Sun's Gravitational Redshift
Most recent answer: 08/01/2012
Q:
time moves slower on the sun because of its massive gravity but by how much. if i spent a day stood on the sun how much more time would have past on the earth
- andrew (age 40)
lincoln england
- andrew (age 40)
lincoln england
A:
Let me work this one out very roughly. You can refine the numbers if needed. The redshift fraction is GM/Rc2 where G is the universal gravitational constant, M is the mass of the sun, R is the Sun's radius, and c is the speed of light. How to work that out, given that I don't remember G, M, or R and don't want to Google around?
I do remember that it takes light about 8 min to get here from the sun. So the distance D to the sun is ~8 min*c. And you can see by eye that R is about D/220.
We also have that the acceleration of the earth toward the sun is 4π2D/year2=GM/D2.
So GM/Rc2= 220*GM/Dc2= 240*4π2D2/year2c2= 220*4π2(8min/yr)2. So there you have it, ~ 2*10-6. That's around a fifth of a second per day.
Mike W.
I do remember that it takes light about 8 min to get here from the sun. So the distance D to the sun is ~8 min*c. And you can see by eye that R is about D/220.
We also have that the acceleration of the earth toward the sun is 4π2D/year2=GM/D2.
So GM/Rc2= 220*GM/Dc2= 240*4π2D2/year2c2= 220*4π2(8min/yr)2. So there you have it, ~ 2*10-6. That's around a fifth of a second per day.
Mike W.
(published on 08/01/2012)