Homework #3: Solutions
Homework #3: Solutions
This Greenhouse isn't so good for the Plants
- Contrary to popular belief, some amount of greenhouse effect is useful; Earth would be a much colder planet without it. What, then, is the fear we have of dumping lots of carbon dioxide (CO2) and other "greenhouse gases" into Earth's atmosphere?
The fear is that Earth's greenhouse effect could become a runaway process like the one that has occurred on Venus. If we artificially dump alot of greenhouse gases into the atmosphere, such as carbon dioxide, this could slightly raise the temperature of Earth. This slight temperature rise could heat Earth enough so that some of the carbon dioxide dissolved in the oceans and locked away in rock will be released into the atmosphere. What we have now is a positive feedback loop. The released CO2 will further heat the planet which will lead to more of it being released from the rocks and oceans and so on. Eventually this will melt the polar ice caps causing major flooding on coastal regions, The ice caps are also very reflective and reflect some of the Sun's energy back into space. With them gone the planet will heat even more. Dramatic climatic changes will occur, there will be great storms, crop failure, famine, drought and lots of other very bad things.
- Suppose a planet had an atmosphere that was opaque at optical wavelengths but transparent in the infrared. Describe how the effect of this type of atmosphere on the planet's temperature differs from the greenhouse effect.
Most of the Sun's energy resides in the optical wavelengths of light that it radiates which heat the Earth's surface during the day (since Earth's atmosphere is transparent to optical photons). At night, the warm surface of Earth radiates the day's collected heat in the form of Infrared photons. Since the Earth's atmosphere is relatively opaque to infrared, most of these photons remain trapped within the atmosphere, and the Earth stays warm. If we have a hypothetical planet with an atmosphere opaque to optical photons and transparent to infrared photons, however, the situation would be reversed. Not much heating of the planet's surface would occur during the day since optical photons cannot reach the surface. Moreover, when the surface radiates at night in the form of infrared photons the meager amount of heat collected during the day can more easily escape. Thus the planet would be a very cold one.
Terrestrial vs. Jovian Planets
Compare and contrast five properties of the Jovian planets with those of the terrestrial planets.
| ||Jovian Planets ||Terrestrial Planets |
|Distance from the Sun ||Far ||Near |
|Size ||Large ||Small |
|Composition ||Mostly Gas ||Mostly Rock |
|Density ||~1 g/cm3 ||~5 g/cm3 |
|Moons ||Many ||Few |
|Rings ||Yes ||No |
|Rotation ||Fast ||Slow |
I know Io
What sort of activity has been seen on Jupiter's moon Io, and what is Io's heat source thought to be?
Volcanism (volcanic activity) has been observed on Io. In fact, it is the most volcanically active body in the Solar System. This is caused by the weakening of Io's crust and the large amount of heat produced in Io's interior, both of which result from the action of Jupiter's gravity on Io. Since Jupiter is such a massive object, it exerts tidal gravitational forces on Io. As Io moves in its elliptical orbit (its orbit is pulled into an ellipse by the gravity of Jupiter's other massive moons), these tidal forces change, leading to Io being alternately being squashed and distended. This results in a huge frictional heating of Io's core and a buckling of its surface, resulting in rampant volcanism.
When Snowballs Attack
The kinetic energy of a moving body is given by KE = (1/2)mv2, where m and v are the mass and speed of the body, respectively. If m is given in grams and v in cm/s, the units of energy turn out to be ergs. 1 megaton of TNT is equal to about 4 x 1022 ergs.
- Suppose a comet is heading on a collision course with Earth. If the comet is a sphere of radius 1 km, with a density (mass per unit volume) of 1 g/cm3 (like water ice), what is its mass?
density = m/V. So that means if we want the mass and we know the density and volume, V, we write
mass = density x V
In this case the comet is a sphere, and for a sphere of radius R: V = (4/3)R3. So we have
m = (1 g/cm3) x (4/3)(1 km * 105 cm/1 km)3 = 4.19 x 1015 g
- If the comet's speed relative to Earth is 40 km/s, calculate its kinetic energy in ergs (be sure to keep track of your units).
The velocity is v = 40 km/s * (105 cm/1 km) = 4.0 x 106 cm/s. So we plug that into the formula for kinetic energy
KE = (1/2)*(4.19 x 1015 g)*(4.0 x 106 cm/s)2 = 3.35 x 1028 g*(cm/s)2 = 3.4 x 1028 ergs
- Convert your answer to megatons of TNT. This is the energy that will be deposited by the comet when it collides with Earth.
KE = 3.4 x 1028 ergs * [(1 megaton of TNT)/(4x1022 ergs)]
KE = 8.4 x 105 megatons of TNT
- For comparison, very powerful nuclear weapons typically release an energy of 50 megatons of TNT. By what factor is your answer in part (c) larger than this? Are you likely to have a good day when the comet hits?
Take the ratio of energies released by an H-bomb and the Comet collision:
Ecomet/Ebomb = (8.4x105 megatons)/(50 megatons) = 1.7 x 104
So the comet will unleash the same devastating energy as 17,000 H-bombs!! That would not be a good day.
Suppose Star Cabra is approaching you with a speed of 150 km/s.
- Will absorption lines in the spectrum of Star Cabra be blueshifted, redshifted, or unshifted from their laboratory wavelengths?
The star is approaching you so the wavelengths of light will be compressed to shorter wavelengths as they approach, hence they will be blueshifted.
- Calculate the expected wavelength of H line (the second hydrogen Balmer line, n = 2 to 4 transition), whose laboratory wavelength is 0 = 4861.3 Å. (Recall that /0 = v/c and c = 3 x 105 km/s.)
Since the wavelength is blueshifted the amount of shift, , will be negative. Now plug in the appropriate values to the doppler shift formula.
- /(4861.3 Å) = (150 km/s)/(3x105 km/s) = 5 x 10-4
Multiply both sides by 4861.3 Å and get
- = (5 x 10-4)*4861.3 Å = 2.4Å
- = - ( - 0) = 0 - = 4861.3 Å - = 2.4 Å
= 4861.3 Å - 2.4 Å = 4858.9 Å
- Now you would like to know if Cabra has any planets around it. What would you look for in the spectrum of Cabra?
If Cabra has a planet around it the planet would gravitationally tug it back and forth as it orbits Cabra. So Cabra would sometimes be moving a little faster than 150 km/s toward us and sometimes a little slower than 150 km/s. The change in the blueshift would be periodic (having the same period as the planet's orbital period about Cabra). So we would look for a periodic redshift and blueshift of the H line around the value of 4858.9 Å.
How far, how bright?
Let's consider parallax and the distances of the stars.
- What is the distance to a star whose parallax is 0.4 arc seconds?
dpc = 1/p(")
Therefore the distance in parsecs to this star is
dpc = 1/0.4(") = 2.5 pc
- What is the parallax of a star whose distance is 25 parsecs?
Ooo tricky. We need to invert the equation. Thus,
p(") = 1/dpc
Therefore the angle in arc seconds is
p(") = 1/25 pc = 0.04"
- Star Buffy has a parallax of 0.3 arc seconds as measured from Earth. A mad scientist sends a probe into orbit around the Sun at a distance of 3 AU from the Sun. The probe measures the parallaxes of stars with the same general technique used on Earth. What is the corresponding parallax of Star Buffy measured by the probe?
For this you need to recall the formula in its original form before defining a parsec and measuring the angles in radians:
d = 1 A.U./p
Because the baseline of the triangulation method was 1 A.U., if we were to increase the baseline to 3 A.U. as the problem suggests, the formula would thus read:
d = 3 A.U./p
So, d = 1 A.U./p1 and d = 3 A.U./p3 therefore,
1 A.U./p1 = 3 A.U./p3
p3 = 3p1 = 3(0.3") = 0.9"
Now let's consider star Angel who is at a distance of 20 parsecs
- if Angel is suddenly moved to a distance of 200 parsecs, what will his new apparent brightness be, relative to his old apparent brightness?
Well, brightness is given by
b = L/(4d2)
and so if we move the star from d1 to d2 = 10d1 then we can make the ratio of the brightnesses.
b2/b1 = L/(4d22) * (4d12)/L = (d1/d2)2 = (1/10)2 = 0.01
b2 = 10-2b1
- How will his luminosity be affected by the change in distance?
Luminosity doesn't change. Just because you physically move a star doesn't mean that there is going to be any change in the amount of energy it is giving off every second. Luminosity is an intrinsic property of the star.
What's the Flux?
Flux is a name we give the for a quantity such as apparent brightness of an object, such as a star. b = L/(4d2), where d is the distance to the object, L is the luminosity of the object. It is the amount of energy received (at a given distance) per unit area per unit time.
- calculate the brightness of the Sun at the distance of Earth, in units of ergs * s-1 * cm-2, using a luminosity L = 3.83 x 1033 ergs/s and a distance
d = 1.5 x 1013 cm (which is 1 AU).
This is just "plug and chug"...
b = L/(4d2)
b = (3.83 x 1033 ergs/s)/[4(1.5 x 1013 cm)2]
b = 1.35 x 106 ergs*s-1*cm-2
This is the Sun's brightness at Earth.
- What is the brightness of a 100 Watt light bulb that is 10 cm away? (NOTE: 1 Watt = 107 ergs/s)
blb = (100 Watts * 107ergs/Watt)/[4(10 cm)2]
blb = 8 x 105 ergs*s-1*cm-2
- Compare your answers in parts (a) and (b). What can you conclude?
b/blb = (1.35 x 106 ergs*s-1*cm-2)/(8 x 105 ergs*s-1*cm-2) = 1.7
This is somewhat interesting since the two numbers differ by a factor of less than 2. It says that you receive just a little less energy per second from a 100 Watt light bulb that is 10 cm away as you do from the Sun. If you bring the light bulb a little closer they would be nearly identical. You would feel just as hot as you do standing out in the midday Sun and the light bulb would be just as difficult to stare at as the Sun is!
Center of Mass
Suppose Stars Fred and Wilma constitute a double-star system. Fred is 4 times as massive as Wilma. Calculate Fred's distance from the center of mass relative to Wilma's distance from the center of mass.
Recall that for two masses, the ratio of their masses is equal to the inverse ration of their respective distances to the center of mass. In class I wrote this as
m1a1 = m2a2
So for Fred and Wilma we have:
mFaF = mWaW
mF/mW = aW/aF = 4
aF = (1/4)aW
So the center of mass is 4 times closer to Fred than it is Wilma because he is 4 times more massive.
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