Make your own free website on Tripod.com

Astronomy Notes

[Last Updated 27 March 2014]


Measuring Distance
Measuring Time
Celestial Sphere
Telescopes
Moon
Mars
Cosmology
Future Space Missions
Current and Past Space Missions


Measuring Distance

PROBLEM

You plotted the position of a star in relation to the background stars six months ago and again last night. The angle of shift of the star's position is eight-tenths of an arcsecond. What is the distance of the star from the earth?

[from the movie Lambada (1990)]

SOLUTION

      -------<--O earth six months ago (viewed from above ecliptic)
            q / |
            /   |
          /     |L/2 = 1 AU
        /       |
      / q   H   |
star *----------+
      \ q       |
        \       |
          \     |L/2
            \   |
            q \ |
      ----------O earth last night


The positions of the earth six months ago and last night form the endpoints of the base of an isosceles triangle, as shown in the diagram above. The arrow indicates the instantaneous direction of the earth's motion six months ago; the earth is moving counterclockwise in an approximately circular path. The base has length L = 2 astronomical units (AU), where 1 AU = 1.50 x 1011 m is the mean distance from the sun to the earth. Let 2q be the apex angle of the isosceles triangle; this is the angle opposite the base. Then 2q is also the angle of shift of the star's position with respect to the background (i.e., more distant) stars = 0.8 arcsecond. The distance from the earth to the star is approximately equal to the length H of the altitude of the isosceles triangle. Then

tan q = (L/2) / H => H = (L/2) / tan q

Now

q = 0.4 arcsecond = (0.4 arcsecond)[(1 arcminute) / (60 arcseconds)][(1 degree) / (60 arcminutes)] = 4/36000 degree = 1/9000 degree = (1/9000 degree)[(p radian) / (180 degrees)] = p/1620000 radian = 1.939 x 10-6 radian

Thus,

H = (L/2) / tan q = (1 AU) / tan(1.939 x 10-6) = (1 AU) / (1.939 x 10-6) = 5.157 x 105 AU
= (5.157 x 105 AU)(1.50 x 1011 m/AU) = 7.735 x 1016 m

Back to Main Menu

Measuring Time

PROBLEM

On the day of the summer solstice, sunrise in Boston is at 9:07 AM UTC and sunset is at 12:24 AM UTC (http://www.wunderground.com). Show that these times are consistent with Boston's location within its time zone.

SOLUTION

"UTC" is coordinated universal time and is the time on the Greenwich, England, meridian, which is at 0° longitude. Boston, Massachusetts, is five time zones behind Greenwich, so the local sunrise and sunset times are 4:07 AM EST (eastern standard time) and 7:24 PM EST, which occur 7 hr 53 min before noon and 7 hr 24 min after noon. Boston is at longitude 71° 3' 37" W = - 71.060° (World Almanac and Book of Facts). The solar time in Boston is (71.060°) / (15°/hr) = 4.737 hr = 4 hr 44 min behind UTC. Thus, sunrise and sunset in Boston occur at local solar times 4:23 AM and 7:40 PM, which are 7 hr 37 min before noon and 7 hr 40 min after noon, so the amounts of daylight before and after noon, if measured by solar time, are approximately equal.

World Almanac Books 2003, The World Almanac and Book of Facts 2003 (New York: World Almanac Books).

Back to Main Menu

PROBLEM

Chinese New Year begins at the second new moon after the day of the winter solstice (World Almanac and Book of Facts 2003, p. 647), based on Beijing standard time (BST), which is eight hours ahead of univeral time (UT), the time at the prime meridian through Greenwich, England. Easter Sunday is the Sunday after the first ecclesiastical full moon (i.e., the Paschal full moon) that occurs after the ecclesiastical vernal equinox, March 21 (Wikipedia, http://en.wikipedia.org/wiki/Easter_Sunday. Ash Wednesday is the Wednesday which is 46 days before Easter Sunday. Use a table of lunar phases to predict the dates of Chinese New Year, Easter Sunday, and Ash Wednesday from 1996-2019.

SOLUTION

Based on the definitions given above, the dates of Chinese New Year, Easter Sunday, and Ash Wednesday can be determined. The results are given in the table below. Two sample calculations are shown.

Example 1. The winter solstice in 2014 occurs at 2014 Dec 21 22:59 UT, which is 2014 Dec 22 06:59 BST. A new moon occurs at 2014 Dec 22 01:35 UT, which is 2014 Dec 22 09:35 BST. This new moon occurs on the same day as the winter solstice, Beijing time, so Chinese New Year in 2015 begins on the day of the second new moon that occurs after 2014 Dec 22, Beijing time.

Example 2. The Paschal full moon in 2000 occurs on Apr. 18. This is a Tuesday, so Easter Sunday occurs on 2000 Apr 23, and Ash Wednesday occurs 46 days earlier, on 2000 Mar 8.

Year Previous Year Winter Solstice1 2nd New Moon after Winter Solstice2 Paschal Full Moon3 Chinese New Year Easter Sunday Ash Wednesday
1996
1995 Dec 22 08:17 UT
1996 Feb 18 23:30 UT
1996 Feb 19 07:30 BST
1996 Apr 3
1996 Feb 19
1996 Apr 7
1996 Feb 21
1997
1996 Dec 21 14:07 UT
1997 Feb 7 15:06 UT
1997 Mar 23
1997 Feb 7
1997 Mar 30
1997 Feb 12
1998
1997 Dec 21 19:57 UT
1998 Jan 28 06:01 UT
1998 Apr 11
1998 Jan 28
1998 Apr 12
1998 Feb 25
1999
1998 Dec 22 01:46 UT
1999 Feb 16 06:39 UT
1999 Mar 31
1999 Feb 16
1999 Apr 4
1999 Feb 17
2000
1999 Dec 22 07:36 UT
2000 Feb 5 13:03 UT
2000 Apr 18
2000 Feb 5
2000 Apr 23
2000 Mar 8
2001
2000 Dec 21 13:25 UT
2001 Jan 24 13:07 UT
2001 Apr 8
2001 Jan 24
2001 Apr 15
2001 Feb 28
2002
2001 Dec 21 19:15 UT
2002 Feb 12 07:41 UT
2002 Mar 28
2002 Feb 12
2002 Mar 31
2002 Feb 13
2003
2002 Dec 22 01:04 UT
2003 Feb 1 10:48 UT
2003 Apr 16
2003 Feb 1
2003 Apr 20
2003 Mar 5
2004
2003 Dec 22 06:54 UT
2004 Jan 21 21:05 UT
2004 Jan 22 05:05 BST
2004 Apr 5
2004 Jan 22
2004 Apr 11
2004 Feb 25
2005
2004 Dec 21 12:43 UT
2005 Feb 8 22:28 UT
2005 Feb 9 06:28 BST
2005 Mar 25
2005 Feb 9
2005 Mar 27
2005 Feb 9
2006
2005 Dec 21 18:33 UT
2006 Jan 29 14:15 UT
2006 Apr 13
2006 Jan 29
2006 Apr 16
2006 Mar 1
2007
2006 Dec 22 00:22 UT
2007 Feb 17 16:14 UT
2007 Feb 18 00:14 BST
2007 Apr 2
2007 Feb 18
2007 Apr 8
2007 Feb 21
2008
2007 Dec 22 06:12 UT
2008 Feb 7 03:44 UT
2008 Mar 22
2008 Feb 7
2008 Mar 23
2008 Feb 6
2009
2008 Dec 21 12:02 UT
2009 Jan 26 07:55 UT
2009 Apr 10
2009 Jan 26
2009 Apr 12
2009 Feb 25
2010
2009 Dec 21 17:51 UT
2010 Feb 14 02:51 UT
2010 Mar 30
2010 Feb 14
2010 Apr 4
2010 Feb 17
2011
2010 Dec 21 23:41 UT
2011 Feb 3 02:31 UT
2011 Apr 17
2011 Feb 3
2011 Apr 24
2011 Mar 9
2012
2011 Dec 22 05:30 UT
2012 Jan 23 07:39 UT
2012 Apr 7
2012 Jan 23
2012 Apr 8
2012 Feb 22
2013
2012 Dec 21 11:20 UT
2013 Feb 10 07:20 UT
2013 Mar 27
2013 Feb 10
2013 Mar 31
2013 Feb 13
2014
2013 Dec 21 17:09 UT
2014 Jan 30 21:40 UT
2014 Jan 31 05:40 BST
2014 Apr 14
2014 Jan 31
2014 Apr 20
2014 Mar 5
2015
2014 Dec 21 22:59 UT
2014 Dec 22 06:59 BST
2015 Feb 18 23:48 UT
2015 Feb 19 07:48 BST
2015 Apr 3
2015 Feb 19
2015 Apr 5
2015 Feb 18
2016
2015 Dec 22 04:48 UT
2016 Feb 8 14:39 UT
2016 Mar 23
2016 Feb 8
2016 Mar 27
2016 Feb 10
2017
2016 Dec 21 10:38 UT
2017 Jan 28 00:07 UT
2017 Apr 11
2017 Jan 28
2017 Apr 16
2017 Mar 1
2018
2017 Dec 21 16:27 UT
2018 Feb 15 21:06 UT
2018 Feb 16 05:06 BST
2018 Mar 31
2018 Feb 16
2018 Apr 1
2018 Feb 14
2019
2018 Dec 21 22:17 UT
2019 Feb 4 21:04 UT
2019 Feb 5 05:04 BST
2019 Apr 18
2019 Feb 5
2019 Apr 21
2019 Mar 6
1http://www.hermetic.ch/cal_sw/ve/ve.php
2http://aa.usno.navy.mil/data/docs/MoonPhase.html (1996-2013), http://users.hartwick.edu/hartleyc/moon/newmoon6.html (2014-2019)
3http://users.sa.chariot.net.au/~gmarts/easter.htm

Back to Main Menu

Celestial Sphere

PROBLEM

Determine the optimal day of the year for observing the following cosmic X-ray sources, with the indicated RA and Dec coordinates:
(a) Crab Nebula (5h 34m 31s, +22d 01m) (Green 2004)
(b) SN1006 (15h 02m 50s, -41d 56m) (Green 2004)
(c) Ophiuchus Cluster (17h 12m 26.9s, -23d 22m 08.4s) (http://www.astro.isas.jaxa.jp/astroe/proposal/ao1/ao1list.html)
(d) Cygnus Loop (20h 51m 00s, +30d 40m) (Green 2004)

SOLUTION

(a) Convert RA into hours. Then divide by 24 hours and multiply by 365.25 days to determine how many days after the autumnal equinox the observation should be made. Assume the autumnal equinox occurs on Sept. 22, which would be ideal for observing a source with RA = 0.

RA = 5h 34m 31s = 5.575 h => 85 days => Dec. 16

(b) RA = 15h 02m 50s = 15.047 h => 229 days => May 9

(c) RA = 17h 12m 26.9s = 17.207 h => 262 days => June 11

(d) RA = 20h 51m 00s = 20.85 h => 317 days => Aug. 5

Green, D. A. 2004, "A Catalogue of Galactic Supernova Remnants," Bulletin of the Astronomical Society of India, 32, 335-370. (http://www.mrao.cam.ac.uk/surveys/snrs/tables-l.pdf)

Back to Main Menu

Telescopes

PROBLEM

The Meade 114EQ-AST telescope comes with an objective mirror with diameter D = 4.5" = 114.3 mm and focal length fo = 1000 mm and eyepieces with focal lengths fe = 25 mm and 9 mm.
(a) What magnifications are possible?
(b) What is the focal ratio of the telescope?
(c) Estimate the angular size of the Moon without a telescope and as viewed using this telescope with the lower magnification.
(d) Estimate the angular sizes of Mars, Jupiter, and Saturn at opposition, without a telescope and as viewed using this telescope with the higher magnification.

SOLUTION

(a) The magnification is given by

M = fo / fe

So, using the 25 mm eyepiece, the magnification is

M = (1000 mm) / (25 mm) = 40

and, using the 9 mm eyepiece, the magnification is

M = (1000 mm) / (9 mm) = 111

(b) The focal ratio is the ratio between the focal length fo and the diameter D.

FR = fo / D = (1000 mm) / (114.3 mm) = f/8.749

(c) The Moon's orbit has semimajor axis a = 384400 km and eccentricity e = 0.055. The Moon's radius is Rmoon = 1738 km. Its perigee and apogee distances are

rp = a(1 - e) = 363258 km

and

ra = a(1 + e) = 405542 km

The minimum and maximum angular sizes of the Moon as seen from Earth without a telescope are

qmin = tan-1(2Rmoon/ra) = tan-1[(2)(1738 km) / (405542 km)] = 8.571 x 10-3 rad = 0.4911°

and

qmax = tan-1(2Rmoon/rp) = tan-1[(2)(1738 km) / (363258 km)] = 9.569 x 10-3 rad = 0.5482°

With a magnification of M = 40, the minimum and maximum angular sizes of the Moon as seen from Earth are

qmin' = qminM = (8.571 x 10-3 rad)(40) = 0.9523 rad = 54.56°

and

qmax' = qmaxM = (9.569 x 10-3 rad)(40) = 1.063 rad = 60.92°

(d) Mars' orbit has semimajor axis a = 227.9 x 106 km and eccentricity e = 0.093. Mars' radius is Rmars = 3397 km. Its perihelion and aphelion distances are

rp = a(1 - e)

and

ra = a(1 + e)

The radius of Earth's orbit, assumed to be circular, is re = 149.6 x 106 km. The minimum and maximum center-to-center distances from Earth to Mars are

rmin = rp - re = a(1 - e) - re = (2.279 x 108 km)(1 - 0.093) - 1.496 x 108 km = 5.711 x 107 km

and

rmax = ra - re = a(1 + e) - re = (2.279 x 108 km)(1 + 0.093) - 1.496 x 108 km = 9.949 x 107 km

The minimum and maximum angular sizes of Mars as seen from Earth without a telescope are

qmin = tan-1(2Rmars/rmax) = tan-1[(2)(3397 km) / (9.949 x 107 km)] = 6.829 x 10-5 rad = 0.2347' = 14.09"

and

qmax = tan-1(2Rmars/rmin) = tan-1[(2)(3397 km) / (5.711 x 107 km)] = 1.190 x 10-4 rad = 0.4090' = 24.54"

With a magnification of M = 111, the minimum and maximum angular sizes of Mars as seen from Earth are

qmin' = qminM = (6.829 x 10-5 rad)(111) = 7.587 x 10-3 rad = 0.4347°

and

qmax' = qmaxM = (1.190 x 10-4 rad)(111) = 1.322 x 10-2 rad = 0.7574°

Jupiter's orbit has semimajor axis a = 778.4 x 106 km and eccentricity e = 0.048. Jupiter's radius is Rjup = 71398 km. For Jupiter, we get

rmin = 5.914 x 108 km
rmax = 6.662 x 108 km
qmin = 2.144 x 10-4 rad = 0.7369' = 44.21"
qmax = 2.414 x 10-4 rad = 0.8300' = 49.80"
qmin' = 2.382 x 10-2 rad = 1.365°
qmax' = 2.683 x 10-2 rad = 1.537°

Saturn's orbit has semimajor axis a = 1425.6 x 106 km and eccentricity e = 0.054. Saturn's radius is Rsat = 60000 km. For Saturn, we get

rmin = 1.199 x 109 km
rmax = 1.353 x 109 km
qmin = 8.869 x 10-5 rad = 0.3049' = 18.29"
qmax = 1.001 x 10-4 rad = 0.3441' = 20.64"
qmin' = 9.855 x 10-3 rad = 0.5646°
qmax' = 1.112 x 10-2 rad = 0.6371°

Back to Main Menu

Moon

PROBLEM

The Moon rotates about its axis in 27.322 days. Its orbit is approximately elliptic with semimajor axis a = 384,400 km and eccentricity e = 0.055. Because its orbital velocity varies according to Kepler's second law ("The radius vector of a planet sweeps equal areas in equal amounts of time"), the side of the Moon facing the Earth varies slightly during its orbit. Calculate the total fraction of the Moon's surface which is eventually visible from the Earth as a result of this.

SOLUTION

The radial distance r from the center of the Earth to the center of the Moon can be written as

r = 1 / (B + A cos q)

where q is its angular displacement from its perigee position,

A = e / [a(1 - e2)] = 1.43514 x 10-7 km-1 = 1.43514 x 10-10 m-1

and

B = 1 / [a(1 - e2)] = 2.60935 x 10-6 km-1 = 2.60935 x 10-9 m-1

B is also equal to

B = - mK/L2

where

K = - GMem

G = 6.67 x 10-11 N m2/kg2 is the gravitational constant, Me = 5.974 x 1024 kg is the mass of the Earth (Karttunen et al. 1987), m is the mass of the Moon, and L is the orbital angular momentum of the Moon. Thus,

B = mGMem/L2 = GMe(m/L)2 => L/m = sqrt(GMe/B) = sqrt[(6.67 x 10-11 N m2/kg2)(5.974 x 1024 kg)/(2.60935 x 10-9 m-1)] = 3.908 x 1011 m2/s

Now the orbital angular momentum of the Moon is constant, and since the mass of the Moon is also assumed constant, L/m is also a constant. The orbital angular momentum is

L = r x p => L = rmvq = rm r dq/dt = mr2 dq/dt => L/m = r2 dq/dt

where

vq = r dq/dt

is the velocity component perpendicular to the radius vector. It follows that r2 dq/dt is also a constant

r2 dq/dt = L/m = 3.908 x 1011 m2/s

So

dq/dt = C/r2 = C(B + A cos q)2 (I)

where C = 3.908 x 1011 m2/s.

We will now calculate the amount by which the visible side of the Moon varies during its orbit. From the above equation (I) for dq/dt, we know the orbital angular velocity of the Moon as a function of its angular position q. Define t = 0 as the time when the Moon is at perigee. The above expression for dq/dt can be integrated to get the time at which the Moon is at some orbital angular position q0.

t = ∫0q0 dq / [C(B + A cos q)2]

The integral may be done numerically using a step size of 2p/360, for example. Set up a spreadsheet in Excel with the following columns:

Column A: an index number (i = 1, 2, 3,..., 361) representing positions in the orbit. There will be a total of 361 rows in the spreadsheet. The (first, last) row corresponds to q = (0, 2p).
Column B: the constant A = 1.43514 x 10-10 in units of m-1
Column C: the constant B = 2.60935 x 10-9 in units of m-1
Column D: the constant C = 3.908 x 1011
Column E: the angular position qi = (i - 1)(2p/360)
Column F: the angular velocity (dq/dt)i = C(B + A cos qi)2
Column G: the time interval (dt)i elapsed between the (i-1)th row and the ith row, (dt)i = (qi - qi-1) / {(0.5)[(dq/dt)i-1 + (dq/dt)i]} (i.e., the change in q divided by the average value of dq/dt during the time interval), which is set equal to zero for the first row
Column H: the time ti corresponding to the ith row, which is the sum of the values of dt for all rows up to and including the ith row
Column I: the angular orientation ai of the Moon about its rotation axis relative to its orientation at perigee, ai = (i-1)(2p/360)
Column J: the difference qi - ai

Lines 1, 2, 3, and 361 should have an appearance equivalent to the following for Columns A, E, F, G, H, I, and J (i, qi, [dq/dt]i, [dt]i, ti, ai, and qi - ai):

  A     E         F           G           H           I          J
  i     qi     (dq/dt)i     (dt)i         ti          ai       qi-ai
  1 0.000000 2.96141E-06    0.000000    0.000000 0.000000000 0.000000
  2 0.017453 2.96137E-06 5893.611095 5893.611095 0.015609964 0.001843
  3 0.034907 2.96122E-06 5893.798267 11787.40936 0.031220424 0.003686
361 6.283177 2.96141E-06 5893.611123 2372241.406 6.283177198 2.8E-06


The last entry in Column H is the number of seconds elapsed by the time the Moon gets back to perigee, which is 2372241.406 s = 27.4565 days. This is slightly different from the Moon's measured orbital period of 27.322 days because the real orbit is slightly different from an ellipse. The formula for period t as a function of the semimajor axis is

t = sqrt(4p2a3/GMe) = sqrt[4p2(3.844 x 108 m)3/(6.67 x 10-11 N m2/kg2)(5.974 x 1024 kg)] = 2372242.291 s = 27.4565 days

which is consistent with the above.

The minimum and maximum values of qi - ai in Column J are ± 0.11003 rad, which indicates that (2)(0.11003 rad) = 0.22 rad of the Moon's surface eventually comes into view due to the variation in its orbital velocity. Thus, the total fraction of the Moon's surface that eventually comes into view is 0.5 + (0.22 rad)/(2p rad) = 0.535 or 53.5%.

Note: The process discussed here, where the variation in the Moon's orbital velocity allows more of its surface to be seen, is an example of a libration. There are other librations or processes that allow us to see more of the Moon's surface. As a result of these other librations (http://www-spof.gsfc.nasa.gov/stargaze/Smoon4.htm), the total fraction of the Moon's surface that can be observed from the Earth at one time or another is about 59%.

At any instant of time, slightly less than 50% of the Moon's surface can be seen from the Earth. Imagine a cone whose vertex is at the observer and which extends towards the Moon so that its side is tangent to the Moon's surface. Assume that the center-to-center distance between the Earth and the Moon is equal to the semimajor axis a = 3.844 x 105 km. The distance from the observer to a point of contact between the cone and the Moon's surface is

r = sqrt[(a - Re)2 - Rm2]

where Rm = 1738 km is the radius of the Moon. Thus,

r = sqrt[(3.844 x 105 km - 6.37 x 103 km)2 - (1738 km)2] = 3.780 x 105 km

The angle g between the center-to-center line between the Moon and the Earth and the radius from the center of the Moon to a point of tangency is

g = tan-1(r/Rm) = tan-1[(3.780 x 105 km) / (1738 km)] = 1.566 rad = 89.74°

The total solid angle of the Moon's surface which is visible from the Earth is

W = ∫ dW = ∫01.566 sin q dq02p df = 2p(cos 0 - cos 1.566) = 2p(0.995402) ster

which is [2p(0.995402) ster] / (4p) = 0.497701 = 49.7701% of the total solid angle. This is the fraction of the Moon's surface which is visible from the Earth at any given instant, assuming no cloud cover or other obstructions.

Karttunen, H., Kröger, P., Oja, H., Poutanen, M. and Donner, K. J. 1987, Fundamental Astronomy (Berlin, Germany: Springer-Verlag).

Back to Main Menu

Mars

PROBLEM

(a) What is the difference between an opposition of Mars and a closest approach?
(b) From a table of oppositions, determine when Mars made its closest approach to Earth in the past 100 years and when it will make its closest approach to Earth in the next 100 years.
For the closest approach in the next 100 years:
(c) What will be the magnitude and size of Mars as viewed from Earth?
(d) What telescope magnification will be needed so that Mars will appear as large as the Moon viewed with the naked eye?

SOLUTION

(a) An opposition of Mars occurs when the ecliptic longitudes of Mars and the Sun are 180° apart, which happens when Earth is between Mars and the Sun. A closest approach between Mars and Earth occurs when the distance between the two planets is minimized. Because of the shapes and orientations of their orbits, oppositions and closest approaches occur at approximately but not exactly the same time. They could occur on different days.

(b) Tables of Mars oppositions and closest approaches are available at http://www.seds.org/~spider/spider/Mars/marsopps.html. From the table of closest approaches, those occurring between 1906 and 2106 are listed in the table below.

Year Opposition
Date
Closest Approach
Date
Distance (km) Distance (AU)
1924
Aug 23
Aug 22
55,776,939
0.37284581
2003
Aug 28
Aug 27
55,758,006
0.37271925
2050
Aug 14
Aug 15
55,957,000
0.374051
2082
Sep 1
Aug 30
55,883,780
0.373564

Between 1906 and 2006, Mars made its closest approach to Earth on 27 August 2003. Between 2006 and 2106, Mars will make its closest approach to Earth on 30 August 2082.

(c) The diameter of Mars is d = 4212 miles = 6.779 x 106 m. At a distance of r = 5.588 x 1010 m, the diameter of Mars subtends an angle of

q = tan-1(d/r) = tan-1[(6.779 x 106 m) / (5.588 x 1010 m)] = 1.213 x 10-4 rad = 25.02"
= 6.951 x 10-3 ° = q2082

According to Norton's Star Atlas, Twentieth Edition, during the 2005 opposition, Mars subtended an angle of q2005 = 19.9" and had a magnitude of M2005 = - 2.3. A difference of five magnitudes corresponds to a factor of 100 difference in brightness. Thus, one magnitude corresponds to a factor of 1001/5 = 2.511886432 difference in brightness. The brightness of an object is inversely proportional to the square of its distance. The angular size of a distant object is inversely proportional to its distance. Thus, the brightness of a distant object is proportional to the square of its angular size. It follows that the ratio between the brightness of Mars during its 2082 opposition and during its 2005 opposition is

B2082/B2005 = (q2082/q2005)2 = (25.02"/19.9")2 = 1.581

and Mars will therefore appear 1.581 times brighter during its 2082 opposition than it did during its 2005 opposition. The difference in magnitude is proportional to the difference in the logarithm of the brightness. The proportionality constant is - 5/2. Thus,

M2082 - M2005 = - (5/2)(log B2082 - log B2005) = - (5/2) log(B2082/B2005) = - (5/2) log 1.581
= - 0.4972 => M2082 = M2005 - 0.4972 = - 2.3 - 0.4972 = - 2.797

During the 2082 opposition, Mars will have a magnitude of M2082 = - 2.797.

(d) The diameter of the Moon is d = 2160 miles = 3.476 x 106 m. Its mean distance from Earth is r = 2.389 x 105 miles = 3.844 x 108 m. Its mean angular size is

qMoon = tan-1(d/r) = tan-1[(3.476 x 106 m) / (3.844 x 108 m)] = 0.5181°

The ratio between the angular size of the Moon and the angular size of Mars during the 2082 opposition is (0.5181°) / (6.951 x 10-3 °) = 74.54. Thus, a telescope whose magnification is 75 would be needed to make Mars appear as large as the Moon appears to the naked eye.

Back to Main Menu

Future Space Missions

[date unspecified] The Terrestrial Planet Finder will be launched by NASA and consist of two space observatories, one to detect visible light from Earth-like planets and another to probe their chemistry (Muir 2005).
[date unspecified] SIM PlanetQuest will be launched by NASA to search for terrestrial planets around 250 nearby stars (Muir 2005).
[date unspecified] NASA's James Webb Space Telescope, a much larger infrared observatory than the Spitzer Space Telescope, will be launched (Douthitt 2005).
2015 Darwin will be launched by the European Space Agency to search for chemical signs of life from Earth-like planets (Muir 2005).
2008 Kepler, a 95-cm space telescope, will be launched by NASA to simultaneously monitor 100,000 stars (Muir 2005).
2006 COROT will be launched by the French space agency CNES to simultaneously monitor 12,000 stars (Muir 2005).

Muir, Hazel 2005, "Ten Years on, a Rich Haul of Planets," New Scientist, 8 October 2005, 8.

Back to Main Menu

Cosmology

PROBLEM

Estimate the rate at which hydrogen atoms would have to be created, according to the steady-state model, to maintain the present density of the universe of about 10-27 kg/m3, assuming the universe is expanding with Hubble constant H = 80 km/s/Mpc.

[from Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 33.41]

SOLUTION

Assume that the radius of the universe is equal to the distance to the farthest galaxies, d ≈ 1010 LY. Then the volume of the universe is

Vuniv = (4/3)pd3

Since

1 LY = (3 x 108 m/s)(1 yr)(365.25 day/yr)(24 hr/day)(60 min/hr)(60 s/min)
= 9.4673 x 1015 m

and

1 pc = (1.496 x 1011) tan[(1/3600)°] = 3.0857 x 1016 m

the radius of the universe is equal to

d = (1010 LY)(9.4673 x 1015 m/LY)(1 pc / 3.0857 x 1016 m) = 3.0681 x 109 pc
= 3068.1 Mpc

Note: One light year is the distance that light travels in one year in a vacuum. The conversion from parsecs to meters comes from the definition of a parsec. A parsec is the distance from the sun to a star for which the change in the direction to the star, as observed from the earth, from one side of the earth's orbit to the other side (assuming a circular orbit with the sun at the center) is 2 arcseconds. The parallax of such a star, defined as half of the change in direction, is 1 arcsecond = (1/3600)°. Thus, tan[(1/3600)°] = (1.496 x 1011 m) / (1 pc), where 1.496 x 1011 is the mean earth-sun distance.

The rate at which the radius of the universe is increasing, given by Hubble's law, is

v = Hd = (80 km/s/Mpc)(3068.1 Mpc) = 2.4545 x 105 km/s = 2.4545 x 108 m/s

The rate at which the volume of the universe is increasing is

dVuniv/dt = (d/dt)(4/3)pd3 = (4/3)p(3d2)dd/dt = 4pd2v
= (4p)(9.4673 x 1025 m)2(2.4545 x 108 m/s) = 2.7646 x 1061 m3/s

The fractional rate at which the volume of the universe is increasing is

(1/Vuniv)dVuniv/dt = (4pd2v) / [(4/3)pd3] = 3v / d
= (3)(2.4545 x 108 m/s) / (9.4673 x 1025 m) = 7.7778 x 10-18 s-1

For each cubic kilometer, the rate at which the universe is expanding is 7.7778 x 10-18 km3/s. Each ordinary hydrogen (1H1) atom has a mass of 1.007825 u = (1.007825 u)(1.6605 x 10-27 kg/u) = 1.6735 x 10-27 kg. A density of r = 10-27 kg/m3 corresponds to (10-27 kg/m3)(1 H atom / 1.6735 x 10-27 kg) = 0.59755 H atom / m3 or 0.59755 x 109 H atom / km3. The rate at which hydrogen atoms would have to be produced per km3 would be (7.7778 x 10-18 km3/s)(0.59755 x 109 H atom / km3) = 4.6476 x 10-9 H atom / s, or one every (1 s) / (4.6476 x 10-9) = 2.1516 x 108 s = (2.1516 x 108 s)(60 s/min)(60 min/hr)(24 hr/day)(1 yr / 365.25 days) = 6.8181 yr.

Back to Main Menu

Current and Past Space Missions


2005 Venus Express was launched on Nov. 9 by the European Space Agency on a Russian Soyuz-Fregat rocket from Kazakhstan, Russia (Muir 2005). The spacecraft will reach Venus in April 2006 and make global maps of the planet's atmospheric composition, check for active volcanism, and measure weather systems.
2003 NASA's Spitzer Space Telescope, an infrared observatory, was launched in August 2003 (Douthitt 2005). Unlike most space telescopes, Spitzer orbits the Sun instead of the Earth. This prevents the Earth and the Moon from blocking Spitzer's field of view and also prevents heat from the Earth from affecting the telescope's performance. It should operate until 2008, when it runs out of the liquid helium that helps cools it. Spitzer was formerly known as the Space Infrared Telescope Facility (SIRTF).
1998 The first component of the International Space Station (ISS), the Russian Zarya Control Module, was launched on 20 November 1998. The first crew began permanent human habitation of the ISS in October 2000. The ISS is scheduled for completion in 2006.
1978 The Einstein Observatory, High Energy Astrophysical Observatory 2 (HEAO-2), was the first fully imaging X-ray Telescope. It operated from 12 November 1978 to April 1981 and detected X-rays in the 0.2-20 keV energy range.
1969 Apollo 11 was the first manned lunar landing mission. Neil Armstrong and Edwin Aldrin were the first and second humans to walk on the Moon. Michael Collins was the command module pilot on this mission. Neil Armstrong received a B.S. in aeronautical engineering from Purdue University and an M.S. in aerospace engineering from the University of Southern California. Edwin Aldrin received a B.S. from the United States Military Academy at West Point, New York, and a Ph.D. in astronautics from the Massachusetts Institute of Technology. Michael Collins received a B.S. from the United States Military Academy at West Point, New York.

Douthitt, Bill 2005, "Night Vision," National Geographic, December 2005, 110-117. [Spitzer Space Telescope]

Muir, Hazel 2005, "Europe to Venus on a Shoestring," New Scientist, 19 November 2005, 6. [Venus Express]

Back to Main Menu