DimaMazin Posted January 18, 2015 Share Posted January 18, 2015 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 ? Link to comment Share on other sites More sharing options...
elfmotat Posted January 18, 2015 Share Posted January 18, 2015 (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 January 18, 2015 by elfmotat 3 Link to comment Share on other sites More sharing options...
DimaMazin Posted January 18, 2015 Author Share Posted January 18, 2015 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. Thank you very very large. If m=1 kg then R=281374807 m. Less than 1 light second. Link to comment Share on other sites More sharing options...
DimaMazin Posted January 24, 2015 Author Share Posted January 24, 2015 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 ? Link to comment Share on other sites More sharing options...
DimaMazin Posted February 1, 2015 Author Share Posted February 1, 2015 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 Link to comment Share on other sites More sharing options...
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