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Some Math To Start.

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Posted by Alexander on March 19, 2001 16:53:45 UTC

Ok, suppose that Earth is moving (in orbital motion) with the speed v via Sun's atmosphere with the atmosphere density being D. Resistance force is F=dp/dt=m(dv/dt)+v(dm/dt), where dm is the mass of Sun's atmosphere being hit by orbiting Earth during time dt. Assuming slow spiraling (dv/dt=0), we get F=v(dm/dt)=vD(dV/dt)=vDS(dl/dt)=v2DS, where dV is volume of Sun's atmosphere being hit by Earth in dt time, S = Earth's cross-section.

So, ma=v2DS, or a=v2DS/m, and therefore characteristic spiraling time is about t=v/a=m/vDS

Now it is time to plug numbers in.

1. Very dense atmosphere model. Assume that ALL Sun is spread uniformly over a ball with the radius = 1 AU = 1.5x10^11 m (Earth's orbit radius). Then the density of Sun's atmosphere (and the density of core) is about D=MS/Vorb= 2x10^30 kg/1.4x10^34 m^3 = 1.4x10^-4 kg/m^3 (about 10^4 less than air density), and then t=350 years. Because in such star nuclear reactions would not proceed (too low density of core), let's then consider:

2. More realistic Sun - compact dense He-burning core with the size less than current Sun and thin hydrogen envelope. In this case most of Sun's mass is concentrated in the core and not much is left for the atmosphere. If a core has, for example, 90% of Sun's mass, then atmosphere is 10 less dense and the spiraling time is about t=3500 years. Most likely red giant's outer atmosphere is even less dense, making spiraling time much longer. Does anybody know how thin is atmosphere of Sun-massed red giant?

Also, if some mass of Sun's atmosphere happen to be outside of Earth orbit (i.e.- Earth is well inside of Sun), then due to the reduced gravity of remaining (inside) portion of Sun the Earth's orbit will even increase in diameter.

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