Solar Energy Q: Mass of Sun Decrease/s?

AI Thread Summary
The discussion revolves around calculating the mass loss of the Sun due to solar energy output, specifically how to approach the problem using the solar flux of 1.4 kW per square meter. To find the total power output of the Sun, one must first calculate the surface area of a sphere with the radius of Earth's orbit and multiply it by the solar flux. The relationship between energy and mass is then applied using Einstein's equation E=mc² to determine the mass decrease per second. Participants express interest in the physics involved and seek clarification on the analytical approach to the problem. The conversation highlights the importance of understanding power as energy per unit time in this context.
asdf1
Messages
734
Reaction score
0
In this question,
"Solar energy reached the Earth at the rate of about 1.4 kW per square meter of surface perpendictular to the direction of the sun. By how much odes the mass of the sun decrease per second owing to this energy law?"

What is your first reaction to this question? How would you make the first move?
 
Astronomy news on Phys.org
First find the surface area of a sphere with the radius of the Earth's orbit. Multiply by that 1.4kW flux and you'll get the total power output of the sun. Then, apply E=mc^2...
 
What is your first reaction to this question?
That it's a homework question, and has been posted in the wrong part of PF?
 
it's not homework~
i'm trying to self-study modern physics...
 
russ_watters said:
First find the surface area of a sphere with the radius of the Earth's orbit. Multiply by that 1.4kW flux and you'll get the total power output of the sun. Then, apply E=mc^2...

@@a
how did you think of why to "analyze it" that way?
 
asdf1 said:
@@a
how did you think of why to "analyze it" that way?
The question posed - a power flux - 1.4 kW per square meter of surface.

So to get the total power (in kW), one simply multiplies the flux (energy/unit area) by the total area, as Russ indicated. This application assumes that the flux is constant in all directions.

Conversely, to get the flux at some distance, simply divide the total power (again assuming the energy is isotropically distributed) by the area of a sphere at that distance (radius of the sphere).

Remember power = energy per unit time.
 
thank you very much! :)
i think physics is interesting, but i often have trouble making the first step...
thanks so much for explaining!
 
Back
Top