Find gravitational acceleration through luminosity

[astrofunk21]

Homework Statement

Use equation 1.3 to estimate the solar radius R_o from its luminosity and effective temperature. Show that the gravitational acceleration g at the surface is about 30 times larger than that on Earth.

Homework Equations

(1.3) L=4∏R²σ_SBT⁴

The Attempt at a Solution

Well we know the gravity of something can be determined through g=\frac{-GM}{r*r}. Solving for r in the gravitational equation will allow us to substitute that into the luminosity equation. With all of those known, we can solve for g and see how it is 30 times larger. This however did not work. My answer was -27.79 m*s^-2.

Now since this number is pretty much off by a factor of ten, I feel that I'm on the right track. Maybe I messed something up in my calculation that you guys will see. My substitution will be below if it helps.

1. L=4∏R²σ_SBT⁴

2. g=\frac{-GM}{r*r}

3. r²=\frac{-GM}{g}

4. L=4∏\frac{-GM}{g}σ_SBT⁴

5. g=\frac{-GMT^{4}4∏σ}{L}

6. Then solve for g!

 

[Simon Bridge]

Well we know the gravity of something can be determined through g=−GM/r∗r. Using this we can first find the effective temperatures for both the Earth and the sun. Solving for r in the gravitational equation will allow us to substitute that into the luminosity equation.

I'm not clear on your reasoning here - doesn;t the question expect you to do this the exact opposite way around?

 

[astrofunk21]

I'm not clear on your reasoning here - doesn;t the question expect you to do this the exact opposite way around?

Hey sorry about that, meant to delete that sentence about the effective temperature. It was a method I was using before. Does it make more sense now?

 

[Dick]

It looks to me like you solved correctly for g in terms of L, T and M. Now they must have given you some numbers to plug in for those variables. What were they?

 

[Simon Bridge]

It's the reasoning that puzzles me - wouldn't it be more direct to solve for R² in eq. 1.3 and sub into the equation for g?

$$g_{Sol} = \frac{M_{Sol}}{m_E} \frac{r^2_{E}}{R^2_{Sol}} g_{E}\; : \; R^2=\frac{L}{4\pi\sigma T^4}$$

Anyway - you got the same equation so ... provided you have not made some arithmetic error, or used bad numbers, what does the mismatch tell you about the estimate? What assumptions are made to get equation 1.3?

 

[astrofunk21]

In terms of some numbers that were used I had:
M = 2x10³⁰kg
T = 5780K
σ_SB = 5.67x10^-8Wm^-2K^-4
L = 3.86x10²⁶W
G = 6.67x10^-11m³kg^-1s^-2

 

[astrofunk21]

Simpler problem than I thought. Got the answer and it's right.

I solved for R like the problem says and like Simon said. Then plugged that into the accel due to gravity equation. This gave me a number around 280 which is close to 30 times that of Earth's gravity.

Thanks guys!

 

[tms]

What assumptions are made to get equation 1.3?

The primary assumption is that the sun is a black body.

 

[Simon Bridge]

I had wondered if the teacher wanted to provide an example of a model that did not work well :) If the numbers were correct, and there was no finger-slippage on the calculator, then the result disagreeing with reality would mean that the model contained a flaw.

Occasionally someone will do that to you just to see who's alert.
You get marks by identifying the flaw - hence the suggestion to look at the assumptions.

But that was not the case this time.