Astronomy 301 – Spring 2010 Midterm, April 28th - Solutions

Name: Useful equations and constants are listed on the last page of the exam. Multiple choice section, circle the correct answer (1 point each): 1) Which of the following correctly lists speeds from slowest to fastest?

A. Earth's speed of revolution about the Sun, Earth's speed of rotation on its axis, the speed of our solar system orbiting the center of the Milky Way Galaxy, the speeds of very distant galaxies relative to us

B. Earth's speed of rotation on its axis, Earth's speed of revolution about the Sun, the speed of our solar system orbiting the center of the Milky Way Galaxy, the speeds of very distant galaxies relative to us

C. the speeds of very distant galaxies relative to us, Earth's speed of rotation on its axis, Earth's speed of revolution about the Sun, the speed of our solar system orbiting the center of the Milky Way Galaxy

D. the speed of our solar system orbiting the center of the Milky Way Galaxy, Earth's speed of revolution about the Sun, Earth's speed of rotation on its axis, the speeds of very distant galaxies relative to us

E. Earth's speed of revolution about the Sun, Earth's speed of rotation on its axis, the speeds of very distant galaxies relative to us, the speed of our solar system orbiting the center of the Milky Way Galaxy

2) If we reduce the Sun to be about the size of a grapefruit then which of the following describes the size and distance of Earth on the same scale? A. Earth is the size of the tip of a ballpoint pen about 1 meter away from the Sun. B. Earth is the size of a golf ball about 1 meter away from the Sun. C. Earth is the size of the tip of a ballpoint pen about 15 meters away from the Sun. D. Earth is the size of a golf ball about 15 meters away from the Sun. E. Earth is the size of a marble about 25 meters away from the Sun.

3) If the first quarter Moon is at your zenith then approximately what time of day is it? A. midnight B. 3 pm C. noon D. 6 pm E. 9 am

4) How did the Ptolemaic model explain the apparent retrograde motion of the planets? A. It held that sometimes the planets moved backward along their circular orbits. B. It placed the Sun at the center so that the planets' apparent retrograde motion was seen as Earth

passed each one in its orbit. C. It varied the motion of the celestial sphere so that it sometimes moved backward. D. It held that the planets moved along small circles that moved on larger circles around the Sun. E. It held that the planets moved along small circles that moved on larger circles around

Earth.

5) We can learn a lot about the properties of a star by studying its spectrum. All of the following

statements are true except one. Which one? A. The peak of the star's thermal emission tells us its temperature: Hotter stars peak at shorter

(bluer) wavelengths. B. The total amount of light in the spectrum tells us the star's radius. C. We can identify chemical elements present in the star by recognizing patterns of spectral lines

that correspond to particular chemicals. D. We can look at Doppler shifts of spectral lines to determine the star's speed toward or away

from us.

6) Consider two future observatories in space. Observatory X consists of a single 50-meter telescope. Observatory Y is an interferometer consisting of five 10-meter telescopes, spread out over a region 100 meters across. Which observatory can detect dimmer stars, and which one can see more detail in its images? (Assume all else is equal, such as quality of optics, types of instruments, and so on.) A. Observatory X can detect dimmer stars and Observatory Y reveals more detail in images. B. Observatory Y can detect dimmer stars and Observatory X reveals more detail in images. C. Observatory X both detects dimmer stars and reveals more detail in images. D. Observatory Y both detects dimmer stars and reveals more detail in images.

7) What is the purpose of adaptive optics? A. It allows several small telescopes to work together like a single larger telescope. B. It is a special technology that allows the Hubble Space Telescope to adapt to study many

different types of astronomical objects. C. It reduces blurring caused by atmospheric turbulence for telescopes on the ground. D. It allows ground-based telescopes to observe ultraviolet light that normally does not penetrate

the atmosphere.

8) The asteroid belt is located A. between the orbits of Mars and Jupiter B. between the orbits of Earth and Mars C. between the orbits of Jupiter and Saturn D. beyond the orbit of Neptune

9) Which of the following statements is not true about the planets so far discovered around other stars? A. Most of them are much more massive than Earth. B. Images reveal that most of them have atmospheres much like that of Jupiter. C. Many of them orbit closer to their star than Jupiter orbits the Sun. D. Many of them have been discovered by observing Doppler shifts in the spectra of the stars they

orbit.

10) Sunspots are cooler than the surrounding solar surface because A. they are regions where convection carries cooler material downward. B. strong magnetic fields slow convection and prevent hot plasma from entering the region. C. magnetic fields trap ionized gases that absorb light. D. there is less fusion occurring there. E. magnetic fields lift material from the surface of the Sun, cooling off the material faster.

11) If the distance between us and a star is doubled, with everything else remaining the same, the luminosity A. is decreased by a factor of four, and the apparent brightness is decreased by a factor of four.

B. is decreased by a factor of two, and the apparent brightness is decreased by a factor of two. C. remains the same, but the apparent brightness is decreased by a factor of two. D. remains the same, but the apparent brightness is decreased by a factor of four. E. is decreased by a factor of four, but the apparent brightness remains the same.

12) What do we mean by the main-sequence turnoff point of a star cluster, and what does it tell us? A. It is the point in a star cluster beyond which main-sequence stars are not found, and it tells us

the cluster's distance. B. It is the spectral type of the hottest main-sequence star in a star cluster, and it tells us the

cluster's age. C. It is the luminosity class of the largest star in a star cluster, and it tells us the cluster's age. D. It is the mass of the most massive star in the star cluster, and it tells us the cluster's size.

13) The vast majority of stars in a newly formed star cluster are A. very high-mass, type O and B stars B. red giants C. about the same mass as our Sun D. less massive than the Sun

14) According to our current understanding, what is a nova? A. an explosion on the surface of a white dwarf in a close binary system B. the explosion of a massive star at the end of its life C. a rapidly spinning neutron star D. an explosion on the surface of a neutron star in a close binary system

15) What is a pulsar? A. a high-mass star undergoing thermal pulses due to new shell fusion layers in the interior B. a white dwarf in a binary system periodically going nova as it accretes mass from the binary star C. a white dwarf in a binary system with a hot accretion spot that periodically comes into view as

the stars orbit each other D. a rotating neutron star beaming radiation along its magnetic axis

Essay answer section (5 points each):

16) (a) Briefly summarize the nebular theory that describes the formation of our solar system. The nebular theory states that the solar system formed from a collapsing cloud of gas and dust billions of years ago. The cloud that formed the solar system began by collapsing under its own gravity and as it did so, it spun faster to conserve angular momentum. When clumps of gas and dust collided, both took velocities nearer the average. Since the average velocity tends to point along a flattened disk, the solar nebula eventually took on a disk shape. The terrestrial planets are thought to have been formed by solid bits of silicates and metals colliding and sticking together in the solar nebula. The silicates and metals were able to condense in the hotter inner part of the disk where the temperature was too high for the hydrogen compounds to be solids. As the silicates form larger clumps, their gravity begins to allow them to gather up mass more efficiently. The biggest bodies, now called planetesimals, are able to collect mass the fastest. Eventually, the biggest planetesimals gather up the others and become planets. The formation of the jovian planets is similar to the terrestrial planets in the early stages with the major exception that the jovian planets are able to use solid hydrogen compounds (“ices”) to build up their masses. Since these ices are far more abundant than the silicates and metals, the jovian

planets were able to build faster and larger than the terrestrial planets. Eventually, they became large enough to even hold on to nebular gases (hydrogen and helium). In this respect their formation is quite different from that of the terrestrial worlds. Once the jovian planets could collect gas, it was easy for them to grow very large in size.

(b) Discuss at least three observed patterns of motion in our solar system and how they are consistent with the nebular theory.

Observed patterns that are consistent with the nebular theory based on conservation of angular momentum during the collapse of the nebular cloud are: all planets orbit the Sun counterclockwise when seen from above Earth's North Pole, all planetary orbits lie nearly in the same plane, almost all planets travel on nearly circular orbits, most planets orbit in the same direction in which they rotate (counterclockwise), almost all moons orbit their planet in the same direction as the planet's rotation and near the planet's equatorial plane, and the Sun rotates in the same direction in which the planets orbit. 17) (a) What are the two main categories of planets in our solar system? The two main categories of planets are terrestrial and jovian.

(b) List at least three properties of the planets that differ for the two categories and briefly state how they differ.

Differences that exist in the planets from the two categories include: Distance from the Sun: terrestrial planets are closer Radius: terrestrial planets are significantly smaller Mass: terrestrial planets are significantly less massive Average density: terrestrial planets are denser Orbital period: terrestrial planets take less time to orbit the Sun Rotation period: terrestrial planets take more time to complete a rotation Composition: terrestrial planets are composed of rocks and metals whereas jovian planets are composed primarily of H, He, and hydrogen compounds Moons: terrestrial planets have fewer moons, if any at all Rings: terrestrial planets have no rings whereas all jovian planets have rings

(b) State which category each of the 8 planets in our solar system falls into. Terrestrial Planets: Mercery, Venus, Earth, Mars Jovian Planets: Jupiter, Saturn, Neptune, Uranus

(c) Describe why these two categories of planets exist in our solar system.

The two categories exist because the jovian planets formed beyond the frost line in the nebular cloud where ices could form allowing for larger planetesimals to form. The jovian planets were then able to accumulate substantial amounts of material into the planets themselves or into the disks surrounding them that formed many of their satellites. Their larger mass also allowed them to capture additional satellites. 18) (a) What are the three primary methods by which we detect extrasolar planets? The three primary methods are: Doppler technique, astrometric technique, and the transit method.

(b) Choose two of these methods and describe how each works and what properties of the extrasolar planet can be measured via each method.

Doppler technique: this method uses the Doppler effect to detect the small wobble in a star’s position due to the planet orbiting it. This technique can measure the planet's mass, orbital distance, and the eccentricity of the planet's orbit. Astrometric technique: similar to the Doppler technique, except that rather than looking for motions toward and away from us, it looks for tiny shifts in position. This technique can measure the planet's mass and orbital distance. Transit method: when a planet appears to cross in front of the star as seen from Earth, the transit method can be used to detect the planet’s presence since the star’s light is dimmed slightly. This method can measure the planet's radius, orbital distance, density, and composition. 19) Describe the distinguishing features of the four innermost layers of the Sun. The four innermost layers of the Sun are the core, radiation zone, convective zone, and the photosphere. The innermost layer of the Sun is the core. Distinguishing features of the core include the fact that it is this layer that produces the Sun's energy through nuclear fusion of hydrogen into helium. This is possible because the core has a temperature of 15 million K and a density 100 times that of water. In the radiative zone, the Sun’s energy is carried outward by photons of light that slowly make their way through the radiative zone via radiative diffusion due to the extreme density of the plasma which has a temperature up to 10 million K. In the convective zone the energy is transported upward by hot gas rising and cooler gas sinking. The photosphere is the visible surface of the Sun and has a typical temperature of 5,800 K. It appears granulated due to the convection cells of the convection zone below. This is also the layer where sunspots exist. 20) (a) Briefly describe the life history of a protostar from its beginnings as a part of a molecular cloud

to the moment hydrogen fusion begins. In cold, dense molecular clouds, gravity brings material together. As gas moves inwards it converts gravitational potential energy to thermal energy and warms up. The surface temperature of the protostar remains around 3,000 K as long as convection remains the dominant mechanism for energy transport. During this stage the protostar's radius becomes smaller and therefore its luminosity decreases. When the dominant energy transport mechanism switches to radiative diffusion then the temperature rises and the luminosity increases while the protostar continues to contract. Once the cloud becomes so dense that the thermal radiation cannot escape, the temperature rises rapidly, and nuclear fusion begins. As the cloud has collapsed from a large size to a small size, it must spin very fast to conserve angular momentum. This results in the formation of a protostellar disk around the protostar. Planets may form in this disk as the star continues to grow. Eventually stellar winds and jets clear away the surrounding gas and a newly formed star emerges.

(b) How would this life story be different if the protostar formed in a cloud that had absolutely no angular momentum?

If there were exactly zero rotation in the star then fusion would still begin in the stellar core but there would be no rotation of the star nor the nebular gas surrounding the star. This would result in the lack of the following phenomena: no protostellar disk, no jets, and probably no strong stellar winds. Also, close binaries would be much less likely. 21) (a) Briefly summarize the stages of life for a low-mass star starting with its existence on the main

sequence. The star will fuse hydrogen in its core for around 10 billion years while on the main sequence. When the core hydrogen is used up, the core contracts until it is supported by electron degeneracy in what is now called a subgiant star, hydrogen fusion continues in a shell outside the

core, and as the outer layers expand and cool the star becomes a red giant. Helium fusion begins in the core, but since the core is degenerate a helium flash takes place and rapidly spreads throughout the core. Helium fusion stabilizes in the now helium burning star, and the star moves left on the H-R diagram. Core helium is used up and helium begins fusing in a shell outside the core, with hydrogen still fusing in a shell above it, which is called "double shell" burning. The outer layers expand, and the star again becomes a red giant. The star undergoes thermal pulses and loses its outer layers through a stellar wind. The core shrinks and heats up but is not able to fuse any more elements. The star becomes a planetary nebula as heat from the core blows away and heats up the gas left over from the red giant phase. Only the naked degenerate core is left behind, which we call a white dwarf.

(b) Draw a Hertzsprung-Russell diagram, label the axes and the direction of increasing values, and mark the location of the main sequence. Then draw the life track of a star with a mass equivalent to the Sun's mass and identify at least 4 of the life stages discussed above on this life track.

Each phase in italics in the paragraph above is labeled on the H-R diagram below.

22) (a) Briefly describe how the stages of life for high-mass stars differ from low-mass stars and why. The first stages are similar to those of a low-mass star, except that they happen over much shorter time periods. While on the main sequence, the star fuses hydrogen by the CNO cycle (rather than the proton-proton chain that is used in low-mass stars) and remains at this stage only for several million years. In addition to helium fusion, high-mass stars also undergo alpha-capture, which creates heavier elements by fusing a helium nucleus with an existing atom. After helium is used up in the core, the core contracts while helium and hydrogen fusion continue in outer shells. Unlike in low-mass stars the core contracts enough for carbon ignition to occur, and the star moves left again on the H-R diagram while carbon fusion occurs in the core. The process continues for stars of still higher mass, zigzagging across the H-R diagram as heavier elements are fused in the core and used up as fuel. Each fusion stage requires less time until iron is finally produced in the core. Iron cannot be fused to produce energy, so the core collapses and, unlike in low-mass stars, the pressures increase so much that electrons and protons are converted to neutrons. A high quantity of neutrinos is released, which may help force the outer layers violently outward in an explosion called a supernova. Elements heavier than iron are created, the outer layers move away from the core at great velocities, and only a neutron star or black hole is left as a remnant.

(b) How do the lifetimes of high-mass stars compare to low-mass stars and why?

High-mass stars have a lifetime that is significantly lower than low-mass stars. This is because high-mass stars have 10 to 100 times more mass (fuel) than a typical low-mass star. This greater mass produces a much higher downward gravitational pressure, leading to much higher core temperatures and higher rates of fusion (which is achieved in part because the CNO cycle can be used rather than just the proton-proton chain). The luminosity of such stars is therefore 1,000 to 1 million times greater than in low-mass stars. So, although high-mass stars have more fuel to burn, they burn it at a much higher rate and therefore run out of fuel much more quickly.

(b) What are the two possible remnants left behind at the end of the life of a high-mass star and what dictates which is produced?

The two possible remnants of the supernovae that occur at the end of the life of high-mass stars are neutron stars or black holes. The mass of the remnant dictates which will be left behind. A black hole is produced if the remnant is greater than about 3 solar masses. Quantitative section (5 points each): Be sure to show your work and make sure your final answer is clearly indicated.

23) If the surface of a star, which emits approximately thermal radiation, is 12,000 K then: (a) How much power does it emit per square meter? We can use the Stefan-Boltzmann law to compute the energy emitted by the object from every

square meter: 4emitted power = Tσ

where T is the temperature and 84 2

watt5.7 10K m

σ −= × .

For our 3,000 K object:

emitted power = (5.7 x 10-8 watt/K4m2) x (12,000 K)4 = 1.2 x 109 watt/m2

(b) What is its wavelength of peak intensity? We find the wavelength of the peak emission with Wien’s law;

peak2,900,000 nm

Tλ =

where T is again the temperature, measured in Kelvins. For a 3,000 K object, we get:

λpeak = (2.9 x 1026 / T) nm = (2.9 x 106 / 12,000) nm = 241.7 nm

(c) What is the frequency of peak intensity?

Using the fact that c = fλ we find

f = c/ λ = (3 x 108 m/s) / (241.7 x 109 m) = 1.2 x 1015 Hz

24) Suppose that two stars in a binary star system are separated by a distance of 100 million kilometers and are located at a distance of 20 light-years from Earth.

(a) What is the angular separation of the two stars? Give your answer in both degrees and arcseconds.

Converting the distance of 20 light-years to seconds:

20 light-years x (1013 km / light-year) = 20 x 1013 km

We can then proceed with the angular separation formula:

physical separationangular separation = 206,265distance

′′ ×

angular separation = 206265 arcseconds x (100 x 106 km) / (20 x 1013 km) = 0.1 arcseconds

Converting to degrees:

0.1 arcseconds x (1 arcminutes/60 arcseconds) x (1 degree/60 arcminutes) = 2.9 x 10-5 degrees

(b) If one of these stars has a physical diameter of 107 km, then what is its observed angular diameter? State your answer in both degrees and arcseconds.

Using the following relation:

2 (distance) (angular size)physical size360°

π × ×=

we find an angular diameter of

angular size = 360 degrees x (107 km) / (2 π x 20 x 1013 km) = 2.9 x 10-6 degrees.

Converting this to arcseconds:

2.9 x 10-6 degrees x (3600 arcseconds / 1 degree) = 0.01 arcseconds

Alternatively, if we notice that the physical diameter is one tenth the physical distance of the two stars then we can simply infer from the calculation from part (a) that the physical diameter of the one star is 0.01 arcseconds and 2.9 x 10-6 degrees.

25) You want to observe a binary stellar system with an angular separation of 0.025 arcseconds.

(a) What diameter telescope would you need to resolve this system at the optical wavelength of 500 nm?

Using the equation 5 wavelength of lightdiffraction limit (2.5 10 arcseconds)diameter of telescope

= × , where the

angular separation of 0.025 arcseconds is taken as the diffraction limit, we can solve for the telescope diameter. By converting the wavelength to meters the answer will be given in meters.

diameter = 2.5 x 105 arcseconds x (500 x 10-9 m) / 0.025 arcseconds = 5 meters

(b) What diameter telescope would you need to observe this system at the infrared wavelength of 2200 nm?

Using the same approach as above:

diameter = 2.5 x 105 arcseconds x (2200 x 10-9 m) / 0.025 arcseconds = 22 meters

Alternatively, we can use the simple fact that the diameter is proportional to the wavelength and is therefore a factor of (2200nm/500nm) larger than the answer calculated in part (a) and more easily arrive at the answer of 22 meters.

(c) Using the technique of interferometry you observe this system with two radio telescopes separated by 2100 meters and observing at a wavelength of 21 cm. Can you resolve the binary system?

Calculating the diffraction limit of the system using the same equation as in part (a) and converting the observed wavelength to meters to have the units cancel out correctly:

diffraction limit = 2.5 x 105 arcseconds x (21 x 10-2 m) / (2100 m) = 25 arcseconds

This diffraction limit is significantly greater than the 0.025 arcsecond angular separation of the two stars, therefore the binary system can not be resolved by these telescopes.

26) Charon completes its orbit around Pluto in 6.4 days with a semimajor axis of 19,700 km.

(a) Calculate the combined mass of Pluto and Charon.

Using Newton's version of Kepler's 3rd law 2

2 3

1 2

4( )

p aG M M

π=

+ we can determine the

combined mass of the two objects if we convert the orbital period to seconds and semimajor axis to meters:

p = 6.4 days x (24 hrs/1 day) x (60 minutes / 1 hr) x (60 s / 1 minute) = 5.5 x 105 s

Mtotal = 4 π2 / G x (a3/p2) = 4 π2 / (6.67 x 10-11 m3 / kg s2) x (19700 x 103 m)3 / (5.5 x 105 s) 2

= 1.5 x 1022 kg

(b) If the semimajor axis of the orbit of these two objects were to double, by how much would the gravitational force between these objects change?

Based on Newton's universal law of gravity , doubling the distance would reduce the gravitational force to 0.25 of what it was originally.

27) Imagine you are on a planet that has a mass of 200 MEarth and a radius that is 10 REarth.

(a) Calculate the acceleration of gravity on the surface of this planet. The easiest way to find the accelerations with the given data is by comparison to the acceleration of gravity on Earth using the relationship below.

planet22

planet planet planet Earth

EarthEarth Earth planet2Earth

MGa R M R

Ma M RGR

⎛ ⎞= = × ⎜ ⎟

⎝ ⎠

where aEarth = 9.8 m/s2. Therefore,

aplanet = aEarth x (200 MEarth / MEarth) x (REarth / 10 REarth)2 = aEarth x 200 / (102)

= 2 aEarth = 19.6 m/s2

(b) How much would a person who weights 100 lbs on Earth weigh on this planet?

Since weight = mass x acceleration, a person's weight on another world is:

Wplanet = aplanet/aEarth x WEarth.

Therefore, Wplanet = 2 aEarth / aEarth x 100 lbs = 200 lbs

28) The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 3 months. The hydrogen absorption line that at rest has a wavelength of 656.285 nm is observed at a wavelength of 656.460 nm in star A and 656.110 nm in star B.

(a) What is the radial velocity of each star A and star B? Be sure to indicate whether each star is moving away or toward us.

We use the Doppler shift formula:

vrad / c = (λshift – λ rest) / λrest

Star A: vrad = 3 x 108 m/s (656.460 nm – 656.285 nm) / 656.285 nm = 80,000 m/s

The radial velocity of Star A is therefore 80,000 m/s away from us.

Star B: vrad = 3 x 108 m/s (656.110 nm – 656.285 nm) / 656.285 nm = - 80,000 m/s

Alternatively, we observe that the wavelength shift of this star is equivalent to that of Star A, except in the opposite direction (to a shorter rather than longer wavelength, implying it is moving toward us rather than away from us). The radial velocity of Star B is therefore 80,000 m/s towards us.

(b) Assume the stars have equal orbital speeds (take the average from your answers for star A & B in part a). What are the masses of the two stars? Assume that each of the two stars traces a circular orbit around their center of mass.

To solve this problem we need to make use of Newton's version of Kepler’s third law (alternatively one could also make use of Kepler's law):

22 3

1 2

4( )

p aG M M

π=

+

We already know the orbital period is 3 months, but this must be converted to seconds:

3 months = 0.25 year x 3.1 x 107 seconds / year = 7.8 x 106 seconds

To get the semimajor axis of the orbit we use the fact that each of the stars in the binary completes an orbit once every 3 months, at a velocity of 80,000 m/s. The circumference of that orbit is thus the time it takes to complete an orbit multiplied by the speed at which the stars move through the orbit, or:

Orbital distance = velocity x time = 80,000 m/s x 7.8 x 106 seconds = 6.2 x 1011 m

Since we are assuming a circular orbit, the average semimajor axis a of that orbit is thus:

a = circumference / 2π = 6.2 x 1011 m / 2π = 1 x 1011 m

Plugging the orbital period and semimajor axis into Newton's version of Kepler's third law and solving for the total mass of the binary system:

M1 + M2 = (4π2/G) x a3/p2 = (4π2/(6.67 x 10-11 m3 / kg s2)) x (1 x 1011 m)3/(7.8 x 106 s)2

= 9.7 x 1030 kg

Converting this to solar masses we get M1 + M2 = 4.9 Msun.

Because the stars have equivalent orbits, they must have equal masses, so each one has a mass of 2.45 Msun.

29) You've just discovered a new X-ray binary. The system contains a bright, G2 main-sequence star orbiting an unseen companion. The separation of the stars is estimated to be 12 million kilometers, and the orbital period of the visible star is 3 days.

(a) Use Newton's version of Kepler's third law to calculate the sum of the masses of the two stars in the system. Give your answer in both kilograms and solar masses.

We first convert the orbital period to seconds and separation to meters:

p = 3 days = 3 days x (24 hours/day) x (60 min/hr) x (60 s/min) = 2.6 x 105 seconds

a = 12 x 106 km x (103 m/km) = 1.2 x 1010 m

Using Newton's version of Kepler's third law we then find:

MTOT = (4π2/G) x a3/p2 = (4π2/(6.67 x 10-11 m3 / kg s2)) x (1.2 x 1010 m)3/(2.6 x 105 s)2

= 1.5 x 1031 kg

Converting this to solar masses: MTOT = 7.6 Msun.

(b) Knowing that a G2 main-sequence star has a mass of 1 Msun, what is the mass of the unseen companion?

Since MTOT = Mcompanion + MG2 = 7.6 Msun we find that Mcompanion = 6.6 Msun.

(c) Assuming this unseen companion is a black hole, what is its Schwarzschild radius in kilometers?

Converting the black hole's mass into units of kilograms:

MBH = 6.6 Msun x (2.0 × 1030 kg / Msun) = 1.3 x 1031 kg

Using the Schwarzschild radius formula:

Rs = 2 G M / c2 = 2 x (6.67 x 10-11 m3 / kg s2) x (1.3 x 1031 kg) / (3 x 108 m/s)2

= 19,268 m = 19 km

Useful Equations & Constants

2 (distance) (angular size)physical size360°

π × ×=

physical separationangular separation = 206,265

distance′′ ×

Stefan-Boltzmann's Law: 4emitted power = Tσ

Wien's Law: peak2,900,000 nm

Tλ =

vrad / c = Δλ / λo or vrad / c = (λshift – λ rest) / λrest

5 wavelength of lightdiffraction limit (2.5 10 arcseconds)diameter of telescope

= ×

2

2 3

1 2

4( )

p aG M M

π=

+

Rs = 2 G M / c2 Constants:

c = 3 x 108 m/s h = 6.626 x 10-34 joule seconds σ = 5.7 x 10-8 Watts/K4m2 G = 6.67 x 10-11 m3 / kg s2 g = 9.8 m/s2

Msun = 2.0 × 1030 kg 1 light-year ~ 1013 km