Midterm 1 Review
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School
Rutgers University *
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Course
01:750:109
Subject
Astronomy
Date
Apr 3, 2024
Type
Pages
4
Uploaded by ChefHawkPerson1459
Practice Exam
1.
D c.
2.
B
3.
E c. 4.
A
5.
D
6.
D
7.
B - Why??
8.
D
9.
D
10. B
11. A d. 12. C - Why??
13. A
14. B
15. C
16. B
17. D b.
18. B
19. A
20. B
21. A IGNORE BAD QUESTION
22. A c.
23. D e.
24. E a.
25. D
26. E
27. B
28. E
29. C
30. B?? e.
Midterm Review
Q. 22 About every two earth years, there will be a night when mars exhibits the fastest apparent retrograde motion in the sky. On that night, which of the following motions will contribute to a shift in the wavelengths of spectral lines from the Martian atmosphere that we can detect from Earth?
-
Apparent retrograde motion will be greatest when mars is in opposition and earth is at the exact moment of passing mars.
-
Shifts in wavelengths of spectral lines is due to the Doppler e
ff
ect, which requires relative motion towards us or away from us from along the line of sight.
-
The moment when earth is passing mars, their orbital motions are parallel, no line-of-sight component of their relative motion means no contribution
to the doppler e
ff
ect.
-
This means we can rule out any possibility with ORBITAL MOTION
-
The only changes in wavelength will be redshift and blue shift from the rotation of mars about its axis
Q. 3 Imagine that you are a rutgers astronomer who studies supernovae. A supernova has appeared in a part of the sky that currently is above the horizon in New Jersey ONLY during the day time. As a result, you can’t study the supernova from new jersey at visible wavelengths. What can you do to get an observation of the supernova at visible wavelengths as quickly as possible
?
-
if only above the horizon during the daytime, where does the supernova have to be?
-
Supernova has to be on the FAR side of the sun from the earth
-
By the time the supernova is above the horizon in any other part of the world, it will be daytime there too. Q. 7 The planet Endor is famous for its beautiful eclipsing moons an inner red moon and an outer green moon that have the same angular size when viewed from the planet’s surface. The two moons are known to have the same mass; one has an orbital period of three Earth days, while the other has an orbital period of 24 Earth days. What conclusion can you draw about the moons’ mean densities? -
Kepler’s third law: larger orbital period will have larger distance from orbit center
-
Means outer moon will need to be the one with 24 day orbit
-
To know density, you need volume (we know the masses are the same)
-
Remember volume of sphere is -
We would need the ratio of the radiuses to get the volume to get density
-
Moons have the same angular size when viewed from the planet, which means radius of red moon/distance to planet should be same as radius of green moon/ distance to planet
-
-
Mean density of the red moon is 64 times higher than that of the green moon
Q. 24 A rogue planet is moving at a constant speed into our solar system. Its path bends a little bit as it passes by the sun, but it is traveling so fast that it continues to move at a nearly constant speed until it exits the solar system. What is the best description of the force of the sun on the rogue planet?
-
Does the rogue planet accelerate? Yes, it changes direction
-
The sun’s gravitational pull is always toward the sun
-
The sun exerts a force on the planet that is always directed along a line connecting the sun and the planet
Q. 23 If the mass of Jupiter was doubled, which distances would need to change for the orbital periods of Jupiter around the sun and europa around Jupiter to remain the same as they are today? -
REMEMBER: Kepler’s 3rd law depends ONLY on the mass of the orbitee, NOT THE ORBITER, so Jupiter’s orbit around the sun WOULD NOT CHANGE
-
We could replace Jupiter with a potato and the potato would take the same time to orbit the sun that Jupiter does now
-
Orbit of the moon WOULD BE AFFECTED because it orbits Jupiter and the object it’s orbiting has changed -
Distance between Europa and Jupiter would have to increase with the mass so that the orbit remains the same
(nothing else would have to change)
Q. 17 New Brunswick is at 40.5 degrees north latitude. What is the altitude of Polaris, the pole star, when seen from New Brunswick?
-
North Star is where the north axis hits the celestial sphere
-
Polaris is ALWAYS at that point, no matter what time of year or time of day
-
Not 90 degrees! That would be straight above you, which would only be possible at the north pole
-
North Pole is at 90 degrees latitude, north star would be 0 degrees at the horizon (0 degrees latitude)
-
Angle of the star is the same as the angle of latitude
4
π
r
3
3
r
green
/
r
red
= (
a
green
/
a
red
)
3
= (
P
green
/
P
red
)
2
= (24/3)
2
= 8
2
= 64
-
40.5 degrees
Q. 6 The planet Venus is nicknamed the morning or evening star because..? -
Venus’ orbit is closer to the sun than Earth’s Q. 12 Professor Baker is observing a distant galaxy with a telescope and detects a spectral line of the carbon monoxide molecule at an observed frequency of 30 GHz. What is the observed wavelength of this spectral line?
C = 3 x 10^8 m/s
Wavelength = (3 x 10^8) / 3 x 10^10 s^-1
= 10^-2 m or 1cm
Q. 30 Pluto completes two orbits around the sun in the same time that Neptune completes three orbits around the sun. Is a collision possible?
-
resonance keep Pluto out of Neptunes way, so Pluto and Neptune will never cross paths
-
No, because Neptune will never be in Pluto’s path when the orbits cross
Q. 9 Why does the Earth have a tidal bulge on the side opposite the moon?
-
Lunar gravity is weaker there than at the center of the Earth
Example Questions
Assume that on a certain day, the moon rises at sunrise and sets at sunset. Seven days later, what would the phase of the Moon be?
-
If moon is “rising” with the sun, the phase must be a New Moon because the moon and sun must be in the same position in the sky
-
7 days is one quarter of an orbit
-
First quarter
Suppose that on the day after your nephew’s birth, Jupiter (whose mass is about 2 x 10^27 kg) has a distance of 10^12m from the Earth. How strong is the gravitational force of Jupiter on your nephew compared to the force he feels from his parents’ 2000kg car, which is 1 m away from him in the hospital parking lot. -
the forces are the same!!
E = hv where E is in Joules, v is in Hz, and h is the “Planck constant” whose value is 6.6 x 10^-34 J Hz^-1. What is the energy of a photon whose wavelength is 660nm? λ
=
c
ν
F
due to gravity
=
Gm
1
m
2
/
r
2
m
1
=
nephew; other is jupiter or the car
Jupiter:
m
2
/
r
2
= (2
×
10
27
)/(10
1
2
×
10
1
2) = 2000
car:
m
2
/
r
2
= (2000)/(1
×
1) = 2000
E
=
h
ν
E
=
h
(
c
/
λ
)
E
= (6.6
x
10
−
34
JHz
−
1
)
3
×
10
8
m
/
s
)
6.6
×
10
−
7
m
E
= 3
×
10
−
19
J
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In which of the following scenarios would the gravitational force of the Earth on the Moon be the same as it is today?
- if the moon is four times more massive and two times farther from the earth than it is today
F
0
=
Gm
1
m
2
/
r
2