Astronomy 102 Problem Set #2 Solutions

Equations needed, Please note the changes!
(1.) Wien's Law:  T= 3 x 106nm/l
The temperature is in Kelvin (Degrees Celsius +273.15).

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,

where 0.003 K m is often referred to as the Wien Displacement constant (the "K" and "m" are units). We're given the star's temperature, so Note that the Kelvin units on the top and bottom cancel out, leaving meters as the units for our answer. This makes a lot of sense, since our answer is supposed to be a wavelength.

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,

A quick consultation with your book or class notes (or perhaps your memory from the spectroscopy lab) should lead you to conclude that light of this wavelength is red. In fact, it's so red it might be called infrared, since our eyes can only see light of wavelengths 400-700 nm or so. However, even though we can't see the light at the peak of the spectrum, we can still see this star, since blackbodies emit over a broad range in wavelength. Furthermore, since the peak occurs in the infrared portion of the spectrum, we should expect to see more red light than blue light, and so our consclusion that this star appears red still holds.

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

where T is the temperature of the object and sigma is the Stefan-Boltzmann constant, which is 5.67 x 10-8 W/(m2 K4). So, if we want the ratio of the intensity of the big sphere to the intensity of the little sphere, we just use this expression twice where the subscript B denotes quantities related to the "big" sphere and the subscript L denotes quantities related to the "little" sphere. The sigma's are constants and appear on the top and bottom, so they immediately cancel out, leaving which says that the ratio of intensities is equal to the fourth power of the ratio of the temperatures. So we can now plug in the two temperatures and get Does this make sense? I mean, "shouldn't the bigger object be brighter?" you might say. Ah, it sounds to me like you're confusing intensity with power. Remember that the intensity tells you how much each bit of the sphere is emitting, not the total emission from the sphere. The surface of little sphere will appear brighter than the surface of the big sphere because it's hotter. We don't yet know which of the objects generates more power.

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.

where P is the power emitted by an object, I is the intensity of the object, and S is the surface area of the object. Using the Stefan-Boltzmann relation for the intensity and the standard formula for the surface area of a sphere, we get where R is the radius of the sphere. Now let's set up the ratio of powers just as we set up the ratios of intensities above: where the B and L subscripts again refer to the "big" and "little" spheres. Once again the sigma's drop out, as do the 4's and pi's, so we end up with From above, we learned that TB4/TL4 = 0.586 so the ratio expression reduces further to Now we can resort to numbers, and putting in the values given in the beginning of the problem, Note that since this is a ratio of quantities, there are no units. The big sphere emits 3.66 times the power that the little sphere emits.

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:

Only this time, the big sphere is the star, and the little sphere is our sun! For RB we can substitute RL*2.5, since its 2.5 times the radius of the sun. So the above equation becomes: The radius of the sun cancelled out! So what's the temperature of the big sphere? 7300 K. And the little sphere? Here you would have to consult the 1st textbook, or your CDROM Redshift 2 (Hey, its good for something!): The surface temperature of the sun is 5800 K (pg 331 FETU, or click on the sun in the CDROM). You get:


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.
 

This time we have the left hand side all figured out: PB/PL = (1 x 103 Lo) / Lo (Power of the sun is its luminosity, right?).
We also know the temperatures of the sun and the B0 star.


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:
 


Problem #5: (For extra credit only, so that you may get 12.5/10 for this HW assignment) A spherical planet of radius 6 x 106 m and surface area 4.5 x 1014 m^2 is 1.5 x 1011 m from a G2 star, with 1 Lo. One side of the star is always facing the sun. Assume that the albedo is 100%, that is: all light from that star is absorbed by the planet, and then re-emitted as black-body radiation.
a. What is the intensity (power/area) of light from the star at the surface of the planet?
b. What is the surface temperature of that planet?
c. What is the total luminosity of the reflected light in units of Lo?
d. What may be the implications of this result for the search for extra-solar planets similar to earth?

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.