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Posted

Distance between two bodies = R

mass of first body = m

mass of second body = m

kinetic energy of any body of the two bodies relative to each other = 0

cosmological force of space expansion = F

gravitational force between the bodies = F

There is no another force.

How to define R ?

Posted (edited)

In the static approximation the gravitational force between two masses is given by:

 

[math]F_{G} = \frac{G M m}{r^2}[/math]

 

where G is the gravitational constant, M is the mass of one body, m is the mass of the other body, and r is the distance between the two masses. The force of expansion between any point in space and a mass is given by:

 

[math]F_{\Lambda} = \frac{\Lambda c^2 m r}{3}[/math]

 

where [math]\Lambda[/math] is the cosmological constant, c is the speed of light, m is the mass in question, and r is the distance between the mass and the point in question. So, setting these equal we get:

 

[math]r^3=\frac{3GM}{\Lambda c^2}[/math]

 

or:

 

[math]r=\sqrt[3]{\frac{3GM}{\Lambda c^2}}[/math]

 

For reference, the value of the cosmological constant is of the order ~10-52 m-2, so the value of r is going to be very very large no matter what mass value you decide to give the bodies.

Edited by elfmotat
Posted

In the static approximation the gravitational force between two masses is given by:

 

[math]F_{G} = \frac{G M m}{r^2}[/math]

 

where G is the gravitational constant, M is the mass of one body, m is the mass of the other body, and r is the distance between the two masses. The force of expansion between any point in space and a mass is given by:

 

[math]F_{\Lambda} = \frac{\Lambda c^2 m r}{3}[/math]

 

where [math]\Lambda[/math] is the cosmological constant, c is the speed of light, m is the mass in question, and r is the distance between the mass and the point in question. So, setting these equal we get:

 

[math]r^3=\frac{3GM}{\Lambda c^2}[/math]

 

or:

 

[math]r=\sqrt[3]{\frac{3GM}{\Lambda c^2}}[/math]

 

For reference, the value of the cosmological constant is of the order ~10-52 m-2, so the value of r is going to be very very large no matter what mass value you decide to give the bodies.

:rolleyes: Thank you very very large. If m=1 kg then R=281374807 m. Less than 1 light second. ;)

Posted

Let's imagine infinite quantity of identical objects, which are in identical intervals located in space. Where forces of cosmological expansion are neutralized by forces of gravity. And forces of gravity are neutralized by cosmological force.

Where is mass of all the objects in big volume more:

when mass of of one object = m

or when mass of one object = m/2 ?

  • 2 weeks later...
Posted

Let's imagine infinite quantity of identical objects, which are in identical intervals located in space. Where forces of cosmological expansion are neutralized by forces of gravity. And forces of gravity are neutralized by cosmological force.

Where is mass of all the objects in big volume more:

when mass of of one object = m

or when mass of one object = m/2 ?

Then, I think, Gravitational Force=5Gm2/r2

 

r1=[5Gm/(c2*cosmological constant)]1/3

and r2=[5G(m/2)/(c2*cosmological constant)]1/3

r1/r2=21/3

v=x3r13=2x3r23

Quantity of first objects~x3

Quantity of second objects ~2x3

mass of first objects in the volume=x3m

mass of second objects in the volume=2x3m/2=x3m

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