Jump to content

Gravity In Relative Motion - a new model?


JohnMnemonic

Recommended Posts

Hello! Quite some time has passed, since my last visit here and a lot of things changed... For example, it turned out, that I can in fact do some math and use it, to show you the only correct model of gravity and energy distribution in relative motion. I've spent around 6 weeks and wasted some 8 pages format a5 on calculations, while looking for the right formula - it took me so long, because I did such things for the first time in some 20 years or so and also, this was that part of physics, which as far as I remeber, I've always hated at most... I wonder, what then can explain all those generations of professional physicsts, who didn't even think about trying to calculate such things... If you really want to show me, that theoretical physicists aren't only just a bunch of overconfident snobs, then show me, that mainstream science can actually deal with the problem, which I present below:

Here's a simple scenario: 4 objects with masses:
m1=4, m2=1, m3=4, m4=1
Objects m1 and m3 move in relation to eachother at v=0,2c (1c=1d/1t)
Distances between m1 and m2, just as between m3 to m4 are equal to 2d. Due to gravitational attraction m1 makes m2 to accelerate at a1=1 (where 1a=0,1d/t^2) and attraction between m3 and m4 is just as strong.
Can you calculate the kinetic energies or acceleration (a2) for object m2 in relation to object m3 or for m3 in relation to m2? I can do it, but I had to find my own way...

key.jpg

Frame of m1:

ffg-1.gif

Frame of m3:

ffg.gif

I will wait a day or two for you to make any attempt of solving this problem and then I will begin to show you, how to do it my way... :)

Link to comment
Share on other sites

3 hours ago, JohnMnemonic said:

Here's a simple scenario: 4 objects with masses:
m1=4, m2=1, m3=4, m4=1
Objects m1 and m3 move in relation to eachother at v=0,2c (1c=1d/1t)
Distances between m1 and m2, just as between m3 to m4 are equal to 2d. Due to gravitational attraction m1 makes m2 to accelerate at a1=1 (where 1a=0,1d/t^2) and attraction between m3 and m4 is just as strong.
Can you calculate the kinetic energies or acceleration (a2) for object m2 in relation to object m3 or for m3 in relation to m2? I can do it, but I had to find my own way...

A few questions:
What is the blue ball and the black dots? Are they mass-less and not affecting the model?
How are the the masses constrained in horizontal movement? 

is "c" the speed of light in vacuum? 

 

Link to comment
Share on other sites

9 hours ago, JohnMnemonic said:


Here's a simple scenario: 4 objects with masses:
m1=4, m2=1, m3=4, m4=1
Objects m1 and m3 move in relation to eachother at v=0,2c (1c=1d/1t)

What does this mean? Usually t is a variable. And if the attraction is gravitational, the speed will not be a constant.

9 hours ago, JohnMnemonic said:


Distances between m1 and m2, just as between m3 to m4 are equal to 2d. Due to gravitational attraction m1 makes m2 to accelerate at a1=1 (where 1a=0,1d/t^2) and attraction between m3 and m4 is just as strong.

Again, what does this mean?  a1 = 1 and 1a=0,1d/t^2 are meaningless. 

 

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.