- #1
Poop-Loops
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- 1
I'm taking Astr. 321 right now. We were supposed to read the first 3 chapters. They go through the different planets (some detail, but nothing too complicated), the sun, asteroids, etc. The lectures went over this in more detail and went through things like which elements were created first, how the rest were created, etc. Basic stuff.
Then I get the homework:
Question 1: If your hand is 7.5cm away from a 100 watt light bulb, it gets the same energy as it would get from the sun on a sunny day. Determine the wattage of the sun. Easy ratio (plus some squares).
Next problem: How much mass is it fusing per second (easy, E=mc^2).
3rd: " Estimate how long the Sun can generate this power knowing that mass of 4 protons that effectively fuse to form a He nucleus in the core of the sun is 1.008 times the mass of a helium nucleus. The missing mas is converted to energy. Nuclear fusion is somewhat inefficient in the Sun because it only occurs in the core- so assume that only 10% of the Sun's mass is ever hot enough to burn H into He and 90% of the Sun's hydrogen is untouched. Assume the Sun starts off as pure hydrogen (not quite true) and that the Sun's luminosity is constant during most of its life, its "main sequence lifetime" when it burns H into He. (answer in years) "
While typing this I came up with an idea: do I take the mass of the Sun from the book (it's 1.99e^30 I think) and then only take Mass - (1/1.008) as the mass released as energy, and at 10% of that speed of fusion?
If not, then I am stumped. I've been working on this since like 2 days ago with a friend.
EDIT: My prof. is Donald Brownlee, btw. ;)
Then I get the homework:
Question 1: If your hand is 7.5cm away from a 100 watt light bulb, it gets the same energy as it would get from the sun on a sunny day. Determine the wattage of the sun. Easy ratio (plus some squares).
Next problem: How much mass is it fusing per second (easy, E=mc^2).
3rd: " Estimate how long the Sun can generate this power knowing that mass of 4 protons that effectively fuse to form a He nucleus in the core of the sun is 1.008 times the mass of a helium nucleus. The missing mas is converted to energy. Nuclear fusion is somewhat inefficient in the Sun because it only occurs in the core- so assume that only 10% of the Sun's mass is ever hot enough to burn H into He and 90% of the Sun's hydrogen is untouched. Assume the Sun starts off as pure hydrogen (not quite true) and that the Sun's luminosity is constant during most of its life, its "main sequence lifetime" when it burns H into He. (answer in years) "
While typing this I came up with an idea: do I take the mass of the Sun from the book (it's 1.99e^30 I think) and then only take Mass - (1/1.008) as the mass released as energy, and at 10% of that speed of fusion?
If not, then I am stumped. I've been working on this since like 2 days ago with a friend.
EDIT: My prof. is Donald Brownlee, btw. ;)