# Can the shedding of radiation of stars contribute to the red-shift of the universe?

The sun is losing mass(1.5cm per year) by emitting radiation and the orbits of the planets are widening because of the weakening gravity of the sun.

Are most of the stars shedding mass via radiation also decreasing the total mass of the universe contribute to the increasing redshift of the universe?

Aside from the loss in mass/gravity. How much force is coming from the radiation pushing on itself from inside the universe?

Where does all that lost mass from radiation go from the universe?

• Short answer: It does not. Stellar gas remains locally bound inside galactic halos. The overall mass of the universe that controls universal expansion is dark matter, about which we don't know too much. – AtmosphericPrisonEscape Jan 23 '18 at 20:28
• @AtmosphericPrisonEscape if all emissions by the stars stopped at the galactic halos then we would not be able to see other galaxies. What of the light? Is there a universe Halo? – Muze the good Troll. Jan 23 '18 at 20:35
• What? We see the gas trapped in the disc of other galaxies. Those disc rotate deep in the gravitational wells of their own. Gas cannot escape it. Light that bounces off this gas however, can. – AtmosphericPrisonEscape Jan 23 '18 at 20:48
• The density of radiation is accounted for in lambda cdm model. Therfore the needing of dark energy. Besides that you should clarify why radiation spreading all over should accelerate expansion. But again it is accounted for and very small at this era. – Alchimista Jan 24 '18 at 10:53
• The expansion by mass loss only happens between objects that orbit each other. Mostly galaxies don't orbit other galaxies. – userLTK Jan 24 '18 at 13:05

Stars are converting some of their mass into light, but this is a small fraction: the efficiency of fusion is about 0.5-1% of mass into energy. And many stars will not burn all the fusible matter in any case. Brighter stars also tend to lose mass through stellar winds, reaching rates up to $10%{-6} M_\odot$ per year or above.