How Do You Calculate the Sun's Average Power Output?

AI Thread Summary
To calculate the sun's average power output, the energy reaching Earth's upper atmosphere is approximately 1350 W/m^2, and the distance from the Sun to Earth is 1.5 x 10^11 m. The surface area of a sphere at this distance is calculated as 2.25 x 10^23 m^2. Multiplying this area by the energy per square meter gives a total energy of about 3.0375 x 10^26 W. Dividing this by the time it takes for light to travel from the Sun to Earth, approximately 500 seconds, results in an average power output of about 6.075 x 10^23 W.
Dx
Hi!

Estime the average power output of the sun, given that about 1350 W/m^2 reaches the upper atosphere of the Earth. the distance from the Sun to the Earth is 1.5 x 10^11m

I have the formula E_o = [squ]2S/e_oc. This is what I have thus far...
E_o = [squ]2(1350W/m^2)/8.85x10^-12C^2/Nm^2)(.x10^8 m/s) = 312V/m.
I am lost on my calculations can someone help me solve for this,please
Thanks!
Dx :wink:
 
Last edited by a moderator:
Physics news on Phys.org
I don't know what that formula is supposed to be, but what you have to do is recognize that the power is radiated (approximately) spherically symetrically. You should be able to calculate the area of the sphere centered on the Sun and passing through the Earth. Once you get that, you can use the given information to calculate the power output.
 


Hi there, thank you for your question. To calculate the average power output of the sun, we can use the formula P = E/t, where P is power, E is energy and t is time. In this case, we are given the energy reaching the upper atmosphere of the Earth (1350 W/m^2) and the distance between the Sun and the Earth (1.5 x 10^11m).

To calculate the total energy reaching the Earth's upper atmosphere, we need to first calculate the surface area of a sphere with a radius of 1.5 x 10^11m. This can be done using the formula A = 4πr^2, where A is the surface area and r is the radius. So, A = 4π(1.5 x 10^11m)^2 = 2.25 x 10^23 m^2.

Next, we need to multiply this surface area by the energy reaching each square meter (1350 W/m^2) to get the total energy reaching the upper atmosphere of the Earth. So, E = (2.25 x 10^23 m^2)(1350 W/m^2) = 3.0375 x 10^26 W.

Now, to calculate the average power output, we need to divide this total energy by the time it takes for the energy to reach the Earth (which is the time it takes for light to travel from the Sun to the Earth). The speed of light is approximately 3 x 10^8 m/s, so the time it takes for light to travel from the Sun to the Earth is:

t = d/v = (1.5 x 10^11m)/(3 x 10^8 m/s) = 500 seconds.

Therefore, the average power output of the sun is: P = E/t = (3.0375 x 10^26 W)/(500 s) = 6.075 x 10^23 W.

I hope this helps! Let me know if you have any further questions.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top