Thanks very much i haven't come across the term 'Gaussian surface' before but i completely understand now.
If you take the distance between the star and the Earth as the radius of the sphere and calculate its surface area, then one square meter is receiving 1.7x10^-12 Watts of power so...
1. A star at a distance of 11 light years from the Earth provides us with a power per unit area of 1.7x10^-12 W m^-2. Calculate the power output of the star.
2. 1ly=9.4608x^15 m therefore 11ly=1.040688x10^17 m
3. How do you work it through? I've never understood these questions...
1. Question: In one region of a nebula the mean separation of atoms is 2.3x10^-3 m. What volume of the nebula contains enough atoms to make a star, similar to the sun, of radius 7.0x10^8 m within which the mean distance between atoms is 1.0x10^-10m?
2. Volume of sphere 4/3(pi(r^3 ))...