PERPUSTAKAAN LOKA GHANA SMAN 1 SEDAYU: Agustus 2008

Bagi siswa yang ingin mencoba soal2 olimpiade Fisika dan astronomi dapat melihat sosal berikut

SELEKSI TINGKAT KABUPATEN
OLIMPIADE SAINS NASIONAL 2007
BIDANG STUDI : FISIKA
WAKTU : 2,5 JAM

Selesaikan soal berikut ini dengan singkat, jelas dan benar.

1. Sebuah pesawat dengan massa M terbang pada ketinggian tertentu dengan laju v. Kerapatan udara di ketinggian itu adalah r. Diketahui bahwa gaya angkat udara pada pesawat bergantung pada : kerapatan udara, laju pesawat, luas permukaan sayap pesawat A dan suatu konstanta tanpa dimensi yang bergantung geometri sayap. Pilot pesawat memutuskan untuk menaikkan ketinggian pesawat sedemikian sehingga rapat udara turun menjadi 0.5 r. Tentukan berapa kecepatan yang dibutuhkan pesawat untuk menghasilkan gaya angkat yang sama? (nyatakan dalam v).
2.
R
r
θSebuah silinder dengan jari jari r (r = 0.2 R) berosilasi bolak-balik pada bagian dalam sebuah silinder dengan jari jari lebih besar R seperti pada gambar.
Anggap ada gesekan yang besar antara kedua silinder sehingga silinder tidak slip. Berapakah periode osilasi sistem (anggap sudut q kecil).
Momen inersia silinder
3. Sebuah tangga berbentuk segitiga sama kaki seperti pada gambar, mempunyai massa yang sangat kecil dan bisa diabaikan. Seorang tukang bangunan dengan massa m kg memanjat sampai ketinggian 3 meter dari dasar. Berapa tegangan tali penghubung (pada posisi horizontal di gambar) antara kedua sisi tangga? (nyatakan dalam m dan g, dimana g = percepatan gravitasi bumi).

4.
m
vM
h
θ
MSebuah bola pejal bermassa m mengelinding turun sepanjang bidang miring segi tiga yang massanya M (M = 7m). Jari jari bola = r (r = 0.1 h) . Mula mula sistem diam. Berapakah kecepatan M ketika bola turun sejauh h (nyatakan dalam h dan g , g = percepatan gravitasi bumi) dan sin θ = 0.6 serta ada gesekan yang besar antara massa m dan M cukup besar agar m tidak slip, tetapi tidak ada gesekan antara M dan lantai.
Momen inersia bola pejal
5. Seorang bungee jumper diikatkan pada salah satu ujung tali elastis. Ujung satunya dari tali itu disambung ke suatu jembatan yang tinggi. Kemudian si bungee jumper ini melompat turun dari jembatan itu dari keadaan diam. Massa orang ini adalah m. Panjang tali kalau kendor adalah L dan konstanta pegas tali adalah k. Medan gravitasi bumi adalah g. Berapa panjang akhir tali saat si bungee jumper ini berhenti sesaat? (nyatakan dalam L, m, g dan k)

Semoga sukses

Sistem koordinat ekuator
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Sistem koordinat ekuator
Sistem koordinat ekuator barangkali adalah sistem koordinat langit yang paling sering digunakan. Sistem koordinat ini merupakan sistem koordinat yang bersifat geosentrik. Mirip dengan sistem koordinat geografi yang dinyatakan dalam bujur dan lintang, sistem koordinat ekuator dinyatakan dalam asensio rekta dan deklinasi. Kedua sistem koordinat tersebut menggunakan bidang fundamental yang sama, dan kutub-kutub yang sama. Ekuator langit sebenarnya adalah perpotongan perpanjangan bidang ekuator Bumi pada bola langit, dan kutub-kutub langit sebenarnya merupakan perpanjangan poros rotasi Bumi (yang melewati kutub-kutub Bumi) pada bola langit.
Seperti halnya bujur, asensio rekta dihitung sepanjang lingkaran yang sejajar ekuator. Asensio rekta dihitung ke arah timur mulai dari titik Aries atau titik Vernal Ekuinok yang merupakan salah satu titik perpotongan antara bidang ekliptika dan ekuator langit, tempat Matahari berada pada tanggal 21 Maret (lihat gambar). Asensio rekta dilambangkan dengan "α", kadang-kadang disebut juga RA (dari bahasa Inggris Right Ascension) dan dinyatakan dalam satuan sudut (jam, menit, detik), dengan 1 jam = 360 derajad / 24 jam = 15 derajad. Dalam pengamatan praktis seringkali harga ini tidak diketahui bahkan harus ditentukan sehingga digunakan besaran lain yang bersifat lokal, yaitu sudut jam atau HA (dari bahasa Inggris Hour Angle).
Seperti halnya lintang, deklinasi diukur dari ekuator ke arah kutub. Deklinasi bernilai positif bila benda langit yang diamati berada di belahan langit utara, dan negatif bila benda langit yang diamati berada di belahan bumi selatan. Deklinasi dilambangkan dengan "δ" dan dinyatakan dalam satuan sudut (derajat, menit, detik).

Problems1:
1. The Daedalus Lunar Farside Observatory (established in 2014) is located at the 180th lunar meridian. An observer at Daedalus notes from the shadow of the vertical crater wall that the Sun is directly overhead. A colleague who is exploring the crater Dante, which is due North of Daedalus, reports by radio that the shadow of a vertical crater wall falls at an angle of 31.5 degrees. How far away is Dante from Daedalus? Does it matter that it is due North from Daedalus? (Hint: This is a variation on the way Erathosthenes measured the circumference of the Earth. You can find the radius of the Moon in Appendix 3 of the textbook. You should sketch a diagram to guide your work.)
2. What is the greatest elongation of Mars as seen from Jupiter? Assume circular orbits in the plane of the ecliptic. (Hint: use the Appendix 3 to get the orbital semimajor axes of these two planets. Remember what the definition of elongation is.)
3. How long is it between consecutive inferior conjunctions of Jupiter and the Sun as seen from Saturn? How long between consecutive oppositions of Saturn and the Sun as seen from Jupiter? (Hint: this is a synodic period question. You can find the sidereal orbital periods in Appendix 3 of the text, but do not use the synodic period, as that refers to the Earth.)
4. Suppose that a tenth planet was discovered in our solar system with a perihelion distance of 80 AU and an aphelion distance of 100 AU. Find the eccentricity e and semimajor axis a of this orbit. (Hint: use the location of the focus from the center of an ellipse.)
5. Using the semimajor axis determined in Problem 4, what would the sidereal period of its orbit be (in years)? (Hint: Use Kepler's Third Law, with respect to Earth's orbit.)

Solutions to Problem Set #1
Solutions:
1. The crater Daedalus is on the far side of the moon at lunar longitude 179.4 E and latitude 5.9 S (this has no impact on the problem). If the Sun is overhead at Daedalus, and at an angle of 31.5 degrees from the zenith (as given by the shadow) at Dante, the arc from Daedalus to Dante subtends the fraction of the circumference 31.5/360, and thus
A = ( 31.5 / 360 ) x 2 Rmoon = 956 km
for R_moon = 1738 km (Appendix 3). It does not matter that Dante is due North, since this gives the arc length along the great circle between Daedalus and Dante no matter the orientation, since the Sun is directly overhead at Daedalus (circular symmetry about the Sun-Daedalus axis).
2. You should first draw a picture of the orbits of Mars and Jupiter from above the ecliptic. Remember to make the line of greatest elongation tangent to the circular orbit of Mars from Jupiter, which makes a right angle with the radius from the Sun to Mars (1.52 AU) which is the opposite side of the right triangle from the elongation angle. The Jupiter-Sun line is the hypoteneuse (5.20 AU) and thus
sin = 1.52 / 5.20 = 0.292 => = 17°
is the Greatest Elongation of Mars as seen from Jupiter.
3. The time interval between any consecutive similar configurations between two planets and the Sun is the synodic period S, where
1 / S = 1 / Pin - 1 / Pout
for the sidereal periods P of the inner and outer planets of the pair. For Jupiter and Saturn
Pin = 11.86 yrs (Jupiter), Pout = 29.46 yrs (Saturn) => S = 19.85 yrs.
Since S is the synodic period, it is the time between two oppositions of Saturn from Jupiter, or between inferior conjunctions of Jupiter from Saturn (or any other similar configurations). Furthermore, Saturn at opposition and Jupiter at inferior conjunction are the same from the respective viewpoints of Jupiter and Saturn!
4. The major axis is given by the sum of r_p and r_a
rp + ra = 80 AU + 100 AU = 180 AU = 2 a
so the semimajor axis is a = 90 AU. The eccentricity can be calculated using the aphelion for example,
ra = a ( 1 + e ) => e = ra/a - 1 = 100/90 - 1 = 1/9
and thus the eccentricity of the orbit is e=0.11.
5. From Kepler's Third Law, normalized to the Earth's orbit
( P / 1 yr )2 = ( a / 1 AU )3
which of course only holds for orbits around the Sun. Thus,
P = ( a / 1 AU )3/2 yrs
which for our Planet X
P = ( 90 )3/2 yrs = 854 yrs.
(How hard would this be to detect? It would move a full 360° around its orbit in 854 years, and thus has an angular velocity of 25' per year, which is half the full moon diameter. This would be fairly easy to detect photographically. In a later problem set, we will compute its brightness.)

Problem Set #2 (due Thu 1 Oct 1998 5pm)
Problems:
1. Local noon (12h local solar time) is defined to occur when the Sun is on the observer's local meridian. On approximately what day of the year does local noon coincide with 12h local sidereal time (LST)? One month later, what is the approximate LST at local noon? Use these numbers to work out approximate what the LST is at local noon on your birthday. (Hint: to answer this question you need to know the RA of the Sun at some time during the year, and how it changes.)
2. Deimos is a satellite of Mars with an orbital eccentricity of e=0.003 (and thus in a nearly circular orbit) and semimajor axis a=23520 km and period P=1.26 days. Using Newton's Laws to derive Kepler's Third Law for a circular orbit we found
G M P2 = 4 2 a3
Use this to determine the mass of Mars (in kg, and in relation to Earth's mass). (Remember in this equation that MKS units are assumed. You can find G in Appendix Table A7-2 in the text book, or in the online constant table.
3. The angular velocity
= d /dt
of a planet at some point in its orbit is given by
= v_ / r
where v_ is the component of the orbital velocity perpendicular to the orbital radius vector. Note that this angular velocity also represents the apparent motion of the Sun against the ecliptic as seen from the planet. We had found in class that the angular momentum per unit mass
H = v_ r
is conserved in the orbit. Mars has an eccentricity e=0.093 and so the variations in the Sun's apparent will be more pronounced than that of Earth. Find the ratio of eastward apparent angular velocities of the Sun at perihelion and aphelion as seen from Mars. Also, compute the average eastward motion of the Sun from Mars (in degrees per day). (Hint: by average eastward angular rate of the Sun, I only mean that it moves so many degrees over one full period.)
4. The circular velocity v_circ at distance r from the Sun (or any other body) is the velocity of a circular orbit with radius a=r
vcirc = 2 a / P = [ G Msun / a ]1/2
which we derived using the centripetal and gravitational forces. Calculate the circular velocities for Earth's orbit (1 AU) and Mars's orbit (1.524 AU). Pretend the orbits of Mars and Earth are circular. There is an elliptical orbit that has perihelion at Earth's orbit and aphelion at Mars's orbit - this is the least energy orbit along which a space probe can be sent from Earth to Mars. Find the semimajor axis a and eccentricity e of this orbit.
Then, calculate the perihelion velocity v_p of this orbit and compare this to the circular velocity at Earth's orbit to find the extra velocity vinj needed to be given to a probe to put it on a path to Mars. Assume that the probe is moving along with the Earth in its circular orbit around the Sun (ignore its orbital velocity around the Earth).
Finally, compute the aphelion velocity v_a of this orbit and compare this to the circular velocity of Mars's orbit to find the vret needed to place the probe into circular orbit with Mars. Should this velocity be added or removed from the aphelion velocity? (Ignore its orbital velocity around Mars, consider only solar orbits. You can get expressions for the perihelion and aphelion velocities by using conservation of angular momentum
H = vp rp = va ra
the fact that
vcirc2 = vp va

Solutions to Problem Set #2
Solutions:
1. The important facts given were that local noon is defined as the time when the Sun is on the observer's local meridian, and that at that moment the Local Sidereal Time happened to be 12h LST. Since LST is defined as the Right Ascension (RA) that happens to be on the local meridian, this question is really asking "on what date is the Sun at RA = 12 h?".
In case you forgot, or were confused, the meridian is the line (great circle) the passes through the zenith (overhead) and the North and South points of the horizon, and is thus an extenison of the longitude line of the observer. Since RA circles are concentric with longitude lines, as the Earth turns consecutive RAs pass overhead each observer, and thus define a "clock" - the sidereal clock which keeps Local Sidereal Time.
To figure out when the Sun is at RA=12h, you need to know the zero-point of Right Ascension: the Vernal Equinox is RA = 0h, Dec = 0°. The Sun is at the Vernal Equinox on March 21, and thus the Sun is at RA = 12 h (and also Dec = 0°) 6 months later at the Autumnal Equinox, around Sep 21.
The Sun moves along the ecliptic a full 24 hours in 12 months, at the average rate of 2 hours per month. Since the Earth rotates in the same sense as it orbits the Sun, the Sun appears to advance along in positive RA, and reaches an apparent RA = 6h around June 22 (Summer Solstice) and RA = 18h around Dec 22 (Winter Solstice).
You can thus compute the RA of the Sun on your birthday by offsetting from these points. For example, my birthday is May 22, which is just about 2 months after the Vernal Equinox, and thus the Sun is at
RA (May 22) = 0h (Mar 21) + 2h x 2 (months) = 4 h
which gives the LST at local noon on that date, as we pointed out above.
2. This calculation is extremely straightforward. We use the Newtonian version of Kepler's Third Law
G M P2 = 4 2 a3
where everything should be in MKS units (be sure to convert km to m for example). Thus, we can rearrange this to solve for the mass
M = ( 4 2 a3 ) / ( G P2 ).
For Deimos a = 2.352 x 10^7 m, P = 108864 sec, and as usual G = 6.67 X 10^-11 N m^2/kg^2, so
M = 6.5 x 1023 kg = 0.109 Mearth
for an Earth mass of 6 x 10^24 kg.
3. At perihelion and aphelion, the velocity vector v is perpendicular to the radius vector r, so the conservation of the angular momentum per unit mass gives
H = v r = vp rp = va ra
which implies
vp / va = ra / rp.
Thus,
p / a = ( vp/rp ) / ( va/ra) = ( ra / rp )2
which for
rp = a ( 1 - e ) ; ra = a ( 1 + e )
gives
p / a = ( 1 - e / 1 + e )2 = 1.452
for the eccentricity e=0.093 of Mars.
Now, by the average angular velocity I merely meant that it moved a full 360° along the Martian ecliptic in one full sidereal period (686.98 days from Appendix 3). Thus, the "average" rate of advance is
avg = 360° / P = 360° / 686.98d = 0.524 °/d
which is about half the solar advance rate as seen from the Earth (360° in 365.26 days).
Note: you could also have obtained this by
avg = vcirc / a ( 2 a / P ) / a = 2 / P
radians per day.
4. You should definitely draw a picture for this one. The transfer orbit from Earth to Mars as a perihelion of 1 AU and aphelion of 1.524 AU, and so the semimajor axis of the transfer orbit is
rp + ra = 2 a => a = ( 1 AU + 1.524 AU ) / 2 = 1.262 AU.
The eccentricity is given by
rp = a ( 1 - e ) => e = 1 - rp/a = 1 - 1/1.262 = 0.208.
The relevant circular velocity of the Earth's orbit is
vcirc,E = [ G Msun / a ]1/2 = 29800 m/s = 29.8 km/s
for a = 1 AU = 1.5 x 10^11 km and Msun = 2 x 10^30 kg. Now, it is useful to scale to this
vcirc = 29.8 km/s / (a / 1 AU )1/2
which gives
vcirc,M = 24.1 km/s ( a = 1.524 AU ) vcirc,tr = 26.5 km/s ( a = 1.262 AU )
for Mars and the "transfer orbit" respectively. (Note: we computed the circular velocity at the semimajor axis of the tranfer orbit for ease of computing the next bits. Its good to think ahead.
Now, we know from class (and Eq 1-17 of the textbook) that
vp = ( 2 a / P ) [ 1+e / 1-e ]1/2
but since
2 a / P = vcirc
and
( 1+e / 1-e ) = ra / rp
then we can rewrite this as
vp = vcirc [ ra / rp ]1/2
and so
vp = 26.5 km/s [ 1.524 / 1 ]1/2 = 32.7 km/s
for our transfer orbit. Therefore, we have to "inject" it from Earth's orbit with an extra velocity
vinj = vp - vcirc,E = 32.7 - 29.8 = 2.9 km/s.
For the aphelion, we have
vp = ( 2 a / P ) [ 1-e / 1+e ]1/2 = vcirc [ rp / ra ]1/2
and so
va = 26.5 km/s [ 1 / 1.524 ]1/2 = 21.5 km/s
for the transfer orbit. Thus, we need to add a further "retro" velocity
vret = vcirc,M - va = 24.1 - 21.5 = 2.6 km/s
to move from the transfer orbit into a Martian circular orbit around the Sun.
Note that we need to add velocity both times, since we are going into respectively higher energy orbits
E = - G Msun / 2a
which become less negative (less bound, higher energy) as a increases.
Note also that we saved some work by playing with the equations for v_p and v_a in terms of v_circ and r_a/r_p before pluggin and chugging!

Sample Problems for Midterm #1
Note that some of these questions use the results of previous questions. Be sure to show your work so that if you make a mistake in one question, we can see that you knew how to do the next one and not penalize you twice for one error! Be sure to check your work and see if the result makes sense. These problems are intended to use concepts we have learned and try and lead you to some new insights, not just repeat problems we have already done. Dont panic on these --- relax and think about them!
This sample problem set is not meant to be a "simulation" of an exam, and for some questions you will be asked to look up information in various places. On the real exam I will give you the information you need, including equation sheets and constants.
You will need a calculator for this and for the midterm! You may turn this set in as an extra credit homework assignment (due day of the exam, Thursday 8 October 1998 ).
Problems:
1. A comet is discovered in our solar system. A search through the historical records indicates that it last passed by in 1776! What is the semi-major axis of the comet's orbit (in AU)? If its perihelion is at the orbit of the Earth, what is the aphelion distance (in AU) from the Sun?
2. Suppose a new planet were discovered that was found to have a greatest elongation of 30 degrees from the Sun. What is the radius of its orbit (in AU), assuming it and the Earth are in approximately circular orbits?
3. The asteroid Ceres is found to have a synodic period of 1.278 years as observed from the Earth. What is the Sidereal period of its orbit?
Using the sidereal period of Ceres just derived, calculate the semi-major axis of its orbit in AU.
The eccentricity of its orbit is e=0.077. What are the perihelion and aphelion distances of Ceres (in AU).
4. The brightest Uranian moons, Titania and Oberon, were discovered by William Herschel in 1787. Oberon has an orbital period of 13.5 days and a semimajor axis of 583000 km. What is the mass of Uranus in Earth masses?
5. The escape velocity for a distance R from a body of mass M is given by
vesc2 = 2 G M / R
If you launch a projectile away from the mass with velocity v_esc from radius R, then at an infinite radius the projectile would come to rest (v -> 0). But reversal of the energy equation means that v_esc is the velocity an inward moving projectile started at rest infinitely far away would achieve when it reached radius R from the mass!
1. Calculate the escape velocity from the surface of Jupiter (in km/s).
2. In July 1994, the Comet Shoemaker-Levy 9 astounded astronomers by impacting the planet Jupiter. This comet was disrupted by a previous encounter with the planet, and perturbed into a catastrophic orbit. If we assume, for the want of a better estimate, that it was on approximately a marginally bound orbit, it should have impacted Jupiter at the escape velocity from Jupiter's surface. Calculate the kinetic energy in one metric ton (1000 kg) moving at this speed (in Joules).
3. If the explosion of one ton of TNT liberates 4.2 x 10^9 J of energy, how many tons of TNT is our one-ton cometoid impact equivalent to?
4. I would estimate that the chunks of SL-9 that hit Jupiter were around 1 km in size. Approximating it as a cube 1km on a side, with the density of water ice (1000 kg/m^3), estimate the total mass of a comet fragment (in metric tons). Then, compute the imact energy (in tons of TNT). To put this in perspective, a really big H-bomb produces 40 mega-tons of energy (40 x 10^6 tons of TNT worth).
6. The famous explorer Christina Columbo bas passed the Cape Horn and reached the Pacific Ocean in her vessel, the Pinto.
1. After sailing for some time, she notices that the South Celestial Pole is 40° altitude above the South horizon. What is the latitude of her current location?
2. At this time, she notices that the bright star Fomalhaut is just crossing her meridian. She looks up the coordinates of that star and finds them to be RA = 22 h 56 m and DEC = - 29° 45'. What is the current Local Sidereal Time (LST) at the location of the Pinto?
3. Is Fomalhaut North or South of her zenith when it crosses the meridian?
4. Before she left port in London, she set her sidereal chronometer to the Local Sidereal Time at Greenwich (Greenwich Sidereal Time = GST), whose meridian defines the zero of longitude. She notices that when Fomalhaut was crossing the meridian, the chronometer showed a time of 10 h 26 m GST. (Note: this means that at that moment, RA = 10 h 26 m was on the meridian at Greenwich.) What is her current longitude in degrees (west of Greenwich)?
5. Find a World Atlas, map of the World, or globe, and locate the Pinto's position there. What is the nearest significant landmass and where is she located relative to it?
7. Suppose NASA wants to send a probe to Venus from the Earth with the least expenditure of energy. The way to do this is to launch it slightly retrograde to the Earth's orbit, so that its velocity cancels some of the Earth's orbital velocity, and it will end up in an elliptical orbit with an aphelion of 1 AU and an perihelion at the orbit of Venus (0.723). What is the eccentricity e, semimajor axis a, and period P of this orbit?
Sketch the orbits of the Earth and Venus around the Sun. Indicate the least energy orbit and the relative directions that Earth, Venus and the probe move along their respective orbits to reach Venus? How long will this voyage take?
Use the energy equation for this orbit to calculate the aphelion velocity of the orbit. Compare this to the circular velocity of Earth's orbit. What is the burnout velocity that you want the rocket to have with respect to the Earth directed opposite to the Earth's motion in order to place the probe in this orbit?
8. Mars has a radius of 0.53 times the radius of the Earth and a mass of 0.107 times the mass of the Earth. What is the surface gravity of Mars compared to the surface gravity g of the Earth?
What is the escape velocity (in km/s) from the surface of Mars?
The rotational period of Mars (the Mars sidereal day) is 24 h 37 m 22.6 s (in standard Earth solar time units). At what distance from Mars would a circular orbit be ``synchronous'' with the rotation of Mars, and thus appear to stay fixed in the sky above a specific spot on Mars's equator?
Mars has an inner satellite Phobos, in an orbit with semi-major axis of 9370 km and period of 0.32 days, and an outer satellite Deimos, with semi-major axis of 23520 km and a period of 1.26 days. How are Deimos and Phobos situated relative to the synchronous point (radius calcualted above) and how would they appear to move in the Martian sky? (Hint: Mars, like the Earth, rotates in the same sense as its orbital revolution. The Sun will appear to rise in the East and set in the West on Mars also. Be careful and think about this problem - draw diagrams to help.)
9. If the Earth were to stop in its orbit, how long would it take to fall in to the Sun? You can consider this an orbit of eccentricity e=1 that reaches from aphelion at the Earth to perihelion at the Sun. This time to fall in to the center (assume a point mass for the Sun, is called the free-fall time.
Derive a formula for the free-fall time t_ff from distance R with mass of M assuming spherical symmetry.
Use this relation to calculate the time it would take the planet Jupiter to collapse (like in the movie 2010) if all the gas pressure were removed from it.
10. Write down a formula for the mass M of a spherical planet with density and radius R. Also write down a formula for the surface gravity g of a spherical planet of mass M and radius R. Combine these two equations to get an equation giving the surface gravity g of a spherical planet of density and radius R.
What is the mean density of the Earth, in kg/m^3?
Scale the above equation to the surface gravity, radius, and average density of the Earth to find a relation between the gravity (in g for the Earth) and radius (R earth) with Earth density. Pay attention to the scaling with R, to see how g scales with radius.

Equations and Constants for Midterm #1
Last update: 1 October 1998
Constants:

Mechanics:

Solutions to Problem Set #3
Solutions:
1. The easiest thing to do is to scale from 1 AU
F(r) = F( 1 AU ) [ r / 1 AU ]-2 = 1370 W/m2 [ r / 1 AU ]-2
which gives F=50.7 W/m^2 at r=5.2 AU. Note that direct computation gives
F = Lsun / (4 r2)
which agrees within our rounding of the solar constant 1370 W/m^2. Thus, to get our 500 W of power, we need an area of around
Area = P / F 500 W / 50 W/m2 = 10 m2
or a panel size of 3.3m x 3.3m! This is pretty big (10ft on a side) and is why Galileo carries an onboard nuclear power source.
A flat panel oriented toward the Sun absorbs over the forward half of its surface area only, while it emits thermal radiation over both sides, so the power equilibrium is given by
Pin = Fin L2 = Pout = T4 2L2 => T = [ Fin/2 ]1/4 = 145 K.
Note if you forgot the other side of the panel, you would have gotten just the subsolar temperature
Tss = [ Fin/ ]1/4
which we often use for our solar system in the form
Tss = [ Lsun / (4 r2 ]1/4 = 394 K [ r / 1 AU ]-1/2.

2. A spherical planet absorbs over its cross-section (a flat disk)
Pin = Fin r2 ( 1 - A ) = Tss4 r2 ( 1 - A )
which includes the correction for reflection with albedo A. The planet emits over its full surface area
Pout = Tsurf4 4 r2 ( 1 - G )
where we have included the greenhouse factor G. In equilibrium Pin=Pout and thus
Tsurf = Tss 2-1/2 [ ( 1 - A ) / ( 1 - G ) ]1/4
where for our solar system
Tsurf = 279 K [ r / 1 AU ]-1/2 [ ( 1 - A ) / ( 1 - G ) ]1/4
using the scaling from problem 1. If A=G, we get the relation appropriate to the Earth
Teq = Tss 2-1/2 = 279 K [ r / 1 AU ]-1/2
which we call the "equilibrium temperature". Thus, for our habitable zone, we want
279 K [ rinner / 1 AU ]-1/2 = 373 K => rinner = [279/373]2 = 0.56 AU 279 K [ router / 1 AU ]-1/2 = 173 K => router = [279/173]2 = 2.58 AU
where these are the inner and outer radii of the zone respectively. The habitable zone thus includes Venus (0.7 AU), Earth (1 AU) and Mars (1.5) AU. Note that the asteroids are generally not included, like Ceres (2.8 AU) though some Earth or Mars-crossing asteroids will be.
3. The hottest point of the sunlit side of the Moon, directly beneath the Sun, is the subsolar point. If the lunar soil (or whatever object we were interested in) were perfectly absorbing blackbody, it would then have the subsolar temperature for 1 AU of 394 K! As it is, the lunar surface albedo is A=0.07 on average, so
T = 394 K [ 1 - 0.07 ]1/4 = 387 K
or nearly 114 C, which is above the boiling point! This is why astronauts were air-conditioned white colored space suits. The average temperature of the Earth was 279 K as mentioned above, or only 6 C, and thus the peak temperature of the moon is about 100 degrees higher!
4. At r = 0.723 AU the subsolar temperature is 463 K and the equilibrium temperature is 328 K. Thus, our surface temperature should be given by
Tsurf = Tss 2-1/2 [ ( 1 - A ) / ( 1 - G ) ]1/4 = Teq [ ( 1 - A ) / ( 1 - G ) ]1/4
which for the measured surface temperature of 700 K and albedo A=0.76 gives
( 1 - G ) = ( 1 - 0.76 ) [328/700]4 = 0.0116 => G = 0.988
and so Venus is so blanketed by clouds and greenhouse gases that only 1% of the surface radiation escapes! Zow!
5. The two possibilities correspond to
[A]
Like Neptune A=0.62, G=A=0.62
[B]
Like Pluto A=0.5 G=0
(a) We have from Problem 2
T = 279 K [ 77.2 ]-1/2 [ ( 1 - A ) / ( 1 - G ) ]1/4 = 31.6 K [ ( 1 - A ) / ( 1 - G ) ]1/4
so for our cases:
[A]
T = 31.6 K
[B]
T = 26.6 K

(b) We use the Wien's Law
max = 2.898 x 10-3 m [ T / 1 K ]-1 = 2898 µm / T(K)
which gives
[A]
max = 92 µm
[B]
max = 109 µm
which are in the far-infrared part of the spectrum.
(c) We now assume R = 24800 km (Neptune, appendix A3), and the total blackbody luminosity is just flux times area
L = 4 r2 Tsurf4 = 4.37 x 1014 W [ T / 31.6 K ]4
so
[A]
L = 4.37 x 1014 W
[B]
L = 2.19 x 1014 W
The flux at the Earth (r=77.2-1=76.2 AU at closest approach) is
F = L / (4 r2 = 2.7 x 10-13 W/m2 [ L / 4.37 x 1014 W ]
so we have
[A]
F = 2.7 x 10-13 W/m2
[B]
F = 1.3 x 10-13 W/m2
which is very faint.
(d) You should draw a skinny triangle with base D, height r, and apex angle theta, which approximates an arc of length D on a circle of radius r:
D / r = (4.96x104 km)/(1.14x1010 km) = 4.35 x 10-6 rad x 206265 arcsec/rad = 0.9 arcsec
which is just barely resolvable by most ground-based telescopes.

Problem Set #3 (due Thu 22 Oct 1998 5pm)
Problems:
1. Jupiter orbits the Sun with a semi-major axis of 5.2 AU. The Galileo spacecraft is currently in orbit around Jupiter, and requires around 500 Watts of power to run the instruments and on-board systems. How large a solar panel would be needed (in m^2) to generate this amount of power at Jupiter, assuming all of the solar flux could be converted into electricity at perfect efficiency? If the panel acted as a blackbody and was in thermal equilibrium with the solar radiation, at what temperature would it be (in K)?
2. The habitable zone is defined to be the region around a star in which life-bearing planets might be expected to exist. Roughly, we might define this to be the orbital radii at which the equilibrium temperature of a planet would be within 100° C of freezing water: T_eq = 273 ± 100 K. Calculate the inner and outer radii of the habitable zone (in AU) around our Sun. (You may assume the albedo and greenhouse factors are equal.) Which planets are in the zone?
3. The albedo of the Moon is about 7%, that is, 0.07 of the incident light is reflected back into space. What would you expect the temperature of the sunlit side of the Moon to be? (There is no atmosphere to transport the energy to the dark half!) Compare this to the expected average temperature of the Earth.
4. Venus is swathed in a thick cloud deck, with the real surface completely hidden from our view in visible and IR light. The thick clouds act like insultation, and thus the apparent temperature of Venus (as far as its blackbody spectrum) should be the equilibrium temperature at its distance from the Sun, setting the albedo and greenhouse factors equal. Calculate this temperature (assume the orbital semimajor axis for its distance from the Sun).
On the real surface, however, the clouds act like an immense runaway greenhouse! The Russian Venera lander found a surface temperature of 700K! If the measured albedo of the clouds is A=0.76, and the surface temperature is 700 K, what is the greenhouse factor G? This is an example of a runaway greenhouse effect, where all the available surface water and carbon dioxide in the rocks has gone into the atmosphere.
5. There have been many hypotheses of planets further from the Sun than Pluto. For example, if we apply Bode's Rule to find the next planet out from Neptune, ignoring Pluto, we would expect it to have an orbital semi-major axis of 77.2 AU. As for composition, might assume that it is like (A) a gas giant, or (B) an iceball like Pluto. In the questions below, consider both these possibilities (and thus make appropriate assumptions for the values of A and G, from what we did in class for Mercury and Venus and Earth, and from the values in the appendix for Pluto and Jupiter or Neptune).
(a)What would we expect the temperature of such a hypothetical Planet X to be for these two possibilites?
(b)Using Wien's Law, what would the wavelength of the maximum emission be? What part of the electromagnetic spectrum would this be in?
(c)Assume (optimistically) that this Planet X is the same size as Neptune. What would the total blackbody luminosity from Planet X be? What would the flux be from Planet X at the Earth? (Assume circular orbits at closest approach.)(d)What would the angular diameter of this planet appear to be on the sky from the Earth (in arcseconds, at closest approach)? (Hint: The angular diameter is the angle corresponding to the diameter of the planet at the given distance from the observer, or the angle subtended by the physical diameter. In practice, it is calculated using the small angle approximation to the triginometric functions, and is similar to the parallax calculation.)

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