Can someone please help me with this question!

The intensity of sunlight reaching the earth is 1360 W/m2.

1. What is the power output of the sun?

2. What is the intensity of sunlight on Mars?

The answer that I got is 1.73*10^17. Please tell me which radius I should use sun or earth. And please show me the steps. Thanks is advance.

At a distance of R = 1.50E11 m (the Earth's orbit radius from the Sun), the Sun's radiation intensity is 1360 w/m².
At this distance the Sun's total energy is spread-out over a sphere or radius 1.50E11 m ..

The area of this sphere is A = 4πR² ...
A = 4π(1.50E11)² = 2.83E23 m²

Total power spread over this sphere area equals the Sun's total emission power P (assuming no absorption loss in travelling through empty space) ...

P = Am² * 1360w/m² = (2.83E23)*(1360) .. .... .. ►P = 3.85E26 W


(2) At the orbit radius of Mars (2.29E11 m) the sphere area is ..
A(m) = 4π(2.29E11)² = 6.59E23 m²

sunlight intensity = P/A(m) = (3.85E26 W) / (6.59E23 m²) .. .. ►= 584 W/m²

You want to calculate the radiation reaching the Earth, and you know that each square meter of sunlight reaching the Earth carries 1360 watts of power. So you need to know how many square meters of sunlight the Earth receives. But consider that sunlight does not land on the planet in an even way. We know that at the poles the sun hits the Earth at a 90 degree angle, so the amount of sunlight that lands at the poles is almost zero.

If turns out that if you imagine the amount of sunlight that would fall in a flat circle the size of the earth, you would have exactly the area that the sunlight hits the Earth, even if it actually hits it rather unevenly.

So the area you need, A, uses R, the radius of the earth in meters: Pi * R ^2.

If you calculate that A and multiply that by 1360 watts per square meter, you will find the number of watts the earth receives.