For the rest of the problem set you will need to consider intensity:
Intensity is the power per area.
I=P/A
I, the intensity, is the power (Watts)
per unit area (m2).
Note: Power and Luminosity are the same thing. (We tend to call
the power of a luminous object luminosity)
Intensity from a "black body":
(2.) I=constant T4
note that T4 is TxTxTxT !!!!
(the constant, named sigma, or s in the book pg. 366, is equal to 5.67 x 10-8 Watt / (m2 K4).)
We don't really need to know what the constant is. Rather, for two hot objects, indexed 1 and 2, of different areas with the same temp we get:
(2.1) P1/P2 = Area1 / Area2
And both constant and temp. cancel out.
For the two balls with the same areas, but different temperatures:
(2.2) P1/P2 = T14/T24
And both constant and area cancel out.
If the two objects have DIFFERENT temperature AND DIFFERENT area the ratio of intensities is
(2.3) P1/P2 = Area1T14/(Area2T24).
The area of a sphere is:
(3) A=4pr2
Where r is the radius of the sphere.
Problem #1: The surface of a star has a
temperature of 3800 K and emits as a blackbody.
a) At what wavelength does this star emit the most light?
b) What color would this star appear to be to the human eye?
Solution: We're given a temperature and asked at what wavelength the star emits the most light. Since this star (and all normal stars) emits like a blackbody, this question is equivalent to asking at what wavelength the blackbody spectrum peaks. Using the Wien Displacement Law,
The above answer is correct, but because it's expressed in meters, it's hard to have a feel as to what color the star's emission would be. We would have an easier time estimating the color if our answer were expressed in nanometers, since these are more normal units for optical wavelengths. Since there are 109 nm in a m,
This star has a surface temperature quite a bit lower than that of the Sun (T = 5800 K), and so should have a peak wavelength longer than the Sun's peak (in the yellow), so it sounds like this answer passes a sanity check.
Problem #2: a) I have two spheres,
one with a radius of 0.2 m and the other with a radius of 0.5 m. I heat
the small one to a temperature of 2400 K, and the big one to a temperature
of of 2100 K.
a) What is the ratio of the intensities of the two spheres?
b) What is the ratio of the total power emitted by these
spheres?
Solution: There's an easy way and a hard way to do this problem. The hard way is to calculate the intensities and powers from the Stefan-Boltzmann Law and the surface area of a sphere, stuffing all of the relevant numbers in a calculator and coming up with numerical answers. However, since all we're interested in is the ratios of intensities and powers, we don't really need their actual values, and we can solve this problem with a lot less effort.
Let's set up the ratio of intensities directly. We know that for a heated object, the intensity (I) emitted by a patch of that object will be described by the Stefan-Boltzmann Law
While the intensity of a blackbody depends only on its temperature, the power emitted by each sphere also depends on the amount of surface area available for emission.
Problem #3: Calculate the total power emitted at all wavelengths by a star whose surface temperature is 7300 K, and whose radius is 2.5 solar radii. (You may assume that the star radiates as a blackbody.) Answer in units of solar power and solar radius.
Solution: This problem is essentially the same as the last one. Since we want the power in units of solar power, Psun, we can use the relationship:
What does this answer mean? It means that the star is 15.68 more
luminous (remember: luminosity is power. Same thing.)
Given that the luminosity of the sun is PL= 4 x 1026
Watt, the luminosity of the star is
PB=15.68 x 4 x 1026 Watt = 6.27 x 1027 Watt.
Problem #4: Calculate the radius of a Main Sequence B0 star (which has a surface temperature of 3 x 104 K, and a luminosity of 1 x 103 Lo, where Lo means "solar luminosities"). Answer in units of solar radius.
Again, we may use our ratios method. sphere B is the B0 star (this is
a guess, this time! it could be the smaller sphere!), and sphere L is the
sun.
What's left?
It's the radius! The radius of the star here is unknown. This time around
its best to use the actual radius of the sun here. From Appendix 2A of
your 1st textbook Rsun=696000km = 6.96 x 108m. The equation
above becomes:
Rearranging this equation we get:
Or
That's the answer. But were we justified in our assumption that
B is the bigger sphere? Just barely! By dividing the radius RB by that
of the sun we get:
Answer:
Note that all parameters above fit a planet EXACTLY LIKE EARTH near a star EXACTLY LIKE THE SUN!
a. Imagine that the star is surrounded by a sphere of radius 1.5 x 10^11 m. The area of that sphere is 4 pi x (1.5 x 10^11 m)^2= 2.83 x 10^23 m^2..
That means that the power per area of that sphere is
I= 1Lo/2.83 x 10^23 m^2 ! (See class demo).
Since Lo is the luminosity of the sun, 4 x 10^26 Watt, the intensity is:
I = 1420 Watt/m^2.
b. The surface temperature of the planet, in this case, is 0 K, unless there is an internal source of energy. That's because none of the energy is absorbed by the surface.
If, however, the albedo was 0%, like on a dark surface near earth's
equator, we would get:
I=1420Watt/m^2= 5.67 x 10^(-8) Watt/(m^2 K^4) x T^4.
or
T=397 K, which is not too far from the temperature on a very hot day:
T (celsius) = 397 - 273.15 = 124 Celsius or 255 Farenheit.
(The reason for this high number is that earth has albedo of less than 50%.)
A surface at this temperature emits in the far infrared, and is invisible if not for the light that it reflects from the sun.
c. The total power emitted by this planet is the intensity (Power/area)
x the area. Here the area will be
pi * r_(planet)^2. That's because only that much light is intercepted
from the star. (See class demo).
So , since pi*r_(planet)^2 is 1/4 of the planet's area,
Power = 1420 Watt/m^2 x (4.5 x 10^14 m^2)/4= 1.60 x 10^17 Watt
Note that the m^2 cancel out.
This power is so much smaller that the power of the sun:
Power / Lo= 1.6 x 10^17 Watt / 4 x 10^26 Watt = 4 x 10^(-10)
d. The distance to the nearest stars is hundreds of thousands of times
earth-sun distance. So the planet and its sun will be incredibly close
to each other. Yet, the reflected sunlight from that planet is 40 billion
times dimmer than the light from the star itself. That's a tall order,
right there, to try and find such a planet near such a star! We are not
quite there yet, but we are getting close: The next generation of space
telescope may be able to detect earth-size planets.