# Some big numbers!

### What is the Earth's average velocity as it orbits around the Sun?

According to Kepler’s first law, the Earth’s orbit around the Sun is in the shape of an ellipse with the Sun at one of the foci. The eccentricity of this ellipse is only 0.016; the eccentricity of an orbit varies from 0 (a circle) to 1 (a very flattened ellipse. Therefore, for the purposes of doing a simple estimation, we can approximate the orbit as a circle. Kepler’s second law states that a planet will sweep equal area in equal amounts of time. Since the orbit is elliptical, the Earth is sometimes closer to the Sun (in early January) and moves faster. Again, for our estimation, we will try to just find the average velocity, which is the change in distance over the total amount of time. The distance of the orbit is just the circumference of a circle with the radius r = the average distance from the Earth to the Sun. We know that the Earth is 1AU (astronomical unit) away from the Sun, which is about 1.5 x 10^11 m, and that one complete orbit takes a year. We can then do dimension analysis to figure out this velocity in a number of different units.  That’s 1000x faster than the average NY highway speed limit!

### What is the total energy output of the Sun?

We can start with the average irradiance on Earth, which is about 1400 W/m^2 above the atmosphere and 1000 W/m^2 here on the surface. Note that this is a measure of the intensity - which is the power per unite area. We can picture a sphere of radiation around the Sun with a radius of 1 AU - the distance from the Sun to the Earth (about 1.5 x 10^11 m). So the 1400 W/m^2 refers to the power radiated by the Sun on the surface of this sphere. To find the total amount of energy that the Sun outputs per second, we can multiply the intensity by the area of sphere. So the total power output of the Sun is about 4 x 10^26 watts, or “400 yottawatts, that’s a lotta watts!” as John pointed out.