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1. Apr 1, 2017

### Ryaners

1. The problem statement, all variables and given/known data

2. Relevant equations
Mass-radius relation: $$R \propto M^{-\frac{1}{3}}$$

3. The attempt at a solution
So I've tried the following:
$$R_{D} \propto M_{D}^{-\frac{1}{3}} \Rightarrow \frac {R_{D}} {R_{sun}} = \frac{M_{D}^{-\frac{1}{3}}} {M_{sun}^{-\frac{1}{3}}} \Rightarrow R_{D} = \frac{M_{D}^{-\frac{1}{3}}} {M_{sun}^{-\frac{1}{3}}} R_{sun}$$
$$\Rightarrow R_{D} = \left( {\frac {1.3 M_{sun}} {M_{sun}}} \right) ^{-\frac{1}{3}} R_{sun} = \left( 1.3 \right) ^{-\frac{1}{3}} R_{sun}$$

This gives me an answer of about $0.916~{R_{sun}}$ , which is incorrect. Where am I going wrong here?

Thanks in advance for any help.

2. Apr 1, 2017

### kuruman

Start with $M_{Sun}^{1/3}R_{Sun}=M_{D}^{1/3}R_{D}$ and replace $M_D=1.3M_{Sun}$. The algebra is less confusing when you eliminate the proportionality constant.

3. Apr 2, 2017

### Ryaners

That's a fair point - though I get the same result:

$$\left( M_{sun} \right) ^\frac {1}{3} R_{sun} = \left( M_{D} \right) ^{\frac {1}{3}} R_D$$
$$\Rightarrow R_D = \left( \frac {M_{sun}} {M_D} \right) ^{\frac{1}{3}} R_{sun}$$
$$\Rightarrow R_D = \left( \frac {1}{1.3} \right) ^{\frac {1}{3}} R_{sun} = \left( 1.3 \right) ^{-\frac {1}{3}} R_{sun}$$

4. Apr 2, 2017

### kuruman

At this point you need to question why you think that the answer is incorrect. What you think is the correct answer may be a misprint or a miscalculated answer by whoever gave it to you. The only other thing I can think of is the starting equation which is approximate and may have to be refined.

5. Apr 2, 2017

### Dick

Perhaps the problem is that the sun is not a white dwarf???

6. Apr 2, 2017

### kuruman

Perhaps, but the problem clearly states that you should use "the mass-radius relation for white dwarfs."

I am sorry, but my resources regarding this question have been exhausted. I took a single astrophysics course several decades ago and I have reached the point where I can no longer help you. Perhaps someone else may be able to step in.

7. Apr 2, 2017

### Dick

It's not really an serious astrophysics point. The mass-radius relation gives you a proportionality. To get the constant of proportionality you need an example mass and radius of a white dwarf. The sun isn't one.

8. Apr 2, 2017

### Staff: Mentor

Hint: Sirius B is a pretty well known example of a white dwarf