Solutions For Homework #2 - Stanford University

1 Solutions For Homework #21.(a) The orbital period can be calculated using the equationT= 2 r rgR2ewherer=Re+hwhereRe= 6378 kmis the earth s radius,ris the satellites distancefrom the earth s center andh= 205 kmis the satellite s orbital alti-tude, andg= the gravitational acceleration. With thesegiven values the orbital period isTorbit= (b) To calculate the orbital velocity either of the equationsv= gR2erorT=2 rv v=2 rTcan be used. For the given values the result isv 7786m/s(c) To calculate theminimumnumber of ascending passes needed to coverthe entire equator, divide the perimeter of the equator by the swathwidth.

2 The perimeter of the equator isp= 2 Re= 40074km, so theminimum number of passes ispw=40074 km50 km 802passesThe time needed to acquire the calculated coverage can be calculatedbyTglobal= (#of passes) Torbit= 802 4260625s= (d) To calculate theminimumnumber of ascending passes needed to coverthe entire equator, the perimeter of the equator is divided by theeffec-tiveswath width. The effective swath width is the length along the155oswathequatorswath widtheffective swath widthFigure 1: The effective swath width is the length along the equator covered by asingle crossing of swath at an angle of55oequator covered by the swath crossing it at a given inclination angle,(Figure 1) For the given swath widthw= 50 km, at an inclination of = 55 the effective swath width issin( ) =wh h=50 kmsin(55 )

3 = 61 kmThe perimeter of the equator isp= 2 Re= 40074km, so the mini-mum number of ascending passes isph=40074 km61 km 657passesThe time needed to acquire the calculated coverage can be calculatedbyTglobal= (#of passes) Torbit= 657 3490294s= both ascending and descending passes are used, the satellite swathcrosses the equator twice in every orbit, cutting the time inhalf, 20days.(e) The polar orbit is most useful for studying ice motions inthe 2: The velocity of the nadir point (point on the earth s surface directlyunder the satellite ) is by the radius ratio(r/Re)smaller than the orbit ground-velocity is what matters for coverage The rate of coverage is measured in area covered per unit time.

4 Given theformula for the orbital velocity,v=2 rTorbit= gR2era rate of coverage can be calculated byrate of coverage=v Rer wwherewis the swath width (see also Figure 2). To get equal coverage:v1 Rer1w1=v2 Rer2w2 w2=w1v1r2v2r1=w1 gR2er1r2 gR2er2r1=w1(r2r1)32= 160 km (6655 km6948 km)32 150 kmAlternatively, this result can be arrived at by calculatingthe two orbitalperiods and the coverage per day for each satellite :T1= (15orbits/day) andT2= 5400s(16orbits/day)3 The coverage per orbit:c1= 2 Rew1= 106km2c2= 2 Rew2 Thus, satellite 1 acquires per day.

5 15orbitsday 106km2orbit= 107km2daySatellite 2 should have the same coverage, 107km2day= 16orbitsday 2 Rew2 w2= 150 kmTo compute the data rate of each satellite we first find the number of lines persecond, then multiply by the number of samples per line. Nextwe multiplyby 8-bit per sample and by 3 channels. So,(i) satellite 1 acquiresground-velocity=v1 Rer1= 6953m/s6953m/s30m= 232lines/s232lines/s 160,000m30m= 1,236,000samples/s1,236,000samples/s 3channels 8bits/sample= (ii) satellite 2 acquiresground-velocity=v2 Rer2= 7418m/s7418m/s30m= 247lines/s247lines/s 150,000m30m= 1,236,000samples/s1,236,000samples/s 3channels 8bits/sample= The normalized vegetation index (NDVI)

6 Is an indicator for vegetation den-sity, which uses the ratio between the difference of the reflectance in thenear infrared and in the visible red part of the spectrum and the sum ofthese two reflectances:NDVI =Rir RvRir+RvTo make a similar measurement with the Thematic mapper instrument wecan use those bands which correspond to the near infrared andvisible redwavelengths. Bands4 6fall within the infrared and band3lies in the redpart. One possible TMVI equation would beTMVI =R4 R3R4+R34. Since there areH2 Oabsorption bands in the spectrum on each side of thewavelengths covered by band5there is little to no energy coming fromthose regions.

7 A spectral band within these absorption bands would not beuseful to study surface band5were chosen to look at a slightly different part of the spectrum,it would be less sensitive to minerals which have characteristic signaturesin the areas just above and m m. These include thewater bearing minerals, such as gypsum, montmorillonite and quartz, aswell as the hydroxyl (OH) bearing minerals, such as Muscovite, Kaoliniteand the regions just above or below the band5wavelengths are stronglyabsorbed by the atmosphere, they may be useful in applications where thethematic mapper instrument is mounted on an airborne platform, rather thana satellite .

8 If the amount of atmosphere between the scannerand the groundis reduced, enough energy may reach the scanner to be able to use theseparts of the (a) The purpose of this problem is to show that increasing levels of at-mospheric CO2concentration results in more power being absorbed5than emitted by the Earth. Recall from lecture that if more radiationis absorbed than is emitted by the Earth, then the Earth warmsup. Weneed to compute thenetpower per square meter on Earth to find thecorresponding increase in incident solar radiation is given as 1000 W/m2, of which 25% isreflected by clouds.

9 This gives a1000 (1 ) = 750W/m2in-cident on Earth. Assuming no absorption by atmospheric CO2, thisincident radiation is absorbed by the Earth and re-radiatedresultingin a net power density balance of zero. However, in the presence ofatmospheric CO2absorption, the Earth emits only 750W/m2ofradiation because some of Earth s emitted blackbody radiation is ab-sorbed again by atmospheric CO2. Here, = 1 QCO2 27020000(1)is the model of atmospheric CO2transmission, as described in theproblem statement.

10 Consequently, we find a net power balance,PnetofPnet= (1 ) 750=(QCO2 27020000)750W/m2 Now, referring to Handout 7 of the lecture notes, we find a plotofatmospheric CO2concentrations (in parts per million) from the year1700 to the present. Taking the levels of atmospheric CO2con-centration in the present to be 360 ppm, we find thatPnet= (2)We are given that the temperature on Earth increases by forevery additional Watt per square meter ofnetpower. Thus, the tem-perature increase compared to pre-industrial times ( the year when the atmospheric CO2concentration was at 270 ppm, imply-ing = 1) istemperature increase= Pnet= (3)6(b) The present levels of atmospheric CO2concentration, according toHandout 7 of the lecture notes, is about 360 ppm.