Mass loss in relation to radius of a star

AI Thread Summary
The discussion focuses on the mass loss equation for stars developed by D. Reimers, which indicates that mass loss is inversely related to the radius of a star. As the radius increases, the surface gravity decreases, leading to a lower rate of mass loss, which appears counterintuitive since one might expect greater mass loss with a larger radius. The equation reveals that the mass loss rate is influenced by the star's luminosity, surface gravity, and radius, with surface gravity being a critical factor in the denominator. Consequently, the relationship suggests that as radius increases, the gravitational pull weakens, resulting in reduced mass loss. This highlights the complex interplay between a star's physical characteristics and its mass loss dynamics.
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One of the most popular mass loss equations of a star, developed by D. Reimers, is given by:
dM/dt = -(4x10^-13) * η(L/(gR)) solar masses per year

Where η is a free parameter close to unity and L, g, and R are the luminosity of the star, surface gravity of the star, and the radius of the star, respectively.

What I am curious about is that when R increases, the amount of mass lost decreases. This seems counterintuitive, as when the radius increases, the density will decrease and the pull of gravitational energy will decrease so you would expect there to be greater mass loss.

Is there a qualitative reason why the star loses more mass as the radius decreases?
 
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Markus0003000 said:
Is there a qualitative reason why the star loses more mass as the radius decreases?
In a sense it doesn't. Look at Reimers' law more closely:
$$\frac{dM}{dt} = -4\cdot10^{-13} \, \eta \frac {L_{\ast}} {gR_{\ast}}$$
That g in the denominator is surface gravity relative to that of the Sun: ##g=M_{\ast}/R_{\ast}^2##. Thus another way to write Reimers' law is
$$\frac{dM}{dt} = -4\cdot10^{-13} \, \eta \frac {L_{\ast}R_{\ast}} {M_{\ast}}$$
 
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