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Astronomy 161
Solar System Astronomy
Prof. Scott Gaudi

Jump to the Answers for [ Homework 1 | Homework 2 | Homework 3 | Homework 4 ]


Homework #1

1. The distance between the Earth and the Sun is 1.496 x 108 km, 1.496 x 1013cm. Answer (b)

2. The difference in the angle of the Sun as seen from Syene and Alexandria due only to the finite distance to the Sun is 860 km/1.496 x 108 km = 5.75 x 10-6 radians = 3.29 x 10-4 degrees. Answer (a)

3. The circumference of Velux is 220 km x (360 degrees/4 degrees)=19800 km. Answer (e)

4. The mass and weight of the rock are 10 kg and 2.2 pounds. Answer (c)

5. The only place on Earth where there are no cicumpolar stars is the equator. Answer (d).


Homework #2

1. Since the central sun always passes through Zenith and always takes the same amount of time to travel from horizon to horizon, you can conclude three things: (1) you are on the equator, (2) the angle of Normalia's rotation axis is perpendicular to the plane of Normalia's orbit, and (3) Normalia's orbit is circular. Answer (c) 0 degrees.

2. The normalia day is twelve hours. Since the sun moves one degree per Normalia day or every twelve hours, the year is 360 degrees x 12 hours/degree = 4320 hours = 180 Earth days. Answer (a) 180 Earth days.

3. Simplicia is observed to eclipse the sun every 720 hours. This is also the time it takes for Simplicia to go through one complete set of phases. Thus, every time Simplicia comes between Normalia and the sun there is an eclipse. Therefore, Simplicia's orbital plane must be very close to Normalia's orbital plane. You know that it must be less than one degree, otherwise you'd see partial eclipses. Therefore, it must be aligned to within the angular size of Simplicia, which is 1/2 of a degree. It could be better aligned, but there's not enough information to figure that out. Answer (c)

4. Since the angular size of Simplicia is always the same relative to the angular size of the sun, and given that the orbit of Normalia is circular and so the angular size of the sun is always the same, the angular size of simplicia is always the same. Therefore, it must have a circular orbit. Answer (c)

5. Since the orbit of Simplicia is aligned to within 1/2 a degree, we know that it must undergo an eclipse every full moon, which occurs every 720 hours or 60 Normlia days. Answer (e)

6. In order to determine whether or not you observe penumbral eclipses, umbral eclipses, or both, you need to be able to determine the length of Normalia's shadow relative to the size of Simplicia's orbit about Normalia. Unfortunately, you have no idea what the size of Normalia is, what it distance from the sun is, or what the size of Simplicia's orbit is, so you can't figure this out. Answer (d)


Homework #3

1. According to Kepler's Third Law, the semimajor axis of an orbit measured in AU cubed is equal to the period of the oribt in years squared. Thus, the semimajor axis in AU is just period in years to the 2/3 power. a[AU]=(P[years])^{2/3}=(1,000,000)^{2/3}=10,000 AU. Answer (b)

2. The apparent brightness is invsersely proportional to the distance squared. Since Fred is 10,000 times further away from the Sun than the Earth, the brightness is 10,000 squared times less, or 100 million times fainter. Answer (b)

3. The gravitational force is proportional to the product of the masses and inversely proportional to the distance squared. The Sun is 100 times more massive than Fred, and 10,000 times closer, so the gravitational force of the Sun on the Earth is 100*(10,000)^2=1 x 10^10 times or 10 billion times larger than that of Fred on the Earth. Answer (e)

4. The tidal force is proportional to the mass and inversely proportional to the distance cubed. The Sun is 100 times more massive than Fred, and 10,000 times closer, so the gravitational force of the Sun is 100*(10,000)^3=1 x 10^14 times or 100 trillion times larger than Fred. Answer (e)

5. According to the equation, the circular velocity of objects orbiting the same central mass are inversely propotional to the square-root of the distance from that mass. Since Fred is 10,000 times further away from the Sun than the Earth, its circular velocity is sqrt(10,000)=100 times less than that of the Earth, or 30km/s/100=30,000m/s/100=300 m/s. Answer (e).

6. The escape speed is just square root of 2 times the circular speed. The circular speed of Fred is 300m/s, so the escape speed is just sqrt(2)*300 m/s=424 m/s. Answer (d)


Homework #4

1. Wein's Law states that the peak wavelength of a blackbody is 2,900,000 nm/T, where T is the temperature in Kelvin. Since T=5800K, the peak wavelength is 2,900,000nm/5800=500nm=0.5 micrometers. Answer (d) 0.5 micrometers.

2. If Horace's brightness peaks at a wavelength of 1.5 micrometers, according to Wein's law, it must have a temperature that is 1.5 micrometers/0.5 micrometers = 3 times cooler. Answer (e) Horace is three times cooler.

3. According to the Stefan-Boltzmann Law, the energy emitted per second per area by a blackbody with Temperature (T) is proportional to the temperature to the fourth power, T^4. Since Horace is 3 times cooler, it must emit 3^4=81 times less energy per second per area. Answer (d) 81 times smaller.

4. After one half life, I have 50% left. After two half lives, I have 0.5*0.5=25% left. After three half lives, I have 0.5*0.5*0.5=12.5%. Thus, three half lives must have passed in six days, so the half life is 6 days/3=2 days. Answer (a) two days.

5. The collecting area of a single mirror depends on its area, which is proportional to its diameter squared. Thus one LBT mirror has a collecting area that is (8.4/2.4)^2=12.25 times larger that HST, two mirrors have a collecting area that is twice this, or 24.5 times larger. Answer (e) about 24.5 times more light.

6. The surface area of a sphere is 4*pi*R^2, where R is the radius. According to the Stefan-Boltzmann Law, the energy emitted per second per area E by a blackbody with Temperature (T) is E=sigma*T^4, where sigma is the Stefan-Boltzmann constant. Thus the luminosity L is L= 4*pi*R^2*sigma*T^4. Horace is 3 times cooler, but nine times bigger, thus the ratio of the luminosity of Horace to the Sun is L_Horace/L_Sun = (R_Horace/R_Sun)^2*(T_Horace/T_Sun)^4=(9)^2*(1/3)^4 = 81*(1/81)=1. Answer (d) Horace has the same luminosity as the Sun.


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