Download Physics of Astronomy – Week 3 quiz

90Physics of Astronomy – Winter midterm –.Thus.2.Feb.2006
This is a CLOSED-BOOK exam to be taken in class. You have two hours, so pace yourself.
SHOW YOUR WORK, to receive full credit, and include units wherever appropriate.
Please circle or underline your answers, on problems with considerable calculation.
(sign legibly)____________________ _ZITA’s _ SOLUTIONS_________(70 points total)
I affirm that I have worked this exam with WITHOUT using a calculator, text, HW, quizzes,
computer, classmates, or other resources.
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Keep answers in exact form or make order-of-magnitude estimates
Return your exam to Pat or Ruth in ____ Lab II by 2:30.
Then head to the library for your 3:00 workshop in the Mac Lab.
Feel free to discuss exam questions with classmates who have taken the exam, after you
all have turned them in.
Math-A students – take the Astro-A and Phys-A parts of this midterm. (Parts I-III)
B-parts are optional.
Math-B students – take the whole midterm. The A-parts should be quick and easy for you.
G = 6.67 x 10-11 N m2/kg2
Radius
Mass
Possibly useless data: c = 3x108 m/s
g=9.8 m/s2
parsec ~ 3 lightyears
Sun
7 x 105 km
2 x 1030 kg
Earth
6.4 x 103 km
6 x 1024 kg
Moon
1.74 x 103 km
7.35 x 1022 kg
Distance between centers of Sun & Earth = 1.5 x 108 km
Distance between centers of Earth & Moon = 3.84 x 105 km
Please leave the space below blank:
(I) Astro-A. mostly qualitative
8 + 7 = 15
(II) Astro-A. quantitative (resolution):
funcamentals
6 + 7 = 13
application
(III) Physics-A: kinematics
14
dynamics
_______________________________________________________________________
(IV) Astro-B (qualitative)
Physics-B: (V) fundamentals (orbits)
7
8 + 13 = 21
(VI) application
(I) Astro-A: Astronomy questions: Most of these are taken directly from the online quizzes.
You should be able to answer them easily. If not, move on to the next question.
What is a solar system?
A collection of two or a few suns (stars) orbiting around each other.
A collection of a million to a hundred billion or more suns (stars) in a large system,
often containing spiral arms; for example, our own Milky Way Galaxy.

A collection of planets and other, smaller objects orbiting around a star (e.g., the Sun).
You are standing somewhere on Earth on a bright, sunny day when you notice that a tall,
vertical telephone pole has no shadow. From that observation by itself, what do you know
about where you might be standing?
Exactly on the equator.

Anywhere between the Tropic of Cancer and the Tropic of Capricorn.
Exactly at the North or South Pole in midsummer (24-hour daylight).
It is warmer in summer than in winter because
the Sun is lower in the sky and sunlight passes through a longer path in the
atmosphere, thereby heating it more in summer.

the Sun is higher in the sky and the days are longer.
the Earth is closer to the Sun.
The Earth is closest to the Sun
during winter in the southern hemisphere.

during summer in the southern hemisphere.
when the Sun is directly overhead at the equator.
The Moon shows different phases because
it moves slowly in and out of the Earth's shadow in its orbital motion around the Earth.

we view the sunlit hemisphere of the Moon from different angles as the Moon orbits the
Earth.
it rotates slowly to reveal more or less of its illuminated hemisphere while remaining
relatively fixed against the celestial sphere of stars.
The declination of the Sun on the first day of spring is
variable, depending on the year.
23.5º north.

0º.
Stars on the equator appear to move through a full 360 degrees in about 1 day because of
the Earth's rotation. How fast are they moving in degrees per minute? (Show your work)
4 degrees per minute

1/4 degree per minute

15 degrees per minute

360 day
hr
6
1


day 24 hr 60 min
24 4 min
The constellation Cygnus appears overhead to an observer in the state of New York, USA, at
midnight on a particular night. In which part of the sky will this constellation appear
to an observer at the same latitude in Montana, 30º farther west in longitude, if
observed at precisely the same absolute time?
Overhead, of course.
In the western sky.

In the eastern sky.
As the Moon orbits around the Earth, its path on the celestial sphere
is very close to the celestial equator.

is very close to the path followed by the Sun (the ecliptic).
is very close to a line of right ascension, passing close to each celestial pole once each
orbit.
At what time in its monthly cycle will the Moon be seen high in the sky in the daytime,
from midlatitudes?
Only when it is full
The Moon is never visible in the daytime.

Quarter Moon
When does the new Moon rise, approximately?

Close to sunrise
At midnight
At sunset
Where in the sky will the full Moon appear at sunset, in the northern hemisphere?

Low in the eastern sky
High in the southern sky
Low in the western sky
If you were an astronaut on the Moon, in what position close to the Moon's equator
would you have to stand in order to be in perpetual darkness?
On the side of the Moon that faces away from the Earth.
On the side of the Moon that faces away from the Sun.

You cannot do this. There is nowhere near to the Moon's equator where you can be in
perpetual darkness.
Lunar eclipse: Which of the following statements is true?
A lunar eclipse occurs only at new Moon and can be seen from anywhere on the
nighttime hemisphere of Earth.

A lunar eclipse occurs only at full Moon phase and can be seen from anywhere on the
nighttime hemisphere of Earth.
A lunar eclipse occurs only at full Moon and can be seen from within a narrow strip on
the Earth's surface.
(II) Astro-A: quantitative question: Resolving Venus
Fundamentals:
Find the angle  subtended by an object of diameter D at a distance d.
 = D/d
Specify the units of. radians
Diagram this relationship and label all quantities ()
How many arcminutes are in one radian? (Show your work.)
radian
360 60arc min
 3600arc min
2 rad
1
Application: When Venus is a distance of 0.72 AU from Earth, is it nearer than the Sun or
further? Find the distance in km.
0.72 AU
1.5 108 km
 1108 km
AU
The diameter of Venus is roughly 12 x 103 km. What angle does Venus subtend at the distance
in part (a)? Express your answer in arcmin.
D 12  103 km
3.6 103 arc min
4
 
 10 radians
 0.3 arc min
d
1.5 108 km
rad
A person with good vision can resolve details as small as 1 arminute. Could this person resolve
the crescent of Venus with naked eye? Explain.
Since the size of Venus is smaller than the smallest angle a good eye can resolve,
the crescent of Venus cannot be resolved by naked eye (despite anecdotes to the contrary).
(III) Physics-A: Everyone should be able to answer this in about 10 minutes. (No tricky
integrals on this one.)
1. (a) What is the fundamental relationship between force F and acceleration a? F=ma
(b) Do accelerations cause forces? Y N
Or do forces cause accelerations?
Y
N
2. Write down the fundamental defining relationships between position, velocity, and
acceleration.
(a) Differential relationships:
v = dx/dt
a = dv/dt
(b) Integral relationships:
x   v dt
v   a dt
3. The force acting on some mass m is F = mg - b t2 + ct½, where b and c are constants. Find the
acceleration, velocity, and position of this mass, assuming that it starts at the origin at rest.
F = ma, so
F mg - b t 2  ct
a 
m
m
1
2
g -
b 2
c 12
t 
t
m
m
b 2
c 12 
b 3
c 3/ 2

t 
t dt  gt t 
t
v   a dt so v    g m
m 
3m
3m / 2

b 3
2c 3/ 2 
gt 2
b 4
4c 5/ 2

x   v dt so x    gt t 
t dt 
t 
t
3m
3m
2
12m
15m


(IV) Astro B: These questions should be quick and easy. If not, move on to the next question.
The Ptolemaic model successfully described and predicted the apparent motions of the
planets, by assuming them to orbit the Earth, with the planets moving
in circles around the Earth, the planes of which were precessing slowly around a
direction perpendicular to the ecliptic plane.
in ellipses, with their respective foci at the center of Earth.

in small circles, the centers of which were moving around the Earth more slowly in
larger circles.
An object orbiting the Sun with an orbital eccentricity of 0.1 has an orbit whose shape is
a long, narrow ellipse.
circular, but with the Sun off-center in the circle.

slightly elliptical, but almost circular.
Kepler's laws
apply for all objects orbiting the Sun, such as planets, comets, asteroids and man-made
space probes, but do not hold for objects orbiting any other object in the universe.
apply only to large planets orbiting our Sun, but provide only an approximate
description for smaller objects, such as asteroids, etc. and are not applicable at all for other
situations such as mutually orbiting binary stars.

can be shown to apply universally, whenever two objects orbit each other under
gravitational attraction.
If there is a force of gravity from the Earth on an artificial satellite of mass 1000 kg, what is
the size of the force exerted on the Earth by the satellite?
Essentially zero because the satellite's mass is negligible compared to that of the Earth.

Exactly the same as the force on the satellite.
Much larger than that on the satellite because the Earth's mass is so large.
A wheelbarrow of mass M kg was transported to the Moon during an Apollo mission. Its
weight on the Earth was W. Which of the following conditions is found to hold on the surface
of the Moon?
Mass and weight of the wheelbarrow are reduced by the same fraction, to 1/6 of their
values on Earth.
The mass of the wheelbarrow remains the same while weight is found to be zero
because the wheelbarrow is at a large distance from the center of the Earth.

The mass of the wheelbarrow remains the same, but its weight on the Moon is less.
According to Kepler's third law, an asteroid orbiting the Sun in a circular orbit at twice
the Earth's distance from the Sun would have a sidereal period of (show your work)
4.0 years.

2.8 years.
2
8.0 years.
3
 p   a 
p
3
period ( yr )  radius( AU ) ........  sat    sat    2   8 ........ sat  8  3
pEarth
 pEarth   aEarth 
2
3
(V) Physics B - This is optional for A-group students.
Fundamentals: Consider a small mass (m) orbiting a big mass (M) with a circular orbit of
radius R and period T.
(a) What is the orbital speed v in terms of R and T?
v=2R/T
(b) What is the centripetal acceleration a?
a=v2/R
(c) What is the force responsible for this acceleration?
F=GmM/R2
(d) Put these together to find a relationship between speed v and radius R (in terms of M
and fundamental constants), starting from F=ma.
F  ma
GmM
v2
m
R2
R
GM
v2 
R
(e) SKETCH the speed v versus the radius R.
(f) Briefly describe what this means, in words.
This is a Keplerian velocity curve: The closer planets orbit faster.
Remember that we assumed that the mass of the system is concentrated at the center
(R=0), in this case, at the Sun.
(VI) Physics B – Application:
The Sun orbits with a speed about v1=250 km/sec at distance of about R1=9 kpc from the center of
our Milky Way galaxy.
(a) Use these data to calculate how much mass (M1) is inside Sun’s orbit. Express your answer in
terms of the Sun’s mass:
3
5m
2
GM
Rv 2 9 10  3(ly )(2.5 10 s ) 9 1015 m
v 
so M 

3
R
G
ly
20 1011 kg .m 2
3
s
2
M SUN
999
103
310 15 11
M
 10
kg 
 1039 kg  1041 kg
20
10
2 1030 kg
M1
1
 109  half a billion suns
M SUN 2
Surprisingly, stars near the dim edge of the Milky Way galaxy (say at 15 kpc) orbit the galaxy’s
center at the same speed.
(b) Why is this surprising? If the stars orbiting in the Galaxy followed a Keplerian velocity
curve, as do the planets orbiting in the solar system, we’d expect the outer stars to orbit slower.
(c) How much mass (M2) is inside the distant star’s orbit? Express this in terms of the Sun’s
mass:
Rv 2
M
G
2
so
M 2 R2  v2  15 2
    1
M 1 R1  v1 
9
M2
15 M 1
5 1
5

   109   109
M SUN
9 M SUN 3 2
6
(d) What percent of the Milky Way’s mass lies between 9 and 15 kpc from its center, if 15 kpc
is the outer edge of the galaxy?
inner mass M 1 9 3 6

  
 60%
total mass M 2 15 5 10
So the percent of the mass that lies outside 9 kpc is 40%.
(e) SKETCH the v versus R curve for stars in our Galaxy (based on these two data points).
(f) Compare this to the velocity curve in part (V) above. What does this mean, conceptually?
This means that the galaxy’s mass is NOT
concentrated at the center, but instead is
distributed throughout the galaxy, even where
there is little light. Most of the galaxy’s light
comes from nearer the center.
We’ve discovered evidence for dark matter!