Download ASTR1010_HW06

ASTR 1010 – Spring 2016 – Prof. Magnani
Answer Key – Homework 6
Question 1 – Ch. 7; #26
The Sun and the Solar System originally came from a clump of gas and dust in a
molecular cloud.
Question 2 – Ch. 7; #28
A protoplanetary disk is the accretion disk of gas and dust around a newly-formed
star. It is the protoplanetary disk that gives rise to the planetary system and any
left-over debris-belts (e.g., Asteroid Belt, Kuiper Belt). The inner part of the disk is
hotter than the outer part because 1) the inner part is closer to the protostar that is
forming, and 2) because of Kelvin-Helmholtz contraction, the inner parts collapsed a
longer way than the outer parts and so converted more gravitational potential
energy into thermal energy.
Question 3 – Ch. 7; #30
The law of conservation of angular momentum says that as mass is brought closer to
the spin axis, the velocity of that mass about the spin axis increases. That is exactly
what a figure skater does, she brings in mass (hands and legs) closer to the spin axis
and her rate of spin increases.
Question 4 – Ch. 7; #33
The process by which tiny grains grow into large planets is known as accretion and
begins as small grains collide and stick until 100 meter clumps of solid material are
produced. At this point, collisions continue to increase the size of the clumps, but
now the collisions must occur more gently (think of glancing as opposed to head-on
collisions). When the clumps have grown into planetesimals (objects about a
kilometer in size), gravity begins playing a larger role (though accretion by
collisions continues) until planets are formed.
Question 5 – Ch. 7; #35
We find large amounts of volatiles only in the outer regions of our Solar System
because only beyond the “frost line” at 3.5 AU does it get cold enough for volatiles to
“freeze out” and condense into small, icy particles. The same concept should be true
for other solar systems with the location of the “frost line” moving in or out
depending on how cooler or hotter than the Sun the star in question is.
Question 6 – Ch. 7; #36
One idea is that the giant planets underwent a phase during accretion where they
accreted lots of ices (frozen gases like water, carbon dioxide, methane, and
ammonia). This allowed them to grow much larger than the terrestrial planets and
their strong gravitational fields allowed them to pull in hydrogen & helium and,
thus, collect massive atmospheres. Another idea is that gravitational instabilities in
their parts of the disk allowed large volumes of gas to collapse and form small
versions of the giants planets which I’ll call proto-giant planets. The increase in size
and mass from those spheres of gas occurs via accretion of ices as in the first idea.
Question 7 – Ch. 7; #37
Most of the debris that was left over after the planets formed eventually crashed
into the planets or achieved stable orbits in the asteroid belt or the Kuiper belt.
Question 8 – Ch.7; #38
The four ways that astronomers search for extrasolar planets are the following:
1) Spectroscopic Radial Velocity Method – the “wobble” of a star produced by
one or more planets orbiting the star (this is just the reaction force according
to Newton’s Third Law) is picked up spectroscopically as the star alternately
moves towards and away from us. The ensuing blueshift and redshift of its
spectral lines tells you how the star is being moved by the planet or planets
orbiting it and from this date the masses and distances of the planets can be
determined. (See Figure 6.14)
2) The Transit Method – If the orbital plane of the planet is lined up more or less
with our line of sight from Earth, then when the planet passes in front of the
star, the total light output (luminosity) from the star will decrease for as long
as the planet is in front of the star. Thus, the light curve (plot of luminosity
versus time) for that star will show periodic dips. (See Figure 6.15).
3) The Microlensing Method – The gravitational field of an unseen planet
passing between us and the star (so, just like method #2, the orbital plane
has to line up with respect to us) will cause the star to brighten temporarily
(this is an effect involving the bending of light from the theory of General
Relativity).
4) Direct Imaging – The easiest to understand. Take a picture, and see little dots
around the star. If they move around the star over time following Kepler’s
Third Law, then they are planets. The trick, of course, is to get rid of most of
the star’s light, otherwise you couldn’t see the faint planets. See Figures 6-16
and 6-17 for actual images.
There is a fifth method, not mentioned by the book. This is the astrometric method
and it is like method #1, but instead of detecting the wobble of the star
spectroscopically, you actually notice that the star’s proper motion through space is
not in a straight line, but follows a sinusoidal pattern. This is not as effective as the
others because it takes many years (decades, really) to confirm and it only works for
nearby stars (which have typically the largest proper motions). This was the “oldtimey” way to detect planets which basically never really worked out. Probably why
the book doesn’t mention it. But it is possible, in theory, to detect planets this way.
Question 9 – Ch. 7; #41
1.6 km = 1 mile
(30 km/s)(1 mile/1.6 km)(3600 s/hr) = 67,500 miles/hr
Question 10 – Ch.7; #42
a) see page A-10;
b)
71%
c)
0.2%
447 Earth masses
Question 11 – Ch.7; #43
Lorbital = mvr
Lspin = (4 pi m r2)/ (5p)
Earth’s orbital angular momentum = 2.7 x 1040 kg m2/s
Earth’s spin angular momentum = 7.1 x 1033 kg m2/s
Basically, the spin angular momentum contributes almost nothing to the total
angular momentum.
Question 12 – Ch. 7; Applying the Concepts - #45
MJ / ME = 318
vJ / vE = 0.440
orbit radiusJ / orbit radiusE = 5.2
so LJ / LE = (318)(0.440)(5.2) = 730
Question 13 – Ch.7; #46
Use conservation of angular momentum for a uniform, spinning sphere (see
problem 43). Basically, assume the mass doesn’t change (not quite true, but good
enough for this problem); the size of the cloud goes down by a factor of 1 x 1016/1.4
x 109 = 7 x 106. In the spin angular momentum equation, L is proportional to R2/P,
so
R2cloud/Pcloud = R2sun/Psun
or
R2sun/ R2cloud = Pcloud/PSun
(1/7 x 106)2 = Pcloud/Psun
So, the period of the shrunken cloud goes down by a factor of (1/7 x 106)2 = 2 x 10-14
So, the Sun would have a period of (1 x 106 yr x 3.16 x 107 s/yr) x (2 x 10-14) = 0.6 s
This is why we need accretion disks.
Question 14 – Ch.7; #47
Must convert diameter in km to radius in m:
R = 2.65 x 105 m
Volume = 4/3 pi r3 = 7.8 x 1016 m3
Density = Mass/Volume = 3500 kg/ m3
Question 15 – Ch.7; #48
The question here is: what fraction of the Sun’s surface is covered by Jupiter
(assume that as you view the Sun and Jupiter from afar, they look circular rather
than spherical). Jupiter is approximately 1/10 the radius of the Sun, so its area is
1/100 that of the Sun (remember: area is proportional to the radius squared). Thus,
if 1/100 of the Sun’s surface is covered by Jupiter, the Sun’s light output drops by 1
percent.
Question 16 – Ch.7; #51
Solar mass = 2.0 x 1030 kg
Period = 3.524 days = 3.045 x 105 s
Use Kepler’s Third Law:
p2 = 42a3/(GM)
a = (GMp2 / 4 pi2 )0.333 = 6.8 x 109 m = 0.045 AU
This is 8 or 9 times closer to the star than Mercury is to the Sun.
What this means for Osiris is that it will undergo significant tidal distortion
and heating by the parent star.
Question 17 – Ch.7; #52
Look at figure 6.13. It’s the projected area of the planet onto the projected
area of the star that produces the dimming effect. The projected area looks
circular, not spherical (just look at the figure to see that this is so…). So the
projected area of Osiris is 0.017(area of star) = 0.017(pi(0.7 x 106)2) = 3.9 x
1010 km2
Area of Osiris = 3.9 x 1010 km2 = pi r2
Radius of Osiris = 1.1 x 105 km,
Diameter = 2.2 x 105 km
This is 1.6 times the diameter of Jupiter
Question 18 – Ch.7; #54
V is proportional to the radius cubed.
So, 1.73 = 4.9
The volume is 4.9 times more.
b) Mass = density x volume. If the density stays the same and the volume goes up by
4.9 then so does the mass.
Question 19 – Ch.7; #55
R = 1.43 RJup
(RJup =
M =2.33 MJup
(MJup = 1.9 x 1027 kg)
a) Mass = 4.4 x 1027 kg
7.15 x 107 m)
b) R = 1.02 x 108 m
c) Volume is (4/3)pi r3= 4.4 x 1024 m3
d) 1000 kg/m3
so it’s gaseous.