By how, I mean what is the equation.


If the core of a red giant contracts the outer radius will expand; but I cannot find an equation that says this, but I have equations that say parts of this.

The surface radius, $R$, of a red giant is porportial to the core temperature, $T_c$, and stellar mass, $M$, in this way: $$ R \propto \frac{M}{T_c}. $$ We can assume that the core radius is much smaller than the stellar radius, $R_c \ll R$.


A red giant with a non-fusing helium core contracts from $R_{c,0}$ to $R_{c,1}$, what are the stellar radii $R_0, R_1$?

Answers in terms of proportions to stellar mass, $M$, core mass, $M_c$, and core radius $R_c$, are preferred but any light (hehe stars) on the subject is appreciated.


It comes from hydrostatic equilibrium, gravity, and a determination of the rate at which fusion reactions occur and how that depends on pressure, temperature, and density (more detail here). I, sadly, do not have the equations handy in my mind, but the gist is this: when the inert core is smaller the gravitational field is greater at its surface, increasing the pressure in the fusing part of the core, driving the fusion reactions faster, and thus producing more heat to puff up the outer layers.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.