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Author Topic: My Force of Gravity Equation  (Read 5339 times)

Offline ghostofdavinci

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My Force of Gravity Equation
« on: 05/12/2008 07:07:20 »
Hi, let me introduce myself, My name is Terell Randolph and I am a freshman at University of Washington majoring in physics and mathematics.  Anyways I have been working on a equation to describe the motion on moving bodies caused by gravity.  After two or three weeks I developed one to describe the force of gravity:

((mc^2)/r)*Sin(x)

where m=mass Force gravity is acting upon, c=Speed of light, x=motion of wave, and r=Distance from sun.

My equation and Newton's equation answers match for and mass or radius, but when it comes to black holes for example Schwarzschild radius with my equation you get 2.5km where as Newton equation you get 3km (even though Schwarzschild radius was derived from GR you can use Newton equation).  I need help seeing why my equation only works for long distances.  First I'll show how I derived c*Sin(x) which is not important so you can skip over it but my equation is based of this.  Then I'll show how I derived my force equation.

This derivation is based on Special relativity and Mass-energy equivalence

E(1)/k=E(2)

Where E(1) is energy of mass at rest, E(2) is mass new energy from observer perspective, and     k=sqrt(1-v^2/c^2) c=speed of light v=velocity of moving body (throughout this derivation the velocity is 0)

(hc)/(λk)=mc^2

h=Planck constant c=speed of light λ=wavelength m=mass

h/(λck)=m

h^2/(λ^2k^2c^2)=m^2

h^2/(λ^2(1-v^2/c^2)c^2)=m^2

h^2/(λ^2(c^2-v^2))=m^2

h/(λ*sqrt(c^2-v^2))=m   Equation (1)

The following derivation sets the condition for λ

f=c/λ

f=Frequency, c=speed of light,λ=wave length

f=c/1 λ=1

Then hf=hc/λ equals hf=hc

Using the conditions above

mc^2=hf

hc^2/(λ*sqrt(c^2-v^2))=hf Using Equation (1)

hc^2/sqrt(c^2-v^2)=hf when λ=1

h^2c^4/(c^2-v^2)=h^2f^2

h^2c^4 =h^2f^2(c^2-v^2)

h^2c^4 =(h^2f^2c^2)-(h^2f^2v^2)

h^2f^2c^2 =(h^2c^4)+(h^2f^2v^2) add h^2f^2v^2 to both sides

(h^2f^2c^2)-(h^2c^4) =(h^2f^2v^2) subtract h^2c^4 from both sides

h^2f^2c^2(1-(c^2/f^2)) =h^2f^2v^2 Factor out h^2f^2c^2

c^2(1-(c^2/f^2)) =v^2 Divide both sides by h^2f^2

if f=c/λ then λ=c/f

c^2(1-λ^2)=v^2

c*sqrt(1-λ^2)=v Equation (2)

This equation satisfy the conditions above because if λ=1 then v=0

solving for λ

λ=\sqrt(1-(v^2/c^2))

Since (v^2/c^2)is a ratio, I can place them on a unit circle, y component is v and the radius is c.  Now I can replace (v^2/c^2) with Sin^2(x)

λ=sqrt(1-Sin^2(x))

Cos(x)=sqrt(1-Sin^2x) Replacing λ with Cos(x)

Going back to Equation(2)

c*sqrt(1-λ^2)=v

c*sqrt(1-Cos^2x)=v replacing λ with Cos(x)

Using trig-identities

c*Sin(x) =v Equation(3)

I'll resume in a second post
« Last Edit: 05/12/2008 07:09:27 by ghostofdavinci »


 

Offline ghostofdavinci

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Re: My Force of Gravity Equation
« Reply #1 on: 05/12/2008 07:07:34 »
Replacing v with Newton's equation for velocity caused by gravity

c*Sin(x)=sqrt(GM/r)

Making Sin(x) as a wave function

x=k(x-ct) k=wave number x=describes wave shape c=speed of light t=time

Solving for K

c=sqrt(T/μ) then c^2=sqrt(T/μ)

c=speed of light T=tension μ=Mass/length (Sun's mass/radius where velocity equals c)

Since the angle is in degrees, I used 360deg instead of 2π

c^2*μ=T and T=360/ω

ω=360deg/(c^2*μ)

c=ω/k

k=360deg/λ

λ=ω/(360deg*c)

k=360deg/λ

k=9.9220075474*10^-52

Solving for x

Since velocity in Newton equation is dependent of square root of the mass/length

R*sqrt(mass/radius) Where R is a constant

R=1.5736276814526*10^38

I'll explain how I got R later.  It is independent of the mass and radius, also it depends on what k is.  If k changes then R changes.

x=k(R*sqrt(M/r-ct)

We can rewrite are equation

c*Sin(x) =sqrt(GM/r)

c*Sin(k(R*sqrt(M/r)-ct))=sqrt(GM/r)

Since a body does not feel the effects until the gravity wave reaches it, we can rewrite the equation as

c*Sin(k(R*sqrt(M/r)-r))=sqrt(GM/r)

From this equation we get the following

c^2*Sin((k(R*sqrt(M/r)-ct))^2 =(GM/r)

(c^2/r)*Sin((k(R*sqrt(M/r)-ct))^2 =(GM/r^2)

(mc^2/r)*Sin((k(R*sqrt(M/r)-ct))^2 =(GMm/r^2)

(E/r)*Sin((k(R*sqrt(M/r)-ct))^2 =(GMm/r^2)

(E/r)*Sin(x)^2 =(GMm/r^2)

That's it, like I said before my equation matches Newton's equation lets just say from the surface of the mass to the end of the universe, but when it comes to become a black hole my equation breaks down and I'm trying to figure out why.  Your help will be greatly appreciated.

 

Offline ghostofdavinci

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My Force of Gravity Equation
« Reply #2 on: 05/12/2008 07:37:46 »
My theory is the closer you get to the center of mass the force of gravity decreases and close to the center of mass gravity is not continuous force.

Here is a graph of the Force Gravity caused by the Sun if it was a black hole on a mass with 1 kg from 0 to 5000 meters using my equation(The amplitude looks small but ymin and ymax are -10^15 and 10^15)



Now here is a graph with dots instead of lines connecting



From the graph above we can see that gravity isn't a continuous force until about 1000m. This may explain why gravity is such a weak force and is hard to combine with the other three force.  Also this may explain why Hawking radiation can escape a black hole.

« Last Edit: 05/12/2008 09:46:12 by ghostofdavinci »
 

Offline Chemistry4me

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« Reply #3 on: 06/12/2008 00:01:07 »
Whoa! I've got no idea what you are talking about but it looks good~!
 

Offline Chemistry4me

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« Reply #4 on: 06/12/2008 00:01:28 »
What are you planning to do with it?
 

Offline Don_1

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« Reply #5 on: 06/12/2008 00:44:09 »
I don't know what ghostofdavinci is planning to do, but I know what he/she has already done:-

Got me totally confused But it all looks very impressive and I think it basically means we won't fall off the planet.

ghostofdavinci, I take my hat off to you, I have not the slightest notion as to what all that means, but I would hazard a guess that it was no mean feat.
 

Offline Bored chemist

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« Reply #6 on: 06/12/2008 02:09:13 »
"My equation and Newton's equation answers match for and mass or radius,"
Good.
Now get them to match the rest of the time.
 

Offline graham.d

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« Reply #7 on: 06/12/2008 12:22:38 »
I have not worked through all your maths, but I don't think you should assume a special condition of λ=1, then derive an equation:
λ=\sqrt(1-(v^2/c^2))
which says 1=1 (because v=0)
then proceed to derive general equations from this.

I'm not sure your derivation is valid but it's hard to folow the scripting without writing it out long hand. I will have a go if I get time.

Try to not put any special conditions in, then work though all the equations.
When you get to a solution that would define the parameter you are trying to solve, only then put in your special conditions. It takes out the doubt.

Have you ever seen this? :

y=x

y^2=y*x

y^2-x^2=y*x-x^2

(y-x)*(y+x)=x*(y-x)

y+x=x

2x=x

2=1
 

Offline LeeE

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« Reply #8 on: 06/12/2008 18:09:11 »
Are you trying to use trig to get around the problem of square roots of negative numbers?
 

Offline Bikerman

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« Reply #9 on: 06/12/2008 18:11:42 »
LOL...that is one of the first fallacies I learned at school. The error is obvious [division by zero in the cancellation (y-x)*(y+x)=x*(y-x)] but has been made by many theorists (albeit in a slightly more advanced format).
 

Offline ghostofdavinci

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« Reply #10 on: 06/12/2008 18:20:21 »
Thanks for all the replies.  The thing is Bored chemist should I automatically assume my equation wrong, because when I compared it to General Relativity, it shows the same effects will happen for example,

4*sin(x) is the equation for bending of light
(h/λ)*sin(x)shows how the force of gravity effects momentum of light which explains gravitational redshift/blueshift

graham.d, I would have ended up with the same result even if I didn't make λ=1.  Instead I would have ended up with
1=sqrt(1-v^2/c^2).  The main reason why I made λ=1 was to show that λ is dependent on the objects velocity.  My belief is that all matter are waves not particles and if hc/λ = energy, so when matter has an increase in velocity λ will decrease which will increase the matter's energy(there will only be an increase from the observer perspective, the one moving will see no increase in energy).  I think nature found a way to increase energy without having to physically increase it.  It's kind of like the doppler effect where a race car has a high pitch when its coming towards you and low pitch driving away from you but the actual pitch never changed. Also I haven't seen that derivation before, but I can tell you why it fails.
 

Offline ghostofdavinci

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« Reply #11 on: 06/12/2008 18:35:51 »
Are you trying to use trig to get around the problem of square roots of negative numbers?

Not really, the only way I can have a negative number is if an object moves faster than the speed of light which is impossible.  The main reason was to make a wave function.
 

Offline graham.d

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« Reply #12 on: 07/12/2008 13:11:18 »
Yes Ghost, the error in the simple math example is as Bikerman says, though I'm sorry he revealed it as it can cause amusement to those who do not spot the flaw. I only put it in the post to show that you have to be very careful, even with simple maths, to not inadvertently introduce an error in symbol manipulation. My problem is that I am prone to making mistakes when doing several pages of calculation and they are always in the first few lines ;-). I now often use Mathcad (with Maple) to do the manipulation of the equations as it does not make errors. The problem here though is that all but very simple calculations often end up having hugely complex results that do not give intuitive insight anyway. D'oh!

I will take a better look at your work.
 

Offline graham.d

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« Reply #13 on: 07/12/2008 15:02:52 »
Some words of explanation about what your assumptions are at each step would be worthwhile. You use the energy equation for a photon E=hc/λ, then you equate mc^2 with E. In this case m would be the observed mass of the photon. It would be normal to use this equation to determine this observed (or apparent) mass (m=h/cλ) as a photon has zero rest mass. The use of the factor k here needs explanation. If you use DeBroglie's equation for a particle at velocity v: λ=h/mv, then the relativistic contraction of the wavelength (λ) cancels the increase in the mass (m) and the factor k can be cancelled from each side of the equation. So you can use λ* to represent kλ and m* to represent m/k and the equation is invariant with velocity.

Sorry I have not got too far :-) It would be worth discussing with your tutor and going over your work with a knowledgable post-grad at your university. Face-to-face discussions can be more fruitful because of the speed of the interaction.
 

Offline LeeE

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« Reply #14 on: 07/12/2008 18:22:23 »
Are you trying to use trig to get around the problem of square roots of negative numbers?

Not really, the only way I can have a negative number is if an object moves faster than the speed of light which is impossible.  The main reason was to make a wave function.

Hmm...  just wondered, because c*sin(x) is derived from Pythagorus, which is what the Lorentz equations use.
 

Offline Bikerman

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« Reply #15 on: 07/12/2008 18:52:51 »
Yes Ghost, the error in the simple math example is as Bikerman says, though I'm sorry he revealed it as it can cause amusement to those who do not spot the flaw.
Sorry about that - I didn't mean to spoil the fun. I really thought it was so obvious that anyone would spot it...
 

Offline ghostofdavinci

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« Reply #16 on: 07/12/2008 23:34:44 »
graham.d, thanks for looking at my work, but one thing doesn't k depends on the velocity.  The higher the velocity the smaller k is.  I wanted to show how the increase in mass is dependent on λ itself and not by the factor k.  I'll be working with a tutor after finales or next quarter.  LeeE, yea k is Lorentz factor
 

Offline TECHFACTOR

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« Reply #17 on: 18/12/2008 20:13:58 »
 TECHFACTOR: I think it is incredible that people can read and write math like this not to mention solve and even improve on this equations . I think in pictures about the same subjects and do not have this talent for my on problem solving necessities.  Therefore I wondered if you might check out my liquid theory in this same forum; in new theory's? And maybe just consider this plausible ;and tell me if the there is A way that this theory might add up? MOST CURIOUS? TECHFACTOR;
 

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My Force of Gravity Equation
« Reply #17 on: 18/12/2008 20:13:58 »

 

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