Question

The luminosity of the Sun is 4*10^33 ergs/s, and its distance to the Earth is 1.5 * 10^13 cm.

You are tasked with building a solar power plant in the Arizona desert, using solar panels with 10% efficiency. How large an area (km^2) must your solar panels cover to match the power output of a large nuclear powerplant (about a Gigawatt)? Please enter your answer in units of km^2.

Answer 1

Use SCI units
1 erg / 10^-7 Joules
4*10^33 ergs/s (10^-7 Joules/erg ) = 4*10^26 Joules/s
= 4*10^26 Watts
Radius from sun = 1.5 * 10^13 cm (1 m / 10^2 cm)
= 1.5 * 10^11 meters
find Watts / m^2 at earth distance
surface area of circle around sun
=4 pi R^2 = 4 * 3.14159 * (2.25*10^22)
= 28.3 * 10^22 m^2
so power / m^2 = 4*10^26 Watts / 28.3*10^22 = 0.141 * 10^4 Watts/m^2
0.141 *10^4 Watts/m^2 * (A/10) = 1 gigawatt = 10^9 Watts
A/10 because only 10% eff
0.141 * 10^4 A = 10^10 m^2
(1/0.141)*10^6 m^2 * 1 km^2/10^6 m^2
= 7.09 km^2
not bad but check my arithmetic !