Name the two opposing forces at balance in the sun.

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Did you mean to objects which are in circular orbit around the sun?. If so, we can use Kepler's third law: (GMm)/r^2=(mv^2)/r. Where "M" is the mass of the sun, G is the gravitational constant  and "r" is the radius of the orbit. As you can see we can divide both sides by "m" so there's no dependency in the object's mass.

Gators, you're on the right track but you forgot about photons.

Basically, we're talking a plasmatic/nuclear interaction that goes thus:

/H + P (uud) + P (uud) = (convert U-D)= N+(pion) = PN (UUD) (UDD).  Therefore, PN+PN+N . . . = (^2^He^3,4,5^).

With every U-D conversion, a high-Q photon breaks loose.  Photons mediate the continued reaction between 1/H\1,2 and 2/He\3,4,5.  Note also that low-Q photons also act as gauge bosons, transferring the electrical charge between the nucleon and lepton.

High-Q photons come out of the sun at somewhere close to C.  That is what we see in the Aurora Borealis and Aurora Australis.

http://en.wikipedia.org/wiki/Sun

Even light will bend to do our bidding if we apply force correctly. The lowest servant in Heaven is still in Heaven, whoever rules in hell is still in hell, but they won't rule for long. No man stands taller than when he stoops to help a little child.

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