I like my astronomy class, but i need help with the math. Got questions that are stumping me can anyone help?

Question

I'm an English Literature major, so math's not really my forté. Here they are:
#1: A regular incandescent lightbulb radiates most strongly at a wavelength of 3000 nm.
a) How hot is the filament? give your answer in Kelvins
b) In which part of the spectrum does it emit the most radiation: visible, or infrared (heat)?
#2: Use the orbit of one of Jupiter's large moons to calculate the mass of Jupiter. You will need to use the size of the orbit and orbital period. Using the formulas mass= 4 X Pi-squared/G (universal gravitational constant) X A-cubed/P-squared (A=orbit, P=orbital period) A has to be in meters and P has to be in seconds. What i've figured out so far for this one is that A= 422,000,000 meters, and P= 270,432 seconds, but take in mind my calculations may be incorrect.
I would appreciate any assistance, even if it just points me in the right direction for figuring these out, because right now, i really don't know know where to start.
Thanks!!!

Answer 1

1) a) Look up the blackbody curve. There will be an equation relating the peak wavelength and the temperature.
b) The visible spectrum is 400 - 700 nm. This is a longer wavelength, so it's not visible - it's redward (longer) than visible, so it's infrared.
2) You know that G=m*M/r^2, where G is the gravitational constant, m is the mass of the moon, and M is the mass of Jupiter. You can google G, you were probably given m, and you can get r - because A=2*pi*r, the circumference of the orbit.

So that would leave you with the equation M=G * (A / (2*pi) )^2 / m.

EDIT: Hi Stacie. Yep, as KW.. said, it's algebra. Just rearranging equations. I know it can seem daunting at first, but when approaching a problem like this, ask yourself - what do I have and what do I need to know? Here, we were given the mass of a moon and it's orbital period. We were asked ot find the mass of the planet, so you need an equation that will relate the stuff you have to what you need. Think of what you can get with what you have, and then plug it in to get what you need.

Answer 2

Sorry, I can't improve on eri's answer but I can reply to staceyy. It's algebra.
She started with an equation which is supposed to be known by the student. Then there was a question which asked for one of the factors in that equation. The others are either given or can be looked up. The only factor missing is M for Jupiter, (Jupiter's mass, the thing she's trying to find.).
So since M is what you are solving for, you have to re-write the equation solving for M. This is where the algebra comes in. After she re-wrote the equation to solve for M, the rest is arithmetic.

Here's a simpler example. D=RT, where distance is D, (in feet), speed is R, (in feet per second), and T is time in seconds.
Q: A car goes 500 feet in 10 seconds. What is its speed?

So now the equation would read 500=R * 10. But you want to solve for R, so you re-write the equation for R.
R=D/T or R=500/10 or R=50 feet per second.

Voila!

Answer 3

#1
a)
T = 2.897768551*10^6 nm °K/3000 nm
T = 965.92°K
b) Most of the radiation at this temperature will be in the form of infrared. " . . . some people may be able to perceive wavelengths from 380 to 780 nm." This is a very small range pretty far removed from the peak intensity of 3,000 nm, which is well within the infrared range.

Using Newton's Law of Universal Gravitation and assuming a circular orbit for simplicity,
mv^2/r = GmM/r^2
v = rω
ω = 2π/P
so
v = 2πr/P
v^2 = 4π^2r^2/P^2
mv^2/r = m(4π^2r^2/P^2)/r
GmM/r^2 = m(4π^2r/P^2)
Note that the mass of the moon cancels out:
GM/r^2 = (4π^2r)/P^2
M = (4π^2r^3)/(GP^2)
G = 6.67428*10^-11 m^3/kg-s^2
Plugging all the numbers in,
M = (4π^2(422*10^8 m)^3)/((6.67428*10^-11 m^3/kg-s^2)(270,432 s)^2)
M = (4π^2(7.5151448e+25 m^3)/((6.67428*10^-11 m^3/kg-s^2)(7.3133466624e+10s^2))
Rounding to 7 significant digits for display,
M = 2.966860810^27 kg/4.881132
M = 6.078221*10^26 kg

This does not agree very well with NASA's value of 1.8987*10^27 kg, being off by a multiple of 3.123776, which is close to π. I double-checked my calculations and don't seem to be able to find the error.

Answer 4

Whups; I'm in deeper then you. Post your question at the School of Math and maybe someone feels about Math as you do about Astronomy. Good luck.