General: Please check HW#2 solution carefully!
Question #1
I don't understand where the amswers for number two came from.
I did this:
>
> a) Area 1 = . and Area 2 =
> (Area1)(T1)^4 / (Area2)(T2)^4 and the answer I got was .273
>
> What am I doing wrong here?
>
> b) For power ratio I used the ratio: Intensity 1*Area
1 / Intensity 2 *
> Area 2, solving for P in the equation given. Is it correct
to then
> multiply the ratio of intensities by the Area1/Area2?
when I do this using my number I get .0437, and your nubmer from part
a I
> get .0938
> I don't understand why I am not getting the correct answers.
Thanks!
Answer:
2a)
In order to get the right answer note that I am asking for the intensity
PER UNIT AREA OF THE SPHERE, ON THE SPHERE. So the area is not important.
You got the answer for b in a. (actually, 1/the answer).
2b.) NOW I am asking how to get the TOTAL POWER, and therefore the
intensity at a given distance from either sphere. You will get the
expression of A.
Question #2
Ran, I have a question on the tutorial. On the last problem,
#8, I know =
> there is a 16 point magnitude difference. Do I multiply the
given watts =
> by 10x10x6 to get the new wattage? Or would that be an outrageously
=
> huge number? Thanks!
You actually get a ~37 magnitude difference. Every time you get the
sun
10 times further away (NOT 100!) the magnitude increases by 5. So if
you
get the sun 10 times further the magnitude goes from -26.7 to -21.7.
When
you get the sun 10 times further than that (100 times the original
distance!) the magnitude is -26.7+5+5=-16.7. We have to set the sun
at a
distance in which the magnitude is whatever you found in question 2.
Then
change the distance from 100x100x100x.... x the distance to the sun
units
of meter (or 10x10x10x.....x distance to the sun, if you find that
easier).
Last find the distance in light years.
No need to find, or use, the POWER of the sun.
Question #3
About HW question #4:
I have one more question. In question 4 they give you the power
and
temp and want the radius. However there is no equation that has
power
and temp together that will help me. If you could throw me a
hint I'd
appreciate it.
Here is the hint: You do have an equation that includes the radius:
The area of a sphere is related to the radius of the sphere through:
A=4pir^2.