100 points total
Kepler's 3rd Law states that
Now, solve for the period of Neptune:
PN2 = (aN/a)3*P2
Now take the square root of both sides:
PN = (aN/a)3/2*P
Now plug in the values and do some arithmetic
PN = (30.06 AU/1 AU)3/2*1 year = 164.8 years
Newton's version of Kepler's 3rd Law includes information about the masses of the objects involved.
So again, doing the problem with ratios will save much time and hassle. For the Earth around our Sun the equation reads:
For the planet Bajor the equation reads
So the ratio between the two equations reads
Now, M* = 15M, so we plug in the values we know...
Now solve for aB...
Weight is the force that gravity exerts on an object. Newton's Universal equation for gravity states that
Likewise, on Mercury the goats weight would be
Where, , is the symbol for Mercury. So, let's make a ratio again. Hmm, seems like these ratio things are pretty darned useful...
Recall that the relationship between frequency and wavelength of a wave depends on its speed. For light we write
By the same method as part (a) you will find that for the green laser
All light travels at exactly the same speed through a vacuum. This includes red, green, infrared, radio, X-rays, etc. (i.e. all of the electromagnetic radiation).
Level 1 has is an energy E1 = 0 and level 3 has an energy E3 = 7. So the difference between the two levels is the energy a photon must have to be absorbed and move an electron from level 1 to level 3. That difference we write as E1-3 = 7. ( is the mathematical symbol used to indicate the amount of change in a quantity). So this is just a matter of finding the differences for all transitions in these 8 levels. When you do that you discover that there is only one other transition with the same energy difference, and that is E5-8 = 7
Moving from level 8 to level 4 gives off a photon with energy E8-4 = 17 - 9 = 8, and Moving from level 4 to level 3 gives off a photon with energy E4-3 = 9 - 7 = 2. So the ratio is
Moving from level 3 to level 2 gives off a photon with energy E3-2 = 7 - 5 = 2. If an identical atom nearby had an electron in level 5 its energy would be E5 = 10. If it were to absorb that photon it would have energy E = 12. There are no levels with energy equal to 12. So, No the emitted photon could not be absorbed by an identical atom with an electron in level 5.
The absorbed photon for the transition from 2 to 4 has an energy E2-4 = 9 - 5 = 4. And it has a wavelength, = 10,000 Å (1 x 10-6 m). The energy that a photon would need to boost the electron up to level 6 from level 2 would be E2-6 = 13 - 5 = 8. Now, wavelength and energy of a photon are related by
2-6 = 10,000 Å*(4/8) = 5,000 Å
No, a person wearing a blue shirt is not hotter than a person wearing a red shirt. The light you see that tells your eyes what color a shirt is is not emitted light, but reflected light. Wien's Law is a relation between the temperature and color (wavelength) of emitted light for a blackbody. When you see a shirt to be blue or red it is so because of the chemical dyes in the material that absorb and reflect different colors. Reflected light does not generally give you any information on the temperature of an object.
The last part was just a joke. You didn't have to answer it. On the old Star Trek series whenever there was a landing party it would contain Captain Kirk (gold shirt), Mr. Spock (blue shirt), Dr. McCoy (blue shirt), and Ensign Joe Schmoe (red shirt). The guy in the red shirt was always the first to die.
Star Bart is 100 times dimmer than the 16 cm telescope can see. In order to just be able to see this star we need to collect 100 times more photons. In order to collect 100 times more photons you need 100 times the collecting area. The area of a telescope is related to the diameter: A = (D/2)2. So let's make a ratio to find out how large our new telescope needs to be.
Yes. Photographic film can collect more photons by waiting for them. Your eye can only add up the light for a short (and fixed) period of time (about 1/30th of a second), but film can be exposed (add up the light) for many hours.
If we assume that the film is as sensitive as your eyes then you would need to wait 100 times longer than your eyes can see, about 3 seconds.
The CCD can detect fainter stars because it is much more efficient. Only a few percent of the photons incident on film cause the film to react, whereas 80 to 90 percent are detected on a CCD.
Since the efficiency factor for a CCD is nearly 100 times that of an eye or film, we would only have to expose the CCD for about 1/30th of a second to see star Bart. This is the same frame rate as video (no accident, since speed of video is tailored to the speed of the eye). So, we should be able to see star Bart by putting a CCD video camera at the focus of the 16 cm telescope.
Earth's atmosphere blocks many wavelengths of light detectable by such a telescopes (ultraviolet and near-infrared). Getting above the atmosphere is necessary to see those wavelengths.
Earth's atmosphere also causes incoming light to scatter about or blur. This limits a telescope's spatial resolution no matter how big the primary mirror. In space, better spatial resolution can be achieved with only moderate sized telescopes. HST does about 10 times better than ground based telescopes.
Earth's atmosphere absorbs some light, making faint objects more difficult to detect. The atmosphere also glows from scattered moonlight, city lights, aurorae, sunlight, and molecules in the air that emit at various wavelengths (especially infrared). This makes it even more difficult to see faint objects. In space, a telescope can see much fainter and hence farther into the Universe's past.
Sounds like a ratio is called for. In class we discussed this very thing. We saw that there is a relationship between the temperature of a black body emitter and the amount of energy per unit area per unit time, , it emits. It's called the Stefan-Boltzmann Law:
So if the photosphere's temperature is 6000 K and the sunspot's is 2000 K less, then the sunspot's temperature is 4000 K. So let's take the ratio of energy output for each region:
ss = 0.2*ph
So, the sunspot only puts out 20% the amount of radiation as the surrounding photosphere.
Nothing at a temperature of 4000 K is black. The sunspots do indeed put out a large amount of energy. It looks dark next to the photosphere as a pure consequence of contrast. The photosphere puts out 80% more light per unit surface area per unit time than the sunspot and so appears so much brighter that it makes the spot look dark in comparison.
The seasons would be less extreme, in fact there would be no seasons at all! The entire reason that seasons occur is because any given place on Earth receives different amounts of energy from the Sun throughout the year. This occurs because Earth's axis is tilted 23.5° from perpendicular. When the axis is perpendicular any given place on Earth will receive the same amount of energy all year long. The Sun would rise and set everyday at 6 o'clock, giving 12 hours of light and dark. The Sun would also always rise from due East and set due West, there would be no change in this position throughout the year. Thus there would be no large changes in the climate. Earth's change in distance from the Sun during the year would have such a slight effect that it would not influence the climate noticeably.
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