To FootThis governs Date and Time. See also my Date Miscellany I, Time Miscellany, Leap Year, Leap Seconds, Critical and Significant Dates (ZIP), Gravity 0, Year 2000 pages, and some of their links; I do not myself deal with measurements.
In 1880, by Act of Parliament, the legal time for Great Britain was made Greenwich Mean Time, GMT. See in Date Miscellany II and/or Time Miscellany.
Before 1925-01-01, the Astronomers' GMT day began twelve hours later than the corresponding legal GMT day; since then, it has started at midnight of legal GMT.
"Greenwich Mean Astronomical Time", GMAT, is now used to describe a scale matching legal GMT from 1925 but omitting (presumably) 1924-12-31 12:00 to 24:00. UT was introduced in 1928 as an internationally unambiguous term for GMT with date changing at midnight in Greenwich; but it has developed varieties such as UT0 UT1 UT2 & UTC, and is not generally well-known.
The legal sense of GMT, (i.e. UT) is used in these pages.
The Gregorian Calendar is intended to remove long-term drift from the date of the Vernal Equinox, and does so reasonably well. The average Gregorian year length is 365.2425 days.
Solstices and Equinoxes are nominally 0.2425 days (5h 49.2m) later each ordinary year, and 0.7575 days earlier each leap year (in addition to shifts due to solar system dynamics).
The Earth goes round the Sun once per year, at a distance of around 150 million kilometres; and the Moon goes round the Earth about once per month, at around 400 thousand kilometres.
If the Sun, the Earth, and the Moon are modelled respectively by a lamp, your head, and a hand-held orange, or similar, and if you revolve on the spot while watching the orange - advanced walkers can also go, more slowly, round the lamp - you will see how the phase of the Moon depends on the relation between the angles of illumination and of observation.
Also, when the lamp is prevented from shining on the orange by your head, or vice versa, you will see roughly why eclipses happen. A Solar Eclipse occurs when the Moon's shadow passes across the Earth, at Full Moon; a Lunar Eclipse occurs when the Earth shadows the Moon, at New Moon.
Because of the relative orbital tilt, about 5°, most months have no eclipses.
The distances from Earth of the Sun and the Moon vary somewhat; but the ratio is approximately equal to that of their sizes. This coincidence is why Solar eclipses have the nature that they do.
Via HMNAO - Eclipses
To tell from its appearance whether the Moon is waxing or waning, observe whether it looks more like a "C", with horns to the right, or like a "D", with horns to the left. Ignore the terminator, the line on the Moon between light and dark; concentrate on the illuminated part of the circular edge of the visible Moon. Now remember, firstly, that C is for Crescendo and D is for Diminuendo; and, secondly, that in the Northern part of the world, it is in fact the other way round.
Reference : Bernard Wicksteed, "It's Fun Finding Out" p.159 foot; book pub. The Daily Express (London) 1947; will have been in the newspaper c. 1946.
If a crescent Moon is seen in the evening after sunset, it will be waxing : between New Moon and First Quarter. If it is seen in the morning before sunrise, waning : between Last Quarter and New Moon.
The Ecliptic is the projection onto the Celestial Sphere of the plane of the Earth's yearly orbit around the Sun.
The Equatorial plane is defined by the Earth's daily rotation. It is tilted by about 23.4° with respect to the Ecliptic, and precesses with a period of about 25800 years. Currently, to the North, the axis is near the star Polaris in the tail of Ursa Minor; to the South, it is in the constellation Octans.
Right Ascension is a co-ordinate measured along the Equator from one of its intersections with the Ecliptic, in hours 00:00 to 24:00; the other co-ordinate Declination is measured in degrees perpendicularly towards the North.
The perpendicular to the Earth's orbital plane meets the Celestial Sphere in central Draco, at RA 18h, Dec +66.6°, and in Dorado, at RA 6h, Dec -66.6°, near the LMC).
The plane of the Moon's monthly orbit is tilted by about 5.2° from the Ecliptic, and precesses with a period of about 18.6 years.
So the Moon's orbit varies from being over the Equator by the tilt of the Earth's rotation modulated by the tilt of the Moon's orbit, each with respect to the Ecliptic; by 18.2° to 28.6°.
The axis of rotation of the Moon is not parallel to that of the Earth; it is at an angle of about 1.5° from the perpendicular to the Ecliptic.
The Moon can be overhead if the latitude is no more than about ±28.6°; see, for example, Jules Verne, "De la terre à la lune". Within that band, it can be seen both to the North and to the South.
The Moon was at or near perigee on 2005-07-21 (I don't know how the major axis of its orbit moves); also Full and low (from the UK), so it looked particularly large.
The Earth's radius is 4000 miles, the Moon's is 1000 miles, 240000 miles away; so, when the Moon is near the horizon, its upper edge is depressed by 1/80 radian or 0.7° relative to the line of centres. But atmospheric refraction then raises the apparent Moon by 0.6°.
So there is I think a period each month above latitude (61.4° to 71.8°) - 0.1° during which the Moon does not rise at all, and one above (61.4° to 71.8°) + 0.?° in which it does not completely set.
Let us consider only locations in the Northern Hemisphere, since in the Southern Hemisphere the situation is exactly opposite apart from possible minor corrections.
Trivially, at the North Pole the sub-Lunar point can only be to the South. But, if visible clear above the horizon, the Moon itself must be to the North (further from the Equatorial Plane).
Between 90°N and about 62-72°N, the Moon can sometimes be seen passing through the North, and the Full Moon does so at local noon. Between 90°N and about 85°N, the Full Moon can sometimes be seen in the North during daylight, and the New Moon (by earthshine) during darkness.
Adding the effects of refraction and of topographical height, it seems that the Moon might almost have been visible in the North from the UK some time in the first half of 2006. Does anyone really know?
It appears that, on Wednesday 2006-04-05 at 05:11:36 UTC, the Moon's limb wss nearly visible in the North from Muckle Flugga lighthouse, Shetland, UK, at about 60°51'N 1°W; but that the lighthouse is not likely to have been high enough for the Moon to be truly circumpolar from the UK). See NPEN; earlier months may be better.
I have read : According to Meeus, the next southernmost Moon will take place on 22 March 2006 (geocentric declination -28° 43' 23"); the last such extremum happened on 29 September 1987.
Full and New Moons are historically related to the Month. For the times of Moon phases and of eclipses, see Fred Espenak, Five Millennia Catalog of Phases of the Moon. For the times of Moon phases, see USNO, Phases of the Moon.
New Moon is close to the start of the Hebrew and Islamic Months.
[1] On that basis, the Sun's mass would suffice for 5×1020 seconds, or 1.5×1013 years, a thousand times the age of the Universe (but less than a thousandth of its mass can/will be so used).
Incident solar power is about 2×1014 kW — 200×1012 kW = 200×109 MW = 200×106 GW = 200×103 TW = 200×100 PW (I've seen 1.73×1017 W reliably quoted), about 30% of which is reflected; thus about 120 PW absorbed. The population is over 6×109 now, so that's about 2×104 kW per capita.
UK electricity consumption is at least approaching an average of 1 kW per capita; and that's only one form of energy. US energy consumption has been estimated as 10 kW per capita. I've read that global energy consumption is about 650 GW, which is about 0.1 kW per capita.
Energy consumption in 2020 has been estimated at 180 PWh/yr, which is 20 TW.
Essentially all energy used becomes heat; and a 1% rise in total heat may give an 0.25% rise in temperature - nearly a degree.
Therefore, if during the 21st Century the population all reaches the present US standard of wealth, and if energy (other than direct solar) becomes very cheap and hence more used, artificial heat may well become a significant addition to natural heat.
The present UK rate of electric power consumption is about 1 kW per capita. Insolation available for an SPS is 1.37 kW/m2 so at 75% system efficiency around one square metre of SPS per capita would be required.
There are at present around 6×109 people in the world. That could mean 6000 SPS each 1 km square, or 60 SPS each 10 km square, etc.; and each would be travelling at about 10 km/s in a region traversed by other satellites at the same sort of speed. What might the collision rate be? Of course, SPS in GSO should not collide with each other very often.
Factors of the order of unity may be ignored here.
Hereabouts, the Solar constant is about
1.37 kW/m2 = 1367 W/m2.
For electromagnetic radiation, E = pc, so
Solar pressure =
(1.37×103 W/m2) /
(3×108 m/s)
= 4.7×10-6 N/m2 =
4.8×10-7 kgF/m2,
if the radiation is absorbed; double that if it is fully reflected.
Thus full reflection gives nearly a kilogram of force per square
kilometre.
Sunlight force on the Earth is of the order of
a hundred thousand tonnes.
To cross-check, 4 MT of photons distributed over 4 π (150×109)2 square metres is about 1.5×10-14 kg/s/m2 of photons arriving at 3×108 m/s, giving if not reflected a pressure of about 5×10-6 N or 5×10-7 kgF per square metre.
Atmospheric pressure is 1 kgF/cm2, about 1×1010 times greater.
I read that solar wind pressure is some 3000 times smaller.
From JSR #549, 20050627 : The solar radiation pressure in the vicinity of the Earth is 4.6 microPascals. For comparison, the solar wind pressure is only 0.4 nanoPascals today, and is typically a few nPa. (One Pascal is one Newton per square metre; our figures essentially agree.)
As the inverse square law applies both to light and to gravity, balance is independent of distance from the Sun.
Solar gravity here (balancing ω2r) is (2π / (365.2425×86400))2 × 1.5×1011 ~ 5.94×10-3 m/s2 = 5.94×10-3 N/kg. It is balanced by reflected light-pressure at an areal density (sail + load) of 2 × 4.7×10-6 / 5.94×10-3 kg/m2 ~ 1.6 grams per square metre.
In paper terms, ~1.6 gsm (ordinary paper is ~80 gsm). For a fully-reflecting paper of specific gravity 0.8, the thickness would be ~2 µm.
Units :
1 megatonne = 1,000,000 tonnes = 106 tonnes =
109 kilograms
1 gigatonne = 1000 megatonnes
1 teratonne = 1000 gigatonnes
1 petatonne = 1000 teratonnes
The approximate density of "normal" air is easy enough to calculate. Air is about 80% N2 of molecular weight (MW) 14×2 and 20% O2 of MW 16×2, so the average MW is 28.8. One gram-mole occupies 24 litres at room temperature and pressure. Hence air weighs 28.8/24 = 1.2 grams per litre = 1.2 kilograms per cubic metre.
Air is heavier than is generally realised - have you ever seen people moving a hot-air balloon and its basket with the balloon empty? That indicates the change in weight of the air in the balloon when it is heated.
Knowing the size of the Earth and the mean surface atmospheric pressure, we can now estimate the total mass of the atmosphere (ignoring corrections for its lower surface not being level).
Taking the radius of the Earth as R = 6371 km, the area of the Earth is 4πR2 = 510×1012 square metres ; the mean atmospheric pressure gives near 1.03 kgF/cm2 equivalent to 10.3 tonnes per square metre, so the weight of the atmosphere must be about 5.3×1015 tonnes. The mass of the whole planet is about 6×1021 tonnes.
Alternatively, the mass of the atmosphere is equal to the mass of a 760 mm thick shell of mercury, density 13.55 tonnes per cubic metre, which is 4πR2 × 760×10-3 × 13.55 = 5.25×1015 tonnes = 5.25 petatonnes.
The airspace occupied instead by land above sea level, and the landspace occupied instead by air below sea level, are deemed unimportant.
The weight of the whole atmosphere, from sea level to the top, is near 1 kg/cm2. The weight of a litre of it, at sea level, is 1.2 grams. A litre is 10 dm3, which equals 1 cm2 × 10 m high. Therefore, on rising 10 metres, the pressure falls by about 1.2 grams, which is 0.12%. This relative rate of fall remains more or less constant throughout the flyable region. It is equivalent to a drop of about 1/9 per kilometre, or halving every 5.8 km.
The causes of global warming are of secondary importance.
What really matters is whether undesirable change is occurring, and what can be done about it.
Precedents for making environmental changes are already established.
| The cause is irrelevant, except as a guide to number 3. | |
|---|---|
| Questions | Answers |
| 1 Is the globe warming | certainly |
| 2 Is this a bad thing | almost certainly |
| 3 Can we do anything about it | yes |
| The conclusion is now obvious | |
At present (mid-2007), carbon dioxide - CO2 - is about 383 ppmv (parts per million by volume) of the Earth's atmosphere, rising by nearly 2.0 ppmv per year on average. The increase since pre-industrial times is of the order of 130 ppmv. Those are global annual averages; global variation of annual average is a few ppmv, and local seasonal variation can reach 15 ppmv.
The atomic weights of carbon and oxygen are 12 and 16, so the MW of CO2 is 44. A concentration of 380 ppmv thus represents 380×44/28.8 ppm = 581 ppm by mass for CO2, and 380×12/28.8 ppm = 158 ppm by mass for carbon. That is 3.05 teratonnes and 0.83 teratonnes respectively for the entire atmosphere.
So a yearly rise of 2.0 ppmv CO2 represents 2.0×44/28.8 ppm = 3.1 ppm by mass for CO2, and 2.0×12/28.8 ppm = 0.8 ppm by mass for carbon. The annual rise of atmospheric CO2 is over 16 gigatonnes per year, containing about 4.5 gigatonnes of carbon.
I have read that the mass of the biosphere is about 10000 gigatonnes, and so is the mass of coal.
I did read (BBC 2006) that "The world's trees store 283 gigatonnes of carbon, 50% more than there is in the atmosphere" - implying about 140 gigatonnes in the atmosphere.
Professor Pekka Kauppi, who was named in the article, has confirmed by E-mail that he had been misquoted, writing "Trees contain 250-300 Gt (gigatons), the atmosphere 800-830 Gt, and the annual emissions of fossil carbon are about 7 Gt at present.". Earlier estimates for trees have been up to about three times higher; see Silva Fennica 37(4) 451 ff.
The article now reads "The world's trees store 250-300 gigatonnes of carbon; the atmosphere contains 800-830 gigatonnes". My estimate above agrees with the latter.
I see it stated that current (2006) emissions represent 7 or 8 gigatonnes of carbon per year, rising yearly at 2.5%.
To significantly reduce both existing and near-future global warming due to CO2, we must reduce its atmospheric concentration by of the order of 100 ppmv, 150 ppm by mass, and so its amount by around 800 gigatonnes.
The MW of CO2 is 44, and that of dry plant material, (CH2O)n, is 30 per carbon atom; so the reduction could be achieved with an extra 550 gigatonnes of plant (dry weight).
If the population is 6.5×109, that's about 85 tonnes per capita.
Alternatively, we need 1 kg/cm2 × 150 ppm × 30/44 = 100 mg/cm2 = 660 tonnes/km2 of dry plant, averaged overall; which is about a pound and a half per square yard, or two kilotonnes per square mile, including sea, ice, desert, city and all. I gather that the average forest contains about ten kilotonnes of plants per square kilometre.
Richard Branson has been reported to be offering a large prize for a way of removing "100 billion tonnes" of CO2 a year from the atmosphere; that would be good, but a little less may suffice.
Methane - CH4 - is present in the atmosphere in much lower concentration (1.8 ppm)than carbon dioxide; but per molecule it has a much greater greenhouse effect (by about 25 times). It would be the next thing to worry about.
Atmospheric aerosols have a cooling effect, and vary over comparatively short time-scales.
The Antarctic is 14.4 Mm2 in area, 98% covered by ice, on average 1.6 km thick. The ocean is 361 Mm2 in area. If all that ice were to melt, sea level would rise by, ignoring minor effects, 14.4×0.98×1600/361 ≈ 62.5 metres approximately.
Ice melting from mainly near the poles will cause a general rise of sea level; that should change the moment of inertia of the Earth and hence alter the rotation rate and the length of the day. A simplistic calculation suggests that this could affect the rate of Leap Seconds.
It is often suggested that atmosphere could be collected in bulk by a scoop-craft dipping from and returning from low orbit. This is not as easy as it seems at first sight.
The mass of a scoop-craft does not matter while it is steadily scooping; only when it is changing velocity non-gravitationally. For a reasonable collection rate, a large drive would be needed to maintain speed; it must accelerate everything collected, and overcome aerodynamic drag.
To scoop, one really needs a ramjet.
In scoop-craft co-ordinates, a fraction X of all that is scooped is decelerated from orbital speed to zero and, for conservation of momentum relative to the scoop-craft, the remaining fraction (1-X) must be accelerated from orbital speed to (orbital speed × X/(1-X)).
The energy needs to be provided by a source carried on the scoop-craft (unless atmosphere can be fused); and a supply of chemical fuel will not suffice to collect more than a lesser amount (as orbital speed exceeds chemical exhaust velocity). It must use on-board nuclear, or solar, or beamed, power.
The kinetic energy, with respect to the scoop-craft, of the fraction collected must very largely go into the exhaust, whether as mere heat or in a better-organised manner - otherwise, the collector will rapidly melt.
If the heat question is solved, a craft arriving from infinity might aero-capture without using power, and simultaneously acquire reaction mass for use later, say in orbit circularisation.
ISTM that engineers have over-complexificated the issue in the way that Specific Impulse is used; the only advantages of measuring it in seconds are that its magnitude in Metric is the same as in Imperial and that it is a number of convenient size - a one-second change in specific impulse is neither trivial nor important.
Forget energy; forget thrust; consider just momentum.
The measure of performance of such a vehicle is the dv that it can do. From simple considerations, this depends on the exhaust velocity and on the mass of the exhaust divided by the mass of the vehicle; a simple matter of ratios, until it becomes necessary to allow for the changing mass of the vehicle. The measure of engine performance is the exhaust velocity [average component in correct direction].
Note that, disregarding the effects of external forces acting on the exhaust or the vehicle, a rocket engine cannot alter the motion of the centre of gravity of the system; it can only make the system (vehicle+exhaust) extend.
The Ulysses probe went directly from Earth (late 1990) to Jupiter (Feb 1992), observed a little at Jupiter, and was there gravity-assisted into an elliptical solar-polar orbit (over the South Pole, Summer 1994; perihelion 1.5AU); it must have used initial propulsion at least sufficient for an Earth-Jupiter Hohmann orbit.
It seems to me that there is an alternative route, possibly practicable, with certain advantages; I have no knowledge of its having been suggested or evaluated before, likely as that seems. I can only approximately calculate how long it would take to complete.
Inject the probe from Earth out-of-ecliptic, into a circular solar orbit which is tilted with respect to the solar system plane by as much as launcher fuel permits. Ulysses must have left Earth with near solar-escape speed, √2 times solar-orbit speed; that seems enough to give a tilt of the order of 23°; certainly to give a non-trivial tilt.
After only three months, the probe will have a skew view of a solar pole.
Three months after that, the probe will encounter Earth (itself of some interest), and can be gravitationally deflected into a similar orbit tilted by a few more degrees. A modest (?) burn (or aerobraking?) will adjust the velocity. After another three months, the probe will have a better view of the other pole of the Sun. Three months later ... ... ...
There appears to be no great need for a near-instantaneous burn at perigee, if better specific impulse can be obtained more slowly; a small ion engine might be useful, thrusting perhaps continuously. Or a solar sail.
By what angle can Earth deflect a probe passing LEO height at solar-orbit speed? Is there an excessive speed change? Can the Moon help?
At LEO height and speed, gravity deflects by 360° in 90 minutes, or 4°/min.
The angle should depend only on the ratio of approach speed to LEO speed. Approach speed is 2π × 93000000 mi / yr, LEO speed is 2π × 4000 mi / 90 min ; the ratio is therefore (93000000/365) / (4000×16) = (nearly) 4.0; so the passing speed will differ but little from the approach speed.
So, at LEO height and passing speed, gravity deflects by (about) one degree per minute. It seems reasonable to approximate the total effect of gravity by that of full strength over the nearest 4000 mi of the near-straight path, which takes the same time as 1000 mi of LEO, or 90/25 = 3.6 minutes, which makes the deflection 3.6°.
That is 3.6° every six months.
The actual passing speed will be less when the orbit is little tilted, and √2 times more near the end; so the earlier deflections will be larger and the later ones smaller.
Presumably standard orbit analysis software will easily deal with this setup.
One could do much the same with Venus, of similar size to Earth. The reduced orbital radius increases the frequency of encounters; the greater speed diminishes the effect; jointly, these make the rate of change of plane inversely proportional to the diameter of the orbit. This advantage is slightly diminished by the lesser gravity of Venus. Then there's the question of getting started; a partly out-of-plane, partly retro, Earth departure burn and a faster first arrival at Venus ...
Now with an initial tilt obtained by an Earth-Jupiter-Hohmann size burn, and an increase of tilt every six months, it seems that it will not take unduly long to attain a near-polar solar orbit, with views of alternate poles every six months and improving meanwhile.
After attaining this orbit, a final deflection within the polar plane will alter the period to effectively terminate the Earth encounters, if required. Alternatively, the final encounter of the series could be adjusted to reduce the perihelion; if the new period was harmonically related to Earth's, there would be a further encounter ...
I am not aware of any previous consideration of such a plan, though it seems most probable that someone has thought of it before.
It appears that ESA's Solar Orbiter may be inrended to use a similar technique, with deflection by Venus.