Solve Potential Energy & Radiated Power of Sun: Help Needed

[hobobobo]

[SOLVED] Homework help very urgent! please help

The potential energy of an object due to its own gravity is of order
Potential Energy ~
GM2
r
where M is the body's mass and r is its radius. There is a factor in this equation that depends on the density distribution within the
object (e.g. 0.6 for a uniform sphere) - but for rough calculations we can ignore that.
Using preceeding approximation.

So I need to calculate the potential energy stored in the self-gravity of the sun, assuming a solar radius of
6.96μ108m and solar mass of 2.0μ1030kg.

I also need to calculate the total power radiated by the sun assuming the temperature of the stellar
photosphere is a 6000° K blackbody and estimate the lifetime of the sun if its energy source was purely locked-up gravitational
energy.

I'm stuck on all three of them since my physics and math aren't strong, so can anyone please help me onto how to start off! Thanx!^^

 

[Kurdt]

Do you have to derive the self gravitational potential energy from first principles or have you been given that formula or what?

You know part two is to do with black body radiation, so what equations do you know or can find that will help you there.

Part three will require you to work out how long it would take the sun at its current temperature to radiate away all that gravitational energy.

 

[hobobobo]

No deriving is invloved as far as I know. They just gave potential E = GM^2/r.

so a) calculate potential energy which uses the equation above, so i got that since solar radius is given as 6.96x10^8 m and solar mass is 2x10^30 kg.

But I'm stuck on of how to calculate the total power radiated from the sun assuming the T of the stellar photosphere is 6000 k blackbody. Do I use the blackbody equation for this? and if so what info do i need for the equation.

I also need to to estimate the lifetime of the sun if its energy is purely locked up in gravitational energy. What kind of equation will give you this?

 

[Dick]

Look up the Stefan-Boltzmann law for the dependence of power radiated per unit of surface area. Use that together with the area of the sun to get the total rate of energy loss.

 

[hobobobo]

E= σT4 j/m^2s you mean this one right dick?

 

[Dick]

E= σT4 j/m^2s you mean this one right dick?

That's the one. I wouldn't call it 'E' though, it's not an 'energy', it's energy per meter^2 per second.

 

[hobobobo]

ok so I got 3.43x10^15 as my total power radiated by the sun assuming T=6000k. I calculated area of the sun by using pir^2 since solar radius which is 6.96x10^8 m. So this question wanted j so i moved things around. I sub in E by using the potential energy that I calculated from part a is that right?

 

[Kurdt]

ok so I got 3.43x10^15 as my total power radiated by the sun assuming T=6000k. I calculated area of the sun by using pir^2 since solar radius which is 6.96x10^8 m. So this question wanted j so i moved things around. I sub in E by using the potential energy that I calculated from part a is that right?

The sun is not a circle, so I'd reconsider the power radiated from it. What you need to do next is find how long it takes the sun to radiate the gravitational energy away.

 

[hobobobo]

oops haha 4pir^2 lol. So the rearranging and finding j is right? But i thought after all this i would just do lifetime = energy from part a divided by energy from part b, isn't that right?

 

[Kurdt]

J? You're trying to find how long the sun will last if its powered by gravitational energy.

 

[hobobobo]

hmmm i followed what dick said about the Boltzmann equation and then including the area of the sun so that's wrong then? well E=constant x T^4 x j x area of the sun which I got to be 1.71x10^68

 

[Kurdt]

Ok sorry, wasn't sure what point you were upto. Yes you multiply the surface area of the sun by the power per unit area, to find the total power radiated by it. Then you need to find the lifetime of the sun.

 

[hobobobo]

well lifetime of the sun I try calculating and it seems wrong came up with a number x 10^26 and that can't be right. I used this: lifetime = (energy) / (rate [energy/time] at which sun emits energy), is there another way to do it becuase this seems to be the only way.

 

[Kurdt]

I'd check your calculations again. Specifically your calculation for the total power radiated by the sun. Remember it also helps us homework helpers if you post all the steps you use as we can quickly identify where you go wrong and how best to help you.

 

[hobobobo]

ok srry I'll post my steps:

So calculation for potential stored self stored gravity of sun as follows:

potential Energy = GM^2/r
= (6.67x10^-11m^3kg^-1s^-1)(2.0x10^30kg)^2/(6.96x10^8m)
=3.83x10^41m^2/s

Calculation of total power radiated by sun as follows:

E=(constant)(T^4)(j)(4pir^2)
=(5.67x10^-8 watt/m^2k^4)(6000^4k)(3.83x10^41m^2/s)(6.09x10^18m^2)
= 1.71x10^68

So are these right so far?

 

[Kurdt]

The j in the Stefan-Boltzmann law is just part of the units (i.e. Joules per meter squared per second). Its not a variable. Also what are the units of potential energy?

 

[hobobobo]

m^2/s?... because we have like m^3/m and kg^-1 and kg so they cancel out.

 

[Kurdt]

m^2/s?... because we have like m^3/m and kg^-1 and kg so they cancel out.

The unit of energy is the Joule. Which is $J=\frac{Kgm^2}{s^2}$.

So could you see where you went wrong in your above calculation?

 

[hobobobo]

oohh? hmm I was following my sheet which says potential energy of an object due to its own gravity is of order Potential Energy = GM^2/r

 

[Kurdt]

oohh? hmm I was following my sheet which says potential energy of an object due to its own gravity is of order Potential Energy = GM^2/r

Potential energy is fine. Perhaps I was a bit misleading when I asked you to think about its units. Look again at my previous posts specifically post 14.

 

[Dick]

Kurdt is right. Your value for the total power radiated by the sun is way off. Can you spell out where each of the numbers comes from?

 

[hobobobo]

ok give me a min lol

 

[hobobobo]

this is the constant of stefan Boltzmann (5.67x10^-8 watt/m^2k^4)

temperature of the stellar photosphere (6000^4k) blackbody

Potential Energy due to its own gravity (3.83x10^41m^2/s)

surface area of the sun (6.09x10^18m^2)

 

[hobobobo]

surface area i got it by 4pir^2 where r is 6.96x10^6 m

 

[Kurdt]

this is the constant of stefan Boltzmann (5.67x10^-8 watt/m^2k^4)

temperature of the stellar photosphere (6000^4k) blackbody

Potential Energy due to its own gravity (3.83x10^41m^2/s)

surface area of the sun (6.09x10^18m^2)

Where does potential energy fit into the Stefan-Boltzmann law?

$$\phi = \sigma T^4$$

 

[Dick]

Where does potential energy fit into the Stefan-Boltzmann law?

$$\phi = \sigma T^4$$

Right. So energy/time=$$A \sigma T^4$$.

 

[hobobobo]

oh damn it I keep forgetting XD

 

[hobobobo]

ok so my answer would still look wrong since I got energy/time= 4.47x10^26

 

[Dick]

Ok, so you've got PE=3.7*10^41 J. And you've got the got that the sun is radiating at 4.5*10^26 J/sec. How many years can that energy last? How do you combine those two numbers to find out?

 

[hobobobo]

Well I did the lifetime = (energy) / (rate [energy/time] at which sun emits energy) which I got 8.55x10^14 which is a really really big number, that number is bigger than a trillion lol.

 

[Dick]

Well I did the lifetime = (energy) / (rate [energy/time] at which sun emits energy) which I got 8.55x10^14 which is a really really big number, that number is bigger than a trillion lol.

What are the units on that? Seconds aren't they? How many years?

 

[Kurdt]

Well I did the lifetime = (energy) / (rate [energy/time] at which sun emits energy) which I got 8.55x10^14 which is a really really big number, that number is bigger than a trillion lol.

Remember the units that is in though.

EDIT: damn beaten to it.

 

[hobobobo]

oh >_> haha yea I'm in a hurry and I keep forgetting to look at the units lol thanks a lot you guys, you guys are a big help.