What is the approximate gravitational red shift for light emitted by the sun?

[jk4]

The book gives the equation:
\frac{\Delta\textit{v}}{\textit{v}} = \frac{GM}{c^{2}R}

the problem gives me the mass of the sun, it's radius, and the wavelength of light being emitted.

So basically I can solve the right side of the equation, but I'm just not sure about how to express the amount of "red shift". Is it the entire left side? or is it just the top of the fraction on the left side? Or something else. I've tried both and don't get the books answer.

According to the book:
Sun mass = 2.0x10^(30) kg
Sun radius = 7.0x10^(8) m
wavelength of light = 500 nm (being emitted by the sun)
"Find the approximate gravitational red shift"
The book answer is 1.06 pm

[EDIT]
I'll throw this in too. the constant G = 6.673x10^(-11) N m^(2) / kg^(2)

 

[Dick]

The gravitational redshift is the change in frequency, which is the numerator of the left side. But notice that they aren't asking for the change in frequency but for the change in wavelength. There's a simple relation between frequency and wavelength.

 

[jk4]

ok I tried that and for \Delta\textit{v} I got 1.27x10^(9) m ... Which is obviously very wrong.

I simply calculate the right side and I get 2.12x10^(-6)
Then I multiply that by (c / 500x10^(-9)) and that gives me my ridiculous answer of 1.27x10^(9) m
and if I convert that to wavelength I get nowhere near 1.06 pm I get 0.231 m

 

[Dick]

The original frequency is c/500nm. The change in frequency is (c/500nm)*2.12*10^(-6). The new frequency is the difference. (1-2.12*10^(-6))*(c/500nm). The new wavelength is 500nm/(1-2.12*10^(-6)). How much does that differ from 500nm?

 

[jk4]

When I do it your way I get the correct answer.

But, I don't know why you say the new frequency is the difference (1-2.12*10^(-6))*(c/500nm) shouldn't it be (1+2.12*10^(-6))*(c/500nm) ?
because change in v is vf-(c/500nm) = 2.12*10^(-6) * (c/500nm)
so
vf = (1+2.12*10^(-6))*(c/500nm)
But when I do it my way I get the negative of the correct answer.

 

[Dick]

It's a RED shift. The frequency should decrease and the wavelength should increase. I put in the sign by hand. The formula doesn't really specify if the wave is coming up from the sun or down towards the sun. It just gives you the delta.