Astronomy Exam 2 Homework
An Earth-like planter is orbiting the bright star vega at a distance of 1 AU. The mass of Vega is twice that of the sun. a) How fast is the planet traveling in its orbit? b) What is the period of the planet?
vcirc = [G Mstar/R]^0.5 = 4.2 x 10^4 m/s = 42 km/s For part b) use Newton's version of Kepler's Third Law and solve for p: p2 = 4 pi^2 a^3 /(G Mstar) p = 2.2 x 10^7 s = 0.71 years
Your microwave vibrates molucules at a frequency of 2.45 x 10^9 Hz, what is the wavelength in cm?
wavelength = c/f = 3.0 x 10^8 / 2.45 x 10^9 = 0.12 m = 12 cm
Astronomers define certain celestial objects as red shifted or blue shifted. What do these terms indicate about the objects?
A redshifted object shows spectral lines shifted towards the lower frequency/longer wavelength part of the spectrum (called the "red" part of the spectrum even if you are not in the optical regime). These objects are moving away from you according to the Doppler formula. A blueshifted objects shows spectral lines shifted towards the higher frequency/shorter wavelength part of the spectrum. These objects are moving towards the observer.
Why are the four giant planets able to collect massive gaseous atomospheres, whereas the terrestial planets could not? Explain the source of the atomospehres now surround three of the terrestrial planets.
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.
What is the source of the material that now makes up the sun and the rest of the solar system?
The Sun and the Solar System originally came from a clump of gas and dust in a molecular cloud.
How does the law of conservation of angular momentum control a figure skater's rate of spin?
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.
Describe the process by which tiny grains of dust grow to become massive planets.
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.
When a space craft is sent to mars, it is first launched into an earth orbit with a circular velocity. a) describe shape of this orbit b) what minimum velocity must we give the space craft to send it on its way to mars?
a) circular b) It must escape from the Earth's gravity. At the surface of the Earth (R=6370 km) that would be 11.2 km/s (use the escape velocity formula; the mass of the Earth is 6.0 x 10^24 kg). If you are orbiting the Earth, you are farther from the center than 6370 km. Let's say you are orbiting 300 km above the surface, then your R in the escape velocity formula is 6370 + 300 = 6670 km. Plug that into the escape velocity and you get 10.9 km/s as the escape velocity from a point 300 km above the Earth's surface.
How bright would the sun appear from neptune at 30 AU compared to brightness seen from earth? the space craft voyager 1 is now 124 Au from sun, compare brightness seen from it to that of earth.
Brightness decreases as 1/d^2 , so if Neptune is 30 times the distance of the Earth from the Sun, the Sun will be lower in brightness at Neptune by a factor of 1/30^2 = 1/900. Thus, the Sun looks fainter by a factor of 900 from Neptune compared to how it looks from Earth. The Voyager spacecraft at 124 AU sees the Sun as (124)^2 = 15,376 times fainter.
Sun has a radius of 6.96 x 10^5 km and blackbody temp of 5780 K. Calculate the suns luminosity.
Calculate the Sun's luminosity: Lsun = (4 pi Rsun^2 ) sigma Tsun^4 = 4 pi (6.96 x 10^8 m)^2(5.67 x 10^-8)(5780 K)^4 = 3.85 x 10^26 J/s Notice that I had to convert the Sun's radius to meters.
A distant light blub has the same apparent brightness as a firefly 5 m away. The lightbulb is a million times more luminious than the firefly, how far away is the light blub?
Lbulb = 1 x 10^6 Lfirefly or 1 x 10^-6 Lbulb = Lfirefly Fbulb = Ffirefly F = L /(4 pi r^2) Lbulb /(4 pi rbulb^2) = Lfirefly /(4 pi rfirefly^2) Lbulb /( rbulb^2) = 1 x 10^-6 Lbulb /( (5)^2) 1/( rbulb^2) = 1 x 10^-6 /(25) = 4 x 10^-8 ( rbulb^2) = 2.5 x 10^7 rbulb = 5 x 10^3 The bulb has to be 5000 meters away
Is light a wave or a particle or both?
Light manifests both particle and wave-properties. When you do an experiment that showcases light's particle properties, then it behaves as particles (photons). When you do one that showcases light's wave properties, then it behaves like a wave phenomenon. This is called wave-particle duality and light has both aspects as part of its nature
The speed of light in a vaccum is 3 x 10^8 m/s. Is it possible for light to travel at a lower speed?
Light may travel at a slower speed than c if it is not propagating in a vacuum but, rather, traveling through some medium with non-zero density, like water or air.
Osris passes in front of its star every 3.525 days, decreasing the brightness of the star by 1.7 percent (0.017). a) what is the diameter of Osiris b) compare this to jupiter
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 10^6)^2) = 3.9 x 10^10 km^2 Area of Osiris = 3.9 x 10^10 km^2 = pi r^2 Radius of Osiris = 1.1 x 10^5 km, Diameter = 2.2 x 10^5 km This is 1.6 times the diameter of Jupiter
What happened to all the leftover solar system debris after the last of the planets formed?
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.
A hot jupiter named Osiris was found around a solar mass star, it orbits the star every 3.525 days. a) what is the orbital radius? b) Compare this to mercurys orbit
Solar mass = 2.0 x 10^30 kg Period = 3.524 days = 3.045 x 10^5 s Use Kepler's Third Law: p^2 = 4pi^2a^3/(GM) a = (GMp^2 / 4 pi^2 )^0.333 = 6.8 x 10^9 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.
The international space station travels in a nearly circular orbit 350 km above the earths surface . What is it's orbital speed?
The formula for things moving in circular orbits is vcirc = [G MEarth/R]^0.5 Need MEarth and REarth : MEarth = 6.0 x 10^24 kg REarth = 6371 km; we are told the International Space Station is in an orbit 350 km above the Earth's surface. So R = 6371 + 350 = 6721 km = 6.721 x 10^6 m Plug in and get 7.72 x 10^3 m/s or 7.72 km/s
To an alien astronomer who observers jupiter passing in front of sun, how much does the sun's brightness drop during the transit?
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.
Why do we find rocky materials everywhere in the solar system, but large amounts of volatile materials only in the outer regions?
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.
a) total mass of all planets in solar system expressed in earth masses? b) what fraction of this does jupiter represent? c) earth?
a) see page A-10; 447 Earth masses b) 71% c) 0.2%
In what way can a charged particle create a magnetic field?
A charged particle can create a magnetic field by accelerating. This generates a changing electric field, which generates a changing magnetic field, which generates a changing electric field, etc.. Thus, an accelerating charged particle creates an electromagnetic wave.
What is the protoplanetary disk? What are the two reasons why the inner disk is hotter than the outer part?
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.
convert 30 km/s to mph
1.6 km = 1 mile (30 km/s)(1 mile/1.6 km)(3600 s/hr) = 67,500 miles/hr
Earths avg albedo is .3 a) calculate its avg temp in kelvin b) Does this temp meet expectations? C) how would this change if albedo was lower or higher?
Albedo = 0.3 So T = 279 (0.7)^0.25(1)^-0.5 = 255 K This is too low! It's below the freezing point of water. The Earth's greenhouse effect, which is not included in the calculation, raises the temperature calculated above by 30 - 35 K. If the Earth's albedo was lower, this temperature would go up (because the Earth would absorb more radiation from the Sun and thus reach a higher equilibrium temperature). If the Earth's albedo was higher, the temperature would go down because the Earth was receiving less radiation from the Sun.
m= 5.57 x 10^24 kg, v=29.8 km/s r= 1 Au what is earths angular momentum and spin momentum?
Lorbital = mvr Lspin = (4 pi m r^2)/ (5p) Earth's orbital angular momentum = 2.7 x 10^40 kg m^2/s Earth's spin angular momentum = 7.1 x 10^33 kg m^2/s Basically, the spin angular momentum contributes almost nothing to the total angular momentum.
Jupiter has mass equal to 318 times earth's mass, and orbital velocity of 13.1 km/s. Earth's orbital velocity is 29.8 km/s. What is the ratio of jupiters anuglar momentum to that of eath's?
MJ / ME = 318 vJ / vE = 0.440 orbit radiusJ / orbit radiusE = 5.2 so LJ / LE = (318)(0.440)(5.2) = 730
Explain the difference between weight and mass.
Mass is the quantity of matter in an object and is measured in kilograms. Weight is a force and so there must be an acceleration on the object for the weight to be defined. As a force, the units of weight are Newtons (or kg m/s2).
The asteroid vesta has a diameter of 530 km and mass of 2.7 x 10^20 kg. Find density
Must convert diameter in km to radius in m: R = 2.65 x 105 m Volume = 4/3 pi r^3 = 7.8 x 10^16 m^3 Density = Mass/Volume = 3500 kg/ m^3
Planet found with raidus 1.43 jupiters radius, and 2.33 mass of juptiers a) what is mass in kg b) what is radius in meters c) what is volume d) what is dentisy
R = 1.43 RJup (RJup = 7.15 x 10^7 m) M =2.33 MJup (MJup = 1.9 x 10^27 kg) a) Mass = 4.4 x 10^27 kg b) R = 1.02 x 10^8 m c) Volume is (4/3)pi r^3= 4.4 x 10^24 m^3 d) 1000 kg/m^3 so it's gaseous.
Mars has about one-tenth the mass of earth and about half of earths radius. What is the value of gravatational accelleration on mars compared to earth's? Estimate your weight on mars compared to your weight on earth.
Remember that the acceleration at the surface of a planet is sometimes called "g" instead of "a". gearth = GMearth/(Rearth)^2 = 9.8 m/s2 gmars = GMmars/(Rmars)^2= G(0.1*Mearth)/[0.5*(Rearth)]^2 = [0.1/0.25][GMearth/(Rearth)^2] gmars = 0.4 gearth When you use the actual numbers instead of our 0.1, 0.5 approximations, you get 0.39 instead of 0.4. Let's say your mass is m = 100 kg Weight = mg gMars = 0.39(gearth) = 0.39 * 9.8 = 3.8 m/s^2 Weight on Mars = 100 * 3.8 = 380 Nt, instead of the 980 Nt you would weigh on Earth. Most Hollywood movies ignore the fact that on Mars you would weigh only 40% of your current weight on Earth.
Why is it impossible to know the exact position and the exact velocity of an electron simultaneously?
The Heisenberg Uncertainty Principle states that you can only simultaneously know the position and momentum of a particle only to an accuracy that is approximately the product of the position of the particle and its momentum multiplied by a very small constant. If you wish to improve the precision with which you know one of these quantities past that point, then you will know the other quantity less precisely. This is not an accuracy of measurement issue, it is a statement that, at the subatomic level, nature assumes a granular, pixel-like behavior rather than a continuous, smooth behavior.
Two comets are leaving the vicincity of the sun. One is in a elliptical orbit and the other is in a hyperbolic orbit. What is the future of these two comets?
The comet in the elliptical orbit is in a BOUND orbit and so it goes around the Sun with a given period (dictated by Kepler's Third Law) and "returns" continually. The comet with the hyperbolic orbit is on an UNBOUND trajectory and so it will pass by the Sun and then continue outwards until it leaves the Solar System. It will never come back (like your first love).
Describe 4 methods that astronomers use to search of extrasolar planets.
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 "old-timey" 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.
Two stars appear to have the same brightness, but one star is 3 times more distant than the other. How much more luminous is the distant star?
This is done just like the problem above except that we are solving for the luminosity instead of the distance. Let's call the nearer star, "near", and the farther away star, "far". rfar = 3 rnear Fnear = Ffar F = L /(4 pi r^2) Lnear /(4 pi rnear^2) = Lfar /(4 pi rfar^2) Lnear /( rnear^2) = Lfar /( (rfar ^2) Lnear /( rnear^2) = Lfar /( (3rnear)^2 ) Lnear = Lfar /9 Lfar = 9 Lnear So the farther star is 9 times more luminous than the nearer star.
A planet with no atmosphere at 1 AU from the sun a) if its albedo was .1 what would temp be? b) if its albedo was .9?
Use T = 279[(1 - a)^0.25]dAU^-0.5 Where the distance is measured in AU Albedo = 0.1 So T = 279 (0.9^)0.25(1)^-0.5 = 272 K Albedo = 0.9 So T = 279 (0.1)^0.25(1)^-0.5 = 157 K
Earth's mean radius is and mass are 6370 km and 5.97x10^24 kg. Show that accelleration due to gravity on earth is 9.81 m/s^2.
Use a = GM/R^2 M = 5.97 x 10^24 kg R = 6370 km =6.37 x 10^6 m Plug those values in (along with the value for G = 6.67 x 10^-11 Nt m^2/kg2) and you will get 9.81 m/s^2
The orbit of Eris has max distance of 97.7 AU from sun If albedo is .8, what is avg temp when its farthest from sun?
Use the same formula as you used in the previous two problems, but now make sure you put in the distance in AU: T = 279[(1 - a)^0.25]dAU^-0.5 T = 279[(1 - 0.8)^0.25](97.7)^-0.5 For a = 0.8, T = 19 K
Venus's circular velocity is 35.03 km/s, and its orbital radius is 1.082 x 10^8 km. Calculate the mass of the sun.
Use vcirc = [G Msun/R^]0.5 and solve for Msun: Msun = (vcirc)^2 R/G = (35.03 x 10^3 m/s)2 (1.082 x 10^11 m)/(6.67 x 10^-11 m3/(kg s2) = 1.99 x 10^30 kg
At the surface of earth, the escape velocity is 11.2 km/s. What would the escape velocity be at the surface of a very small asteroid have a radius of 10^-4 that of earth's and a mass 10^-12 that of earths?
Use vesc = [2G Msun/R]^0.5 For Earth: vesc(earth)= [2G (Mearth)/( Rearth)]^0.5 For Asteorid: vesc(asteroid)= [2G (Masteroid)/( Rasteorid)]^0.5 Express the Mass and radius of the asteroid in terms of those of the Earth: vesc(asteroid) = [2G (10^-12 Mearth)/(10^-4 Rearth)]^0.5 = [10^-8 *2G (Mearth)/( Rearth)]^0.5 = 10^-4 *[2G (Mearth)/( Rearth)]^0.5 vesc(asteroid)= 10^-4 vesc(earth) vesc(asteroid)= 10^-4 *(11.2 km/s) = 11.2 x 10^-4 km/s = 1.12 x 10^-3 km/s = 1.12 m/s You could throw a baseball off the surface of this asteroid and it would leave on a hyperbolic orbit - however, it would then be captured by the gravitational pull of the Sun and orbit the Sun (following Kepler's Third Law).
A satellitle detected a planet with 1.7 earth diameters a) how much larger is the volume of this planet compared to earth b) how much more massive is this planet if denisty is the same
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.
Earth speeds along at 29.8 km/s in its orbit. Neptune's nearly circular orbit has a radius of 4.5x10^9 km, and the planet takes 164.8 years to make one trip around the sun. Calulate the speed of neptune in its orbit.
v = 2pR/period = 2p(4.5 x 109 km) / (164.8 years * 3.16 x 107 s/year) = 5.4 km/s You can also use the formula for circular velocity where R is now the radius of the orbit (you will need to know the mass of the Sun which is 2 x 1030 kg): vcirc = [G Msun/R]^0.5 vcirc = [(6.67 x 10^-11 Nt m2/kg^2) *(2 x 10^30 kg)/4.5 x 10^12 m)]^0.5 = 5.4 x 10^3 m/s = 5.4 km/s Note that we had to convert the radius of the orbit into meters before calculating the value of the velocity. Why? Because G = 6.67 x 10-11 Nt m2/kg^2 demands that the lengths be entered as meters.
A radio station is broadcast at 7.90 x 10^5 HZ what is the wavelength? ANother station is broadcast at 9.83 x 10^7 HZ what is its wavelength?
wavelength = c/f = 3.0 x 10^8 / 7.9 x 10^5 = 380 m wavelength = c/f = 3.0 x 10^8 / 9.83 x 10^7 = 3.1 m
