Mass loss in relation to radius of a star

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SUMMARY

The discussion centers on the mass loss equation for stars as formulated by D. Reimers, expressed as dM/dt = -(4x10^-13) * η(L/(gR)). Key variables include luminosity (L), surface gravity (g), and radius (R) of the star. A critical observation is that as the radius (R) increases, the mass loss (dM/dt) decreases, which contradicts intuitive expectations regarding density and gravitational pull. The equation reveals that surface gravity (g) is inversely related to the square of the radius, leading to a decrease in mass loss with increasing radius.

PREREQUISITES
  • Understanding of stellar physics and mass loss mechanisms
  • Familiarity with D. Reimers' mass loss equation
  • Knowledge of luminosity and surface gravity concepts
  • Basic grasp of gravitational dynamics in astrophysics
NEXT STEPS
  • Research the implications of D. Reimers' mass loss equation in stellar evolution
  • Explore the relationship between stellar radius and surface gravity in detail
  • Investigate other mass loss models in astrophysics
  • Examine case studies of stars with varying radii and their mass loss rates
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Astronomers, astrophysicists, and students studying stellar dynamics and mass loss phenomena will benefit from this discussion.

Markus0003000
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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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