Physics 134
Homework Set 8 Solutions
Do Problems 4-6 from HW7.
4. In class, we computed the surface temperature of Earth, assuming it was in equilibrium with Solar radiation and noting the importance of the atmospheric greenhouse effect. In this problem we will use the same kind of calculation to learn about other parts of the Solar System.
In previous years, I asked students to repeat the calculation we did in class. Since the solutions I provided then are a nice (in my oh-so -humble opinion) I insert them here in case they are helpful to anyone.
(a) At a distance of D⊕ = 1 AU (near Earth) the Sun radiates with an intensity (flux)
of b = 1370 W/m2. Since the area of a sphere or radius R is 4πR2 we can relate
this to the Sun’s luminosity. Write an expression relating the two. Note that I have adopted a Sun-centered notation in which the distance between Earth and Sun is denoted D⊕ rather than D . It’s still 1 AU.
The flux at a distance D⊕ from the Sun is given by
F (D ) =
L
4πD2 ⊕
,
(b) The flux at the Solar surface is related to the surface temperature by the Stefan-Boltzmann law F = σT 4. Use this to find an expression for b in terms of T ,
the Solar radius R , and the Sun’s distance from Earth D⊕ = 1 AU.
Considering a sphere right outside the Solar photosphere, we have also F (R ) =
L
4πR2 .
Dividing the two expressions we find the scaling relation F (R ) b = D⊕ R 2 , so that b = R D⊕ 2 F (R ) .
The Stefan-Boltzmann law relates F (R ) to T , and substituting this we have b = R D⊕ 2 σT 4 .
(c) The total power from solar radiation impinging on Earth is the product of the flux b with the planet’s “effective area.” This last is most easily found by replacing
the planet with a disk of the same radius facing the Sun. Clearly this receives the same amount of radiation as does the planet. Find a formula for the total power Pin incident as radiation on Earth, which has a radius of R⊕ = 6380 km. There
is no need to evaluate this numerically.
Replacing Earth with a disk of radius R⊕ which, since it casts the same shadow
as the Earth, would receive exactly the same amount of incoming radiation, and would have a surface area πR2⊕, we find that the total power incident on Earth is
Pin = πR2⊕b =
R
D⊕
2
πR2⊕σT 4 .
(d) A fraction of this incoming radiation simply reflects off Earth back to space. This fraction is called the Bond albedo of the planet, and for Earth it is approximately aB
⊕ = 0.31. The rest of the energy (0.69 of the power you computed in (d)
above) is absorbed by the planet, heating its surface. The rate at which energy is absorbed by Earth is thus Pabs = (1 − a)Pin.
Why does the Earth not get hotter and hotter as it absorbs this energy? The Earth is itself radiating energy to space. The luminosity of the Earth is related to its surface temperature and radius by the same expression we used for the Sun. Write an expression for the total luminosity of Earth L⊕ in terms of R⊕ and the
surface temperature T⊕ of the planet.
Solar radiation is absorbed by the Earth’s crust. Since the crust, like the Sun’s photosphere, radiates as a blackbody, it emits a flux of infrared radiation given by the Stefan Boltzmann law from each square meter of its surface. Thus we have
Pout= L⊕ = 4πR2⊕σT⊕4
(e) If L⊕ > Pabs the planet radiates more than it absorbs, cooling down. Conversely,
if L⊕ < Pabs the planet absorbs more energy than it radiates, warming it up.
Equilibrium obtains when the two quantities are equal, L⊕ = Pabs. Using all
the expressions you have found above, find the surface temperature of Earth at equilibrium in terms of D⊕, R T , and a⊕. Note that R⊕ drops out of your
expression – the equilibrium temperature is the same for a planet of any size. Equating the two expressions we have
4πR2⊕σT⊕4 = (1 − a) R D⊕ 2 πR2⊕σT⊕4 , or solving for T⊕ T⊕ = (1 − a)1/4 R 2D⊕ 1/2 T .
(f) Now, finally evaluate your prediction above numerically for D⊕ = 1 AU = 1.496×
1011m, R = 6.96 × 108m, and T = 5780 K. Does your result agree with the
average surface temperature on Earth which is 282 K? Use Wien’s law to deter-mine the wavelength at which Earth radiates to space. What kind of radiation is this?
Inserting the numbers
T⊕ = (0.69)1/4· 6.96 × 108 2 · 1.496 × 1011 1/2 5780 = 254 K .
This is about 19 degrees below the freezing point (273 K), and if this were the temperature of Earth life would likely never have evolved here. The actual average of 288 is above freezing, and with local and temporal variations allows for nice days as we have had this fall in NC.
Using Wien’s law the ground, at a temperature of 288 K, radiates at a wavelength of
λ = 0.0029 m
T = 1.01 × 10
−5
m = 10.1 µ . At 10 microns, this is radiation in the near infrared.
(g) There are several reasons for the discrepancy, including the ongoing produc-tion of heat in the Earth’s core by both Kelvin–Helmholtz processes (contrac-tion converting gravita(contrac-tional energy to heat) and by nuclear processes (radioac-tive decays). We can measure this, and it turns out that these processes con-tribute about 75 W/m2 to heating the surface. This is not negligible compared to b = 1370 W/m2, but certainly not enough to explain the discrepancy.
The dominant effect, however, is the fact that Earth’s atmosphere, while quite transparent to the incoming visible light, absorbs the radiated light quite effi-ciently. This is subsequently reradiated from the heated air (else atmosphere would heat up), but some of the energy is radiated back toward the ground. We can set up a simplified model of this greenhouse effect by modeling the at-mosphere as a black body with uniform temperature Tatm. We then have two
objects, Earth’s surface, and the atmosphere, each in equilibrium under exchange of radiation energy with each other, with space, and with the Sun. The relevant processes are:
Solar radiation is (we assume) ignored by the atmosphere and impinges directly on Earth’s surface. A fraction a = 0.31 of this is reflected, again passes the atmosphere unimpeded, and disappears into space. The rest is totally absorbed by the surface.
The atmosphere radiates at a uniform (in our simple model) temperature over the entire surface of Earth at a rate determined by its temperature. The atmosphere has two surfaces, so its effective area is twice that of the Earth’s surface (the height of the atmosphere above ground is negligible). Radiation emitted from the upper surface of the atmosphere disappears into space. Radiation emitted towards Earth is completely absorbed and contributes (together with the direct Solar heating mentioned above) to heating the surface.
As a result of all this the surface radiates at its (assumed uniform) temperature T⊕. The radiation is infrared, and the atmosphere is not completely transparent
to this. A fraction g of the emitted IR is absorbed by the atmosphere (this is the only radiation absorbed there in our model) heating the atmosphere. The rest of Earth’s emissions disappear into space.
Based on this picture, write two equilibrium equations. One equates the power absorbed by the atmosphere to the total power it emits. The second expresses the same for the surface of Earth. These two equations determine both Tatm and
T⊕. Find a value for the greenhouse factor g that leads to a surface temperature
of 288 K as measured in reality.
In our model there are two independent systems, Earth’s surface at a temperature T⊕ and the atmosphere at temperature Tatm. They interact with each other and
with the environment through radiative heat transfer as described above. The equilibrium temperature of each system is determined by setting the net ourgoing power equal to the net incoming power.
Let’s start with the atmosphere, this is simpler. Since it is transparent to visible light, sunlight has no impact on this. The only incoming power is the infrared light emitted from Earth that the atmosphere absorbs. This is a fraction g of the Earth’s luminosity which we computed above, so
Pabs,atm= g 4πR2⊕σT⊕4 .
The outgoing power is in the form of blackbody radiation from the surface of the atmosphere. In our model this “surface” comprises two concentric spheres, the “bottom” and “top” of the atmosphere, each of radius R⊕, so that
Pout,atm = 8πR2⊕σTatm4 .
Setting them equal at equlibrium we have Tatm =
g 2
1/4 T⊕ .
It’s worth pausing here to note that we have here the essence of the way the greenhouse effect works to keep our planet habitable. To make things simple let’s imagine that g = 1, meaning the atmosphere absorbs all of the infrared radiation incident on it. Looking from space in infrared light, an observer would see not the radiation from the surface but rather the atmosphere with an effective temperature lower by a factor of 2−1/4. The total power emitted by this would, in equilibrium, equal the total power absorbed from the Sun. The atmosphere, in our model, has no effect on this, so with g = 1 we would necessarily have
Tatm = 254 K; since the atmosphere is cooler than the surface, the equilibrium
surface temperature is increased by the presence of the atmosphere.
Now for Earth’s surface. Incoming power to Earth includes Pabs computed above
as the power absorbed as visible sunlight. Since the atmosphere is transparent to this, our previous calculation is still valid. But there is an additional component of incoming power: the radiation emitted from the bottom of the atmosphere impinges on the surface. Let’s assume for a moment, because this is what I suggested in the problem, that this is completely absorbed by the surface. Then we have a total incoming power to the surface of
Pabs,⊕ = (1 − a)πR2⊕
R
D⊕
2
σT 4 + 4πR2⊕σTatm4 .
The surface emits power as blackbody radiation so we have Pout⊕ = 4πR2⊕σT⊕4 .
At equlibrium these must be equal, leading to
(1 − a) R D⊕
2
T 4 + 4Tatm4 = 4T⊕4 .
Inserting the expression we found above for Tatm in terms of T⊕ we have
T⊕4(1 − g/2) = (1 − a) R 2D⊕ 2 T 4 , or T⊕ = 1 − a 1 − g/2 1/4 R 2D⊕ 1/2 T .
Comparing to the expression we found in (e) we see that the net effect of the greenhouse effect is the additional factor of (1 − g/2)−1/4. We thus have
T⊕ = (1 − g/2)−1/4 T⊕(g = 0) .
To keep the surface at our pleasant 288, we need
1 − g/2 = 254 288
4 ,
or
g = 2 1 − 254 288
4!
= 0.790 .
Inserting this we predict that the atmosphere will have an effective blackbody temperature of Tatm= g 2 1/4 T⊕ = 228 K .
The total power emitted by Earth is the sum of the blackbody radiation from the top surface of the atmosphere and the 1 − g fraction of the radiation from the surface that escapes absorption. Adding these, we would of course find that together they equal the total power absorbed from sunlight.
Before going on, it’s sensible to admit that our model is over simplified. I men-tioned above that we neglected internal heating of the surface. I list here a few other neglected effects; you can probably think of some others. As usual, we are not actually trying to replace the IPCC; our job is to develop an understanding of what they are talking about.
• We treated the atmosphere as a blackbody with a uniform temperature. In fact, of course, temperature varies significantly with altitude, and both emission and absorption are distributed throughout the air column. Accounting for this re-quires complex models.
• The sunlight impinging on Earth is not all visible light. The broad blackbody spectrum of the Sun extends into the IR, so that the atmosphere absorbs some sunlight. Incorporating this in our model is not too hard, try it!
• A significant fraction of the visible light reflected from Earth is in fact reflected from cloud tops (think of images taken from space, what do you see?). In our simple model this is simply folded into the effective Bond albedo of the planet. If we allow for atmospheric absorption we need to consider reflection at cloud level separately.
• There are other, non-radiative, mechanisms of heat transfer between surface and atmosphere. One example of this is latent heat of water. Water evaporates at the surface. As we noted, turning liquid into vapor absorbs energy in the form of
latent heat . When the vapor rises it condenses back to water and ice in clouds, liberating this heat into the atmosphere.
There is a slide (courtesy NASA) outlining these and other effects in the Unit 4 slides posted. In our simplified model, all of these are lumped into an effective value for the greenhouse factor g.
There is one omission that likely confused at least some of you (the more careful ones) and which I would like therefore to address here. It is a bit inconsistent to assume that Earth’s surface reflects 0.31 of the sunlight impinging on it but none of the IR radiation from the atmosphere. The Bond albedo, as discussed below, is in fact an average reflectivity weighted by the Solar spectrum which, while peaked in visible light, includes IR radiation. So let’s see how our answers would change if we were to assume that Earth reflects the same fraction of atmospheric emissions as it does of the incoming sunlight.
This would add a term
ga 4πR⊕2 σTatm4
to the power absorbed by the atmosphere, the fraction of the reflected IR radiation that the atmosphere would absorb. Equilibrium for the atmosphere would then require
gT⊕4 + gaTatm4 = 2Tatm4 , so that Tatm= g 2 − ag 1/4 T⊕ .
In the energy balance for the surface, the contribution of atmospheric radiation would be smaller than we assumed above by a factor of 1 − a to account for the fraction reflected away, so we would find
(1 − a) " R 2D⊕ 2 T 4 + Tatm4 # = T⊕4 ,
and upon inserting Tatm and simplifying
T⊕ = (1 − a)(2 − ag) 2 − g 1/4 R 2D⊕ 1/2 T = (2 − ag) (2 − g) 1/4 T⊕(g = 0) .
Solving for g we see that to get T⊕ = 288 we need g = 2 288 254 4 − 1 288 254 4 − 0.31 = 0.972 .
We see that taking IR reflection into account suppresses the effect, requiring a higher effective g to achieve the same outcome.
(h) This helps make sense of some of the debates about global warming. Adding greenhouse gases – efficient absorbers of radiation like CO2 – to the atmosphere
can increase g, leading to higher equilibrium temperature. Note that we are here neglecting one of the nastiest features of the problem: increasing the surface temperature leads to increased melting of polar icecaps. Where would that en-ter our calculations? Would it mitigate the warming effect or enhance it? Our simplified model neglects other mechanisms for energy exchange between surface and atmosphere. Can you suggest any? Find the value of g for which the sur-face temperature would rise by 2 K – a temperature rise widely considered as a tipping point at which nonlinear processes begin to significantly accelerate the temperature increase.
Melting icecaps replace highly reflective ice with darker more absorptive water. This decreases the albedo, heating Earth and in turn accelerating the melting. This is one of the many feedback mechanisms that make the temperature sensitive to small changes in g in ways that our simple model does not explicitly exhibit. I mentioned some of the effects we neglect above, I hope you can think of others. To get a mean surface temperature of 290 K, we would need a g of
g = 2 1 − 254 290
4!
= 0.823
using our original model (with no IR reflection from Earth) or
g = 2 284 254 4 − 1 284 254 4 − 0.31 = 1.007
in the more careful model that took this into account. Of course g ≤ 1 so our model seems to predict a limit to global warming so long as we neglect the feedback loops of about (set g = 1) of 289.6 K. Presumably feedback is essential.
(a) Mars orbits, as we know, at a distance of about 1.5 AU from the Sun. As we saw, the Martian atmosphere is very thin and we can neglect any greenhouse effects. Mars has a Bond albedo of 0.25, meaning that 0.25 of the total solar radiation impinging on the planet is reflected with the rest being absorbed and heating the surface. Find the equilibrium temperature on Mars.
Neglecting greenhouse effects the equilibrium temperature of a planet with Bond albedo a orbiting at a distance D from a star with temperature T and radius
R was found above to be
T = (1 − a)1/4 R 2D
1/2 T .
We can insert the numbers for Mars directly or divide by our answer for a greenhouse-free Earth and use scaling to predict
TMars =
1 − .25 1 − .31
1/4
(1.5)−1/2· 254 = 212 K .
Wikipedia cites a mean surface temperature of 210 K.
(b) Repeat the same calculation for Saturn, which has a Bond albedo of 0.342 and orbits, as we have found, at a distance of 9.58 AU from the Sun. Since on Saturn the “surface” is the reflective cloudtops, we can again ignore greenhouse effect. Using scaling we find here
TSaturn =
1 − .342 1 − .31
1/4
(9.58)−1/2· 254 = 81.1 K .
(c) The actual temperature of Saturn’s cloud tops is measured to be 93 K. This should be higher than your answer in (b), indicating we have missed something. Find the ratio of the power Saturn radiates to the power the planet absorbs from the Sun (this will be a number greater than one). Saturn radiates more than it absorbs. What we have missed is the contribution of internal heat – in the case of Saturn largely due to Kelvin-Helmholtz processes in which heat is generated as the planet contracts through conversion of gravitational potential energy. In the case of Saturn this largely takes the form, as noted in class, of Helium condensing
and raining out of the atmosphere and onto the liquid mantle deeper within the planet. What fraction of Saturn’s radiated power is due to internal heat?
Saturn is hotter than it should be. Since its emission is determined as we saw above by its temperature, this means it emits more energy than it absorbs from the Sun. We can go back and do the calculation but there is an easy way here. If Saturn’s temperature were 81.1 K it would emit precisely as much as it absorbs. Since it is hotter, it emits more, and our expression for Pout above shows that all
other things being equal, this depends on temperature as T4. Thus we see that for Saturn Pout Pabs = 93 81.1 4 = 1.73 .
Saturn emits 1.73 times as much as it absorbs from the Sun. A fraction
rinternal = 0.73/1.73 = 0.42
of this is due to internal heat. 42% of the light emitted by Saturn as infrared radiation is due to internal processes.
5. A planet’s Bond albedo gives the fraction of total Solar radiation it reflects and thus determines the total Solar heating. The geometric albedo gives the fraction of visible light the planet reflects, thus determining how bright it looks.
(a) Mars has a geometric albedo of 0.17 (the red color of the planet, which means it reflects infrared light well, causes the large difference). When the planet is in opposition, it exhibits a full phase. You can use this and the planet’s radius (3380 km) to determine the total Solar power incident on the planet and hence the total power reflected from it (assume the Solar luminosity is all visible light for this exercise). This power will be distributed over one half of a sphere surrounding Mars at any distance. Use this information to find the apparent brightness of Mars at opposition in terms of the Solar constant b .
The total power incident on a planet of radius R orbiting at a distance D from the Sun, as we computed above, is
Pin = πR2b(D) = πR2
R
D 2
We will assume here this is all in visible light. A fraction aG of this is reflected
(as visible light), meaning the total reflected luminosity in visible light is
L = aGPin = aGπR2
R
D 2
σT 4 .
Since this light spreads uniformly over one-half of the universe, at a distance d from Mars the flux of energy (apparent brightness) will be
b(d) = L 2πd2 = aG 2 R d 2 R D 2 σT 4 .
We want to compare this to the apparent brightness of the Sun (on Earth) which is given by b = R D⊕ 2 σT 4 .
Comparing the last two expressions (or dividing them) we see that
b(d) = aG 2 R d 2 D D⊕ −2 b .
Now we can plug in the numbers for Mars at opposition, where d = DMars−D⊕ =
0.523 AU bMars = (0.17/2) 3380 km 0.523 · 1.496 × 108km 2 (1.523)−2b = 6.84 × 10−11b .
Mars is quite faint compared to the Sun, but rather bright for a star (compare to our calculation for µ Cephei in previus HW that shows that star is 100 times fainter than Mars.
(b) Repeat the calculation for Saturn at opposition, using the radius for Saturn we found in a previous HW of 60,000 km. Saturn’s geometric albedo is 0.47. Which planet should appear brighter at opposition? Note that the internal heat issue is irrelevant here since the thermal radiation from Saturn is certainly not in visible light!
We can use scaling or just plug into our expression
bSaturn = (0.47/2) 60000 km 8.58 · 1.496 × 108km 2 (9.58)−2b = 5.66 × 10−12b .
Saturn at opposition will be more than ten times dimmer than Mars at opposition. Of course, the variation in brightness is expected to be much larger for Mars than for Saturn since the change in distance to Earth (by a factor of about five!) between Mars at opposition and at conjunction is far more significant than for Saturn. Much of the time the two planets appear about equally bright and in fact at its dimmest Mars is slightly dimmer.
6. The early protosun may have had a luminosity 9 times its current luminosity, as you found above. Using this luminosity, and assuming (not quite accurately) that the nebula was as transparent as the interplanetary medium we have currently, find the distance in AU from the center of the Sun at which the ambient temperature of a perfect spherical blackbody in equilibrium with the radiation would be 175 K.
This would be our rough estimate for the position of the snow line since this is the temperature below which water vapor freezes at zero pressure.
From our calculations in class (repeated above) we know that the temperature of a spherical object with zero albedo at a distance D from a star of luminosity L will be given by T = L 16πσ 1/4 D−1/2 .
To use scaling we want the answer for D = 1 AU and L = L . This is what Earth’s’
temperature would be if a as well as g were zero. Looking at our work above this is 254/0.691/4 = 278.7 K . Now we use scaling, as usual the constants drop out and we have T 278.7 K = L L 1/4 D 1 AU −1/2 , or D 1 AU = L L 1/2 T 278.7 −2 = 91/2· (175/278.7)−2 = 7.6 .
1. In this problem we will follow the detection of a multi-planet system orbiting the star Kepler-11. The data here are mostly taken from the paper announcing the discovery, which you can find at http://arxiv.org/abs/1102.0291. You should find you are able to read much of this and understand it. I have simplified several details to make calculations easier.
(a) The planet Kepler-11f orbits its star (Kepler-11) at a radius of 0.25 AU with a period of 46.7 (Earth) days. Use this to estimate the mass of the star it orbits, in terms of the solar mass M .
Using Newton’s version of Kepler’s third law we have
P2 = 4π
2
GMD
3
.
Here M is the total mass of the system and D the distance between the orbiting objects. We will neglect the mass of Kepler-11f compared to that of its star, taking M to be the mass of the star, so that
P11f2 = 4π
2
GM11
D11f3 .
We can compare this usefully to the same equation applied to Earth’s motion about the Sun
P⊕2 = 4π
2
GM
D⊕3 ,
Dividing left and right-hand sides we find a useful scaling relation P11f P⊕ 2 = M11 M −1 D 11f D⊕ 3 .
We can solve this for the ratio of Kepler 11’s mass to that of the Sun M11 M = D11f D⊕ 3 P 11f P⊕ −2 = (0.25)3 46.7 365.25 −2 = 0.956 .
(b) The planet Kepler-11e orbits at a radius of 0.194 AU. Predict the period of this planet’s orbit, in Earth days.
Since the two planets orbit the same star, Kepler’s third law yields P11e P11f 2 = D11e D11f 3 .
We can solve this for
P11e = P11f D11e D11f 3/2 = 46.7 0.194 0.25 3/2 = 31.9 d .
(c) The star Kepler-11 has a surface temperature of 5680 K and a radius of 1.1R .
Find the luminosity of Kepler-11 in terms of the solar luminosity L .
Luminosity, radius, and temperature are related by L = 4πσD2T4 .
Comparing this equation for Kepler-11 and for the Sun we find the scaling relation L11 L = D11 D 2 T 11 T 4 = (1.1)2 5680 5778 4 = 1.13 .
(d) Kepler-11f has radius Rf = 2.61R⊕ and mass Mf = 2.3M⊕. What would you
conclude about its likely composition? Mass, radius, and density ρ are related by
M = 4π 3 R
3ρ .
As usual, comparing to Earth yields a scaling relation Mf M⊕ = Rf R⊕ 3 ρf ρ⊕ , from which we have
ρf ρ⊕ = Rf R⊕ −3 Mf M⊕ = 2.3/(2.61)3 = 0.13 .
Kepler-11f is far less dense than Earth, less dense even than Saturn. It is almost certainly a gas planet.
(e) Kepler-11b, the innermost planet in the system with an orbital radius of only 0.091 AU, has a radius Rb = 1.97R⊕ and a mass Mb = 4.3M⊕. What would you
conclude about its likely composition? Computing as above we have
ρb ρ⊕ = Rb R⊕ −3 Mb M⊕ = 4.3/(1.97)3 = 0.56 .
Kepler-11b is far more dense than Kepler-11f, but less dense than Earth. It is unlikely to be a gas planet. However, if it were similar to Earth in composition,
we would expect the larger, more massive planet’s gravitation to compress it to a larger density. With a density less than that of Earth, Kepler-11b must be made of naturally less dense stuff than is Earth. Thus, the most likely answer is that Kepler-11b is largely composed of ice.
(f) Which of the following would you consider a likely history for the formation of these two planets mentioned above? Why?
(A) These planets formed closer to the star and migrated out to their current orbits.
(B) These planets formed near to the orbits they now occupy. (C) These planets were captured from another star.
(D) These planets formed farther from the star and migrated in to their current orbits.
Kepler-11 is quite similar to the Sun in mass and luminosity. The snow line in its planetary disk, therefore, was likely at a similar distance to its location in the Solar system - around 5 AU from the star. Gas and ice planets form beyond the snow line, so Kepler-11b and Kepler-11f are likely to have migrated to their present orbits after forming farther away from the star. The answer is D.
(g) The six planets conjectured to orbit Kepler-11 were all discovered using the transit method, and their properties were inferred from details of the light curve of the star, as you can read in the paper. What does finding one transiting planet tell us about the likelihood of other planets in the system transiting as well?
(A) There is absolutely nothing to be learned here, we just got lucky with Kepler-11.
(B) It is unlikely that more planets transit, since gravitational interactions with the transiting planet would have deflected them to other orbits.
(C) Since planetary orbits in a system are expected to lie near a common plane, finding one transiting planet makes it more likely that other planets will also transit.
(D) Since planetary orbits in a system are expected to lie in a common plane, finding one transiting planet makes it certain that other planets will also transit.
Planetary orbits usually (but not always) lie near a plane, the analog of the Solar sytem’s ecliplic, determined by the angular momentum of the cloud from which the system formed. Thus when a system has a transiting planet others are likely to transit as well. The answer is C.
Let us consider how we might verify these candidate planets using other methods. 2. Suppose we want to directly image Kepler-11f in visible light.
(a) Find the ratio of the planet’s luminosity in reflected starlight to that of the star Kepler-11. The planet has an albedo of 0.5.
Consider a configuration in which we observe the planet in a full phase (that is, we see the fully illuminated side of the planet). You may assume that Kepler-11 radiates all its power in visible light.
We can find the power in radiation from Kepler-11 impinging on the planet as
Pin = πRf2 L11 4πD2 f = L11 Rf 2D2 f !2 ,
where here Df signifies the orbital radius, to distinguish it from the planet’s
radius. Of this, a fraction a = 0.5 is reflected and the rest absorbed. It is impor-tant to note that the absorbed light is eventually re-emitted, but it is emitted as infrared radiation, and we here propose to image the planet in visible light. In many cases, infrared imaging is a better choice, since hot stars emit less of their power in the infrared whereas planets add to reflected infrared light some of their own blackbody radiation. The planet’s luminosity in visible light is thus
Lf = aPin = L11a Rf 2Df 2 = 0.5 2.61 · 6371 2 · 0.25 · 1.496 × 108 2 = 2.471 × 10−8L11 .
(b) Also find the maximum angular separation, in arcseconds, between the planet and star as seen from Earth. The star Kepler-11 lies about 1.264 × 108 AU away from Earth.
The maximum angular separation α between planet and star will occur when line between planet and star forms a right angle with our line of sight. At this point, the angle between them is related to the distance D11 of the system from Earth
by the small-angle formula as
α 206265” = Df D11 , or α = 206265” 0.25 1.264 × 108 = 4.08 × 10 −4 arcsec .
This problem should give you a sense of the difficulty faced by direct imaging methods. 3. Alternatively, we could try to verify the discovery using radial velocity measurements, since the existence of a transiting planet tells us the system is aligned so that these will be effective. Lets restrict our attention to the innermost planet, Kepler-11b from a previous question. Use the mass you found for the star Kepler-11 in 1(a) and the planet mass Mb = 4.3M⊕.
(a) Find the speed with which the star Kepler-11 moves around the center of gravity as a result of its gravitational interaction with this planet.
We found an expression for the speed with which a star moves due to the presence of a planet in terms of the relevant quantities in class. Assuming the planet mass is small, the star moves about a circle of radius
R11 = (Mb/M11)Rb
with a period identical to that of the planet. We could, as we did then, complete the calculation in terms of masses and orbital radius, but here we know more. Kepler-11b’s period can be found from those of the other planets as usual
Pb = Pf Rb Rf 3/2 = 46.7 0.091 0.25 3/2 = 10.256 d = 8.86 × 105s . The orbital speed of Kepler-11 is thus given by
v11 = 2πR11 Pb = (Mb/M11) 2πRb Pb = 4.3 · 5.972 × 10 24 0.956 · 1.989 × 1030 2π · 0.091 · 1.496 × 1011 8.86 × 105 = 1.35 × 10−5· 9.65 × 104 = 1.30 m/s .
(b) If we were to use the Hα absorption line with a wavelength of λ0 = 656.282
nm in the spectrum of Kepler-11 to detect the motion of its orbit due to the gravitational attraction of Kepler-11b as it circles the center of gravity, what is the longest wavelength at which we would detect this line?
The longest wavelength will be detected when the star is moving away from Earth at the speed computed in the previous question. This is a positive radial velocity in astronomical convention. The observed wavelength λ is given by
λ = λ0(1 + v/c) = 656.282(1 + 1.30/2.998 × 108) = 656.282002854 .
