White dwarf mass-radius relationship

In summary, the person is struggling with combining equations in a physics forum and is seeking help. They are advised to show their work and given general advice on how to approach the problem. The person then shares their progress and is stuck on what to do with a specific exponent. They plan to try again in the morning.
  • #1
PeteWheatstraw
6
0
instead of typing it out, here is the problem

whitedwarf_zps675b25b5.png


I know what to do, my math just isn't good enough to combine the equations properly. Been at it a few hours, brain hurts, please help!
 
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  • #2
Hello PeteWheatstraw,

Welcome to Physics Forums!
PeteWheatstraw said:
instead of typing it out, here is the problem

whitedwarf_zps675b25b5.png


I know what to do, my math just isn't good enough to combine the equations properly. Been at it a few hours, brain hurts, please help!
As part of the forum rules, you need to show your work. If you get stuck, you need to show us where you're stuck. But you at least need to show what you've done so far, in order for help.

But for what it's worth (very general advice), start by substituting mass over volume for the density [itex] \rho_c [/itex], where the volume is the the volume of a sphere, [itex] \frac{4}{3} \pi R^3 [/itex], and the mass is M.

After rearranging to isolate R, apart from the other variables, you'll end up with a bunch of numerical constants, many of which are under exponents. Break these numbers into their prime factors (e.g., 8 → 2·2·2), multiply out the exponents if necessary (e.g., 35/3 → [3·3·3·3·3]1/3) and regroup things so that you can get the numbers to be under the exponents that you want (e.g., [3·3·3·3·3]1/3 → 3·[3·3]1/3 → 3·32/3).
 
Last edited:
  • #3
Thanks for the tips.

some how I end up with (I'd type it all out but it's late)...

2∏R/3G=1/M^-1/3

and I'm stuck, and I know its wrong in the 1st place...I'll try again in the morning. Really my issue is I have no idea what to do with the 5/3 exponent and when to do it.

Thanks again though.
 

FAQ: White dwarf mass-radius relationship

What is a white dwarf?

A white dwarf is a type of star that has exhausted all of its nuclear fuel and has collapsed to a very small size. It is the final stage of evolution for stars with masses similar to our Sun.

How is the mass of a white dwarf related to its radius?

The mass-radius relationship for white dwarfs is described by the Chandrasekhar limit, which states that the more massive a white dwarf is, the smaller its radius will be. This is because the inward pull of gravity increases as the mass increases, making the star more compact.

What is the maximum mass a white dwarf can have?

The maximum mass for a white dwarf is about 1.4 times the mass of our Sun, known as the Chandrasekhar limit. This is because if the mass exceeds this limit, the inward pull of gravity will become so strong that it will overcome the electron degeneracy pressure that supports the star, causing it to collapse into a neutron star or black hole.

How does the mass-radius relationship of white dwarfs compare to other types of stars?

The mass-radius relationship for white dwarfs is unique compared to other types of stars. As mentioned before, the more massive a white dwarf is, the smaller its radius will be. In contrast, for main sequence stars (like our Sun), the more massive a star is, the larger its radius will be.

How does the mass-radius relationship of white dwarfs impact their lifespan?

The mass-radius relationship of white dwarfs is important in determining their lifespan. As white dwarfs have exhausted their nuclear fuel, they no longer have a source of energy to counteract the pull of gravity. This means that the more massive a white dwarf is, the faster it will cool and dim, leading to a shorter lifespan compared to less massive white dwarfs.

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