becomes greater than the force of the magnets

i think it would be best to not just read the introductory paragraph of a wikipedia page and then use what little information it provides to demonstrate the validity of you idea, you mix up terminology because you don't know what you're talking about.

Edited September 30, 2014 by andrewcellini

[Strange]

My reasoning is obvious, magnetic field, opposes EMR

 

Even if that were true you need to quantify the effect.

 

The light, gravity and magnetic field from stars 4 million light years away might have an effect on the Earth's orbit based on just "reasoning". But if you work out how large the effect is, you will see it is so small it can be ignored.

 

Without math your "reasoning", such as it is, is of no value.

 

No one answers my questions, I asked if the diagram of a,b,c,d squares, if the physical process on that was correct, to what I defined with that example, and if that is correct, that does mean my whole gravity concept is possible also.

 

I answered it by pointing out that you need to calculate the sizes of the relative forces to see if there is an equlibrium.

[Relative]

 

Even if that were true you need to quantify the effect.

 

The light, gravity and magnetic field from stars 4 million light years away might have an effect on the Earth's orbit based on just "reasoning". But if you work out how large the effect is, you will see it is so small it can be ignored.

 

Without math your "reasoning", such as it is, is of no value.

 

I answered it by pointing out that you need to calculate the sizes of the relative forces to see if there is an equlibrium.

 

 

Ok, so firstly I want to do maths for the earth been attracted to the sun and vice versus,

 

 

F=MAG force is equal to acceleration by gravity, would that be correct?

[andrewcellini]

F=MAG force is equal to acceleration by gravity, would that be correct?

no, local gravity (like near the surface of the earth) can be approximated using F=mg where g is the field (acceleration due to gravity on a body).

 

F=GMm/r^2 is newtonian universal theory of gravitation.

Edited September 30, 2014 by andrewcellini

[Strange]

F=MAG force is equal to acceleration by gravity, would that be correct?

 

Almost certainly not.

 

What is M?

What is A?

what is G?

[Relative]

My second formula wold need to show the opposing stopping force, that stops the first force from making a collision of planets.

 

 

F=m<-a>MF?

 

Force is equal to mass decelerated by magneic field, or emr, ? or both?

no, local gravity (like near the surface of the earth) can be approximated using F=mg where g is the field (acceleration due to gravity on a body).

 

F=GMm/r^2 is newtonian universal theory of gravitation.

 

Yes that one sounds correct, using both a and b it is the other bit part two im really stuck on

 

Almost certainly not.

 

What is M?

What is A?

what is G?

mass acellerated by gravity

[andrewcellini]

My second formula wold need to show the opposing stopping force, that stops the first force from making a collision of planets.

 

 

F=m<-a>MF?

 

Force is equal to mass decelerated by magneic field, or emr, ? or both?

electromagnetism =/= gravitation.

[Sensei]

Relative, you have no basic knowledge about units. See couple the first posts of this thread and learn from it.

 

ps. Where you have been whole primary school, when kids were learning "what is Force"? etc. kind of things..

If you multiply m*a*G, you certainly won't get Newtons unit..

Edited September 30, 2014 by Sensei

[Relative]

electromagnetism =/= gravitation.

that sounds good thx.

[Strange]

mass acellerated by gravity

 

So not just wrong, but meaningless.

 

You have been given the correct equation for calculating the force. (And agreed that the current maths works.) So use it.

 

The force between two magnets is much more complicated.

[Relative]

Relative, you have no basic knowledge about units. See couple the first posts on this thread and learn from it.

 

ps. Where you have been whole primary school, when kids were learning "what is Force"? etc. kind of things..

If you multiply m*a*G, you certainly won't get Newtons unit..

Well you know maths beyond the standard I struggle with, just because I did not learn it, I am getting there knowing formulas, and soon will pick up the rest...

Relative, you have no basic knowledge about units. See couple the first posts of this thread and learn from it.

 

ps. Where you have been whole primary school, when kids were learning "what is Force"? etc. kind of things..

If you multiply m*a*G, you certainly won't get Newtons unit..

I am sorry if my units do not make sense, I just say what I see, I see matter always want to accelerate to matter, so mass been the term you use, MA, and gravity is the force involved, so MAG,

 

 

I suppose im just giving a formula to the work, force is caused by matter wanting to accelerate by gravity.

''In physics, a force is said to do work when acting on a body there is a displacement of the point of application in the direction of the force

 

Where the earths field meets space, there is a displacement of force at the northern lights?

To me the curvature of time and space, is displacement of magnetic field by momentum and velocity.

 

I think the sun does this?

[andrewcellini]

To me the curvature of time and space

you should learn more fundamentals before trying to incorporate more complicated models.

[Relative]

you should learn more fundamentals before trying to incorporate more complicated models.

There is nothing complicated about it except to make some maths fit,

 

 

 

 

 

This is why we ellipse in my opinion by my logic process.

The sky is so blue.......

[swansont]

To me the curvature of time and space, is displacement of magnetic field by momentum and velocity.

But it's not what it means to physicists, i.e. the ones who made the models, and that's what matters.

 

200px-Dopplereffectsourcemovingrightatmach0.7.gif

 

I think the sun does this?

 

Not with gravity it doesn't. It looks exactly the same as if it were motionless. That's what the models say, and the models work.

[Relative]

I might not know it all and all i need to know , but can you at least see what I am trying to explain now?

 

 

 

[Sensei]

F=m*a

That's correct formula for force, which any kid in the first class of physics in primary school is learning.

 

How do we know mass of object?

Because we can put it on weighting scale, and read value. Old-school weighting scale was comparison of two bodies. One with known weight. With one with unknown (analog weighting scale).

 

How do we know acceleration I showed you post #17

http://www.scienceforums.net/topic/85792-gravity/?p=828512

 

Acceleration at the ground of Earth is

[latex]a=\frac{M*G}{r^2}=9.81 \frac{m}{s^2}[/latex]

M - mass of Earth

G - gravitational constant

r - radius of Earth

 

How do we know mass of Earth?

We can f.e. estimate average density

[latex]p = \frac{M}{V}[/latex]

density p, is mass divided by volume

volume of Earth is

[latex]V = \frac{4}{3}*\pi*r^3[/latex]

 

For r = 6371000 meters radius

[latex]V = \frac{4}{3}*\pi*6371000^3 = 1.083*10^{21} m^3[/latex]

 

Density of Earth is approximately [latex]5514 \frac{kg}{m^3}[/latex]

 

so

 

[latex]M = p*V = 5514 \frac{kg}{m^3} * 1.083*10^{21} m^3 = 5.972*10^{24} kg[/latex]

 

 

Calculate acceleration

[latex]a=\frac{M*G}{r^2}=\frac{5.972*10^{24} kg * 6.67*10^{-11}\frac{N*m^2}{kg^2}}{6371000^2} = 9.81 \frac{m}{s^2}[/latex]

[Relative]

why the sky is blue?

F=m*a
That's correct formula for force, which any kid in the first class of physics in primary school is learning.

How do we know mass of object?
Because we can put it on weighting scale, and read value. Old-school weighting scale was comparison of two bodies. One with known weight. With one with unknown (analog weighting scale).

How do we know acceleration I showed you post #17
http://www.scienceforums.net/topic/85792-gravity/?p=828512

Acceleration at the ground of Earth is
[latex]a=\frac{M*G}{r^2}=9.81 \frac{m}{s^2}[/latex]
M - mass of Earth
G - gravitational constant
r - radius of Earth

How do we know mass of Earth?
We can f.e. estimate average density
[latex]p = \frac{M}{V}[/latex]
density p, is mass divided by volume
volume of Earth is
[latex]V = \frac{4}{3}*\pi*r^3[/latex]

For r = 6371000 meters radius
[latex]V = \frac{4}{3}*\pi*6371000^3 = 1.083*10^{21} m^3[/latex]

Density of Earth is approximately [latex]5514 \frac{kg}{m^3}[/latex]

so

[latex]M = p*V = 5514 \frac{kg}{m^3} * 1.083*10^{21} m^3 = 5.972*10^{24} kg[/latex]


Calculate acceleration
[latex]a=\frac{M*G}{r^2}=\frac{5.972*10^{24} kg * 6.67*10^{-11}\frac{N*m^2}{kg^2}}{6371000^2} = 9.81 \frac{m}{s^2}[/latex]

wow you know your stuff really well and I will read and re-read this over and over until i get it.

is 9.81ms2 applied to the sun and the earth, would the sun accelerate towards the earth at the same V?

 

 

 

and how the eck do you get all those symbols up on a keyboard?

 

meaning this

[Strange]

why the sky is blue?

 

That is a completely different topic. Please start a new thread.

 

is 9.81ms2 applied to the sun and the earth, would the sun accelerate towards the earth at the same V?

 

The force is the same. But the acceleration depends on mass. So the Sun will move less than the Earth. As the Earth orbits the Sun, the Sun moves slightly: they both orbit their common center of gravity.

 

and how the eck do you get all those symbols up on a keyboard?

 

Latex: http://www.scienceforums.net/topic/3751-quick-latex-tutorial/

Edited September 30, 2014 by Strange

[Relative]

And thank you , I have learnt some new interesting things today.

velocity zero

If the right circle was to rotate, the left circle would rotate the right circle?

[Strange]

If the right circle was to rotate, the left circle would rotate the right circle?

 

Why would it?

[Sensei]

is 9.81ms2 applied to the sun and the earth, would the sun accelerate towards the earth at the same V?

Acceleration 9.81 m/s^2 is at Earth's ground. The further you're from the ground, the smaller is that acceleration...

 

It could be approximated to:

 

[latex]a = \frac{M*G}{(r+d)^2}[/latex]

 

r - Earth radius

d - distance to Earth's ground

 

so for example 100 km above there will be:

 

[latex]a = \frac{M*G}{(6371000+100000)^2} = 9.51 \frac{m}{s^2}[/latex]

 

1000 km above Earth:

[latex]a = \frac{M*G}{(6371000+1000000)^2} = 7.33 \frac{m}{s^2}[/latex]

 

and so on further in cosmos..

[Relative]

 

Why would it?

Because of the forces involved, the two circles are connected by an invisible link, the force,

 

There is no friction that can prevent movement

[Strange]

Because of the forces involved, the two circles are connected by an invisible link, the force,

 

es.jpg

There is no friction that can prevent movement

 

It looks like you are talking about the two bodies orbiting each other. In which case, yes: the Sun and Earth both orbit a point in space, their common center of gravity.

 

"Rotation" means to turn around an axis; I though you were imagining the two circles as gears: when you turn one, the other rotates. That won't happen without some sort of mechanical connection.

[xyzt]

would the sun accelerate towards the earth at the same V?

V is speed, not acceleration. You need to learn the basics before you start making "theories".

Edited September 30, 2014 by xyzt

[Relative]

 

It looks like you are talking about the two bodies orbiting each other. In which case, yes: the Sun and Earth both orbit a point in space, their common center of gravity.

 

"Rotation" means to turn around an axis; I though you were imagining the two circles as gears: when you turn one, the other rotates. That won't happen without some sort of mechanical connection.

well actually , for now yes , consider it as gears, and the right circle is static, but spins, there is no friction to prevent the left circle from moving, the linkage is an invisible pink unicorn, For now please just imagine the link has a transparent bar, it will make it easier, I am just checking something in my thinking,

 

 

So the left circle will rotate around the right circle?

 

V is speed, not acceleration. You need to learn the basics before you start making "theories".

 

Sorry , my mistake i appologise.

If the left circle is at an equilibrium and has zero friction, what force would prevent the left circle from rotating around the right circle?

Edited September 30, 2014 by Relative