modeman said:
I dod not ask because I did not think gravity was the force that holds all stars. I asked because I read this article along time ago about astrophysics, where two astrophysicists from India had recently found a force that proves that gravity was not the force that holds planets and stars in orbit, at least the stars. I don't know what to believe so I came to ask here. Here's the address...
http://www.zetatalk.com/usenet/use90089.htm
Like I said, I do not know what to believe, since we have Relativity as the current theory, so please tell me what you think.
Did they forget about the existence of gravitationally bound binary stars?
Further, their statements about how energy is released from fusion is right up there with the Time-cube guy. Simulations of stellar interiors depend on gravity, and no repulsion force, and are accurate to within 1% of helioseismology measurements. A repulsion force like what they are talking about would literally blow stars apart.
For example:
<br />
\frac{dP}{dr} = -G \frac{M_r \rho}{r^2}
<br />
\frac{dM_r}{dr} = 4 \pi r^2 \rho
<br />
\frac{dL_r}{dr} = 4 \pi r^2 \rho \epsilon <br />
Are the basic equations of stellar structure. The first equation is derived from:
<br />
dm\frac{d^2r}{dt^2} = F_g + F_{p,t} + F_{p,b}<br />
Which means that the net force on a mass portion dm is equal the the sum of gravity, downward pressure on the top of the portion, and upward pressure on the bottom of the portion. If we add a repulsive force F_r we get:
<br />
dm \frac{d^2r}{dt^2} = F_g + F_r + F{p.t} + F{p,b}<br />
which becomes
<br />
dm \frac{d^2r}{dt^2} = -G \frac{M_r dm}{r^2} - A dP + F_r & dm = \rho A \dr
<br />
\rho A dr \frac{d^2r}{dt^2} = -G \frac{M_rdm}{r^2} - A dP + F_r
<br />
\rho \frac{d^2r}{dt^2} = -G \frac{M_r}{r^2} \rho - \frac{dP}{dr} + \frac{F_r}{A dr} <br />
Now, the condition for hydrostatic equilibrium, that is stable stars that don't blow apart or implode is:
<br />
\frac{d^2r}{dt^2} = 0<br />
So, we get:
<br />
-G \frac{M_r}{r^2} \rho - \frac{dP}{dr} + \frac{F_r}{A dr} = 0<br />
Now, we know what the pressure gradients inside the sun are due to helioseismology measurements. That means, that for the sun to be stable, it would have to be far, far more dense in order to counteract that mysterious repulsion force (which they never give an expression for, despite 23 years of work). However, if the star were that dense, the planets, governed by gravity, could not be in their current orbits.
Either the star blows apart, or the planets need to start flying through the sky a lot faster. Neither is the case.
: https://www.physicsforums.com/showthread.php?t=102135
Hehe, I will...when I get a well payed job...
)
