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Angle between ecliptic and Milky Way

(Continuation)--About the year 2012

396A Posssibility of Asteroid Hitting Earth (1)
396B Posssibility of Asteroid Hitting Earth (2)

It is cooled or heated by the air passing by it, through conduction of heat.
It is heated by the resistance of the moving air (overcoming resistance absorbs energy).
It radiates heat or receives heat by radiation.

Let me start with conduction. What is the temperature of the air outside? Here in Maryland it is February right now, the air temperature is close to the freezing point and if you stick your hand--which receives blood at about 98 deg. F--out of a car window, it is likely to be cooled, Stick it out on a summer day in Baghdad, when the air is at 120 deg F., and I am not sure what will cool what.

And which side of the car will you be sitting there? All warm objects radiate heat, including your hand, and this cools them down--but meanwhile, they also receive radiation. If you sit on the sunny side (in Baghdad), your hand probably receives more than it gives out. If it is in deep shade, maybe not (though if hot walls are nearby, heated by sunlight, they radiate towards you, too).

And heating need not be evenly distributed. Consider a bullet. A pistol bullet flying below the speed of sound pushes air out of its way and encounters resistance, which heats it up. A rifle bullet is supersonic, and in addition also compresses air in its front, creating much greater heating. But the heating is not evenly distributed, the tip gets much hotter than the rest (in either case). It is not just "what speed", but what part of the object, too.

A practical problem: an airliner at 35,000 feet. The outside air there is very cold, say 40 below zero, and the airplane skin is quite cold--in spite of the friction of air rushing by, at close to the speed of sound. But I read somewhere that in airplanes going at 3 times the speed of sound, their skin can get warm enough to be weakened (at least at the front edge). Sticking out a hand from such a plane would not be a good idea--besides, the force would be so high, even at jetliner speed, your hand would be ripped off before it could get much heating.

Anyway, those are some thoughts. May the Force be with you.

240. One year of continuous sunlight?

I am looking for the answer to a question that am having, and was wondering if you would be able to help. I am not from a scientific background, so I hope my question about the earth and the sun is clear.

From my understanding the North and South poles can spend months without seeing the sun drop below the horizon, that is they have 24 hours of daytime. My question is would it be possible for someone to spend one year in daylight without seeing the sun drop below the horizon? I appreciate that one would have to move from the north to the south or vice versa in order to do this and was hoping that it wouldn't require an aircraft moving faster than Mach 2 or a space shuttle. Basically, could someone with the will and the ability, move around the planet usng conventional travel in such a way as to avoid seeing the sun set for one full year?

I hope that makes sense, and thank you for your time.

Reply

I don's think what you ask for is feasible--not without, say, a jet plane constantly flying westward, or a mad dash from one pole to the other at equinox, along a spiral following the Sun.

It is easier in orbit, and there exists a "sun synchronous" orbit which keeps the Sun in view, permanently. It is described in
http://www.phy6.org/Education/wlopolar.html

There does exist (I understand) a solar observatory at the South Pole which tracks the Sun for half a year, as it circles around the horizon. The North Pole is covered with sea ice, so I am not sure it has a permanent research station.

241. Shielding out radio waves

Really enjoyed finding your site, and as a layman, it was incredibly enlightening and insightful, particularly the sections on electromagnetics and the earth's electromagnetic fields. My question is this:

With the overwhelming increase in cell phone towers and the now constant exposure to radiation and electro magnetic waves, is there any blocker that one can easily create and use in one's home to make it a clear/clean space?

I realize that there is no direct evidence as yet that these microwaves are dangerous (just as some would say there is no evidence of global warming), but I am of the opinion that accumulation and unrelenting exposure could be harmful to humans, and I'd like to err on the side of caution. Since the announcement that Philadelphia will become a "wireless" city in the next year by placing antennas on every street pole, I have been searching for some way to at least minimize the level of waves passing through our homes. Do you have any ideas?

Reply

The answer depends on your definition of "easily": you can shield out EM waves, but it's a hassle.

Electromagnetic waves are kept out of any enclosure which is entirely enclosed in a good conductor of electric current. Such an enclosure is known as "Faraday Cage" after the scientist who first devised it in the middle 1800s, and it needs not be completely enclosed--it can tolerate openings, as long as these are appreciably smaller than the wavelength of the wave.

One example is a microwave oven (it keeps radiation in, not out, but the same principle applies). Inside it a very high density of microwave energy exists, but none escapes, because walls are of aluminum and the window, too, has a mesh of aluminum or copper embedded in it, with openings too small for the waves.

Laboratories use enclosures, frameworks over which everywhere a copper screen is stretched--like window screens, but of copper (insect screen made of iron does not have sufficient conductivity at high frequencies). Doors are tricky--they may need copper frames, with foils of a copper alloy (phosphor bronze?) all around to seal the crack between door and frame. If you walk into such an enclosure and close the door behind you, your cell phone will probably not detect anything. Cell phones do work in cars, because their windows are large enough to let the radiation in--also iron may not be such a good shield, and anyway, the car's frame is not grounded.

But cheer up: I do not think all this radiation has any effect. The reason may be (just a guess) that EM waves only give up energy in "quanta" (see
http://www.phy6.org/stargaze/Sun5wave.htm
and following sections) and those of radio waves are too small to affect molecules. After all, even microwaves in the kitchen heat food but do not otherwise modify its chemistry. I was asked about this before, and you will find the answer at
http://www.phy6.org/Education/FAQs6.html#q78.

There does exist evidence for global warming, though. Lots of it.

242. The way gravity changes with depth

Hello, I was reading one of your answers when i got confused understanding what you mean. You answered the 45th question ("Can Gravity Increase with Depth?") and wrote
"On the other hand, the closer approach to the center adds to the force GmM d(1/ R²) = GmM (2 dR/ R³)

BUT WHY? how can we show all the reasoning in one mathematical equation?
thanks for your time.

Reply

Your question raises an issue answered in question #29. Newton showed that in a hollow spherical Earth, at a distance R from the center, gravity is the same as if all the mass INSIDE radius R were concentrated at the center. The attractions of all masses at distance greater than R cancel each other.

Newton had an elaborate proof. However, if you know more advanced math, the use of the potential function –GM/r (developed by Gauss, Laplace, Poisson, Green and others) gives you the result immediately.

So suppose you stand on the Earth's surface, your R is one Earth radius, and you moved to a location at distance R/2. The attractive force, inversely proportion to the distance squared, should be 4 times larger, since you are twice as close to the center. However, if the density of the Earth is uniform throughout, the amount of attracting matter is only what is contained in a sphere of radius (R/2), and is therefore only 1/8 as much. So the force of gravity is weaker, only half the one at the surface (4 times 1/8).

However, the Earth gets denser towards the middle, due to all the weight piled up on top--also, much of its core is iron, denser than rock. So it is just possible (mathematically, at least) that the matter inside the sphere of (R/2) has twice the average density of the Earth as a whole. In that case, the attraction is doubled, and as a result, gravity is the same at R/2 as on the surface.

If the average density is more than double, the gravity is stronger. I don't think that happens, but in theory at least, it is possible.

Continuation of the question

Our instructor posted for us the question following below. However, I am having some difficulty solving it, using the reasoning I learned from you.

The Earth has a radius of 6370 km, and a mass of 5.98 x 10²⁴ kg.

---The thickness of the crust is 25 km and it weighs 3.94 x 10²² kg.
---The thickness of the mantle is 2855 km and it weighs 4.01 x 10²⁴ kg.
And finally:
---The radius of the core is 3490 km and it weighs 1.93 x 10²⁴ kg.

Show that g has a local minimum within the mantle; find the distance from the Earth's center where this occurs, and the associated value of g.

Reply

I make it a rule not to solve homework of other people--but let me help you a little here. Please don't trust my calculatinos--I make mustakes too. Recalculate everything.

Forget about the crust: the point you seek is in the mantle, and the crust exerts no gravity pull there--it's a hollow sphere, and you are inside it.

The volume of a sphere of radius R is taken as 4.189(R³) The outer radius of the mantle is then 6.375 10⁶ meters, and the volume is 1.07 10²¹ cubic meters.

The radius of the core is 3.49 10⁶ meters and its volume 0.178 10²¹ cubic meters, so the mantle contains 0.892 10²¹ cubic meters

The density of the mantle is then DM = 4.5 ton/cubic meter (or kg.liter) The density of the core is DC = 10.84 ton/m³ , and the ratio is K=DM/DC = 0.415

Suppose the radius of the core is R and we are observing gravity at a distance R+r.

4.189 G DC (R³) / (R+r)²

All the core contributes to gravity where you are--but only that part of the mantle which is closer to the center. For the rest of the mantle, you are inside a hollow shell and its pull adds up to zero. Therefore the pull of the mantle is

4.189 G DM [(R+r)³ – R³]/(R+r)² =
4.189 G K DC [(R+r)³ – R³]/(R+r)³

Combined

F = {4.189 G DC}{K [(R+r)³ – R³] + R³}/(R+r)² =
{4.189 G DC}{K (R+r)³ + (1-K)R³}/ (R+r)²

The factor {4.189 G DC} is just a number (call it Q or whatever), it plays no role. What follows it must be differentiated by r. I trust you can do the rest. Note I like to work in ratios--one way of keeping numbers small. In the end, don't look for r but for (r/R), which anywhere in the mantle is a relatively small number.

243. The Sun's Axis

Hi Mr. Stern,

My sister and I have been surfing the web to find out whether or not the sun has an axis. So far we've found nothing concrete. If it does have an axis, we want to know how it turns, how long is it's revolution, and if it's not the center of this galaxy, how it and our planets turn in our own rotation?

Thank you!

Reply

The Sun does have an axis, and rotates in about 25 days (27 days as seen from Earth, because it too moves during each rotation). But it is not solid, and flows inside it make the rotation period depend on latitude. This also happens with Jupiter, for instance, but there the difference is much smaller. For a table see my site

http://www.phy6.org/earthmag/mill_5.htm

(scroll down until you reach the table). The rotation axis is inclined by about 7 degrees to the line perpendicular to the plane of the ecliptic, which is also (more or less) the plane in which other major planets orbit.

Don't mix in the galaxy here: the galaxy is huge compared to the solar system and may have something like 100 billion suns, with a whopping black hole at the center--see

http://www.phy6.org/stargaze/Sblkhole.htm

Viewed from far north of the solar system, the Sun rotates counterclockwise, so do the major planets and most moons, and the orbits are also counterclockwise. All this suggests a common origin, in a cloud of dust and gas which originally rotated in the same sense. All bodies that condense from such a cloud retain its sense of rotation.

But why the 7-degree tilt, I don't know.

244. "Gravity particles"?

Hi! Great website, by the way!! I'm in Plymouth, in the UK. Please, email me back the answer to my question, and thank you!

I am doing a piece of coursework on orbital planes of the planets, about the creation of the solar system and how the planets came to be in the same plane. How do the planets (apart from Pluto) stay in the same plane? I found something about repulsion and gravity particles coming out of the sun at

http://www.grantchronicles.com/astro10.htm

but am not sure if this is just one man's theory or the best theory scientists have today. (Sorry, but I'm sometimes a bit skeptical about website theories.)

Could you also tell me why Pluto orbits the sun as it does? I found some strange theory about a 10th planet that entered our solar system ages ago and brought into our solar system Pluto and its moon Charon, but then broke off and became part of our solar system. Is this a reasonable theory? And if so why doesn't Pluto re-adjust into the same orbital as the other planets?

Reply

I visited Plymouth in 1989, on a side-trip from a scientific conference in Exeter, it is a very pretty place, with old alleys and modern stuff--I think it was armed forces day then, with extra festivities. We visited a geomagnetism lab and ended our visit in an excursion by boat up the Tamar river, all the way past a high arched stone bridge. And we got to taste clotted cream. You live in a lovely place.

As for orbital planes... they are of course fixed by conservation of angular momentum. Those planes may be very slightly affected by the attraction between planets, but they seem all roughly follow the plane of the original cloud from which the solar system was formed. Their inclinations differ slightly, perhaps because at the time of their formation, collisions between big clumps of matter (which later joined up to form planets) were still occuring. Planets do not share exactly the same plane, they only do so approximately.

However, be careful about "repulsion." I looked up the web page you cited. Its mention of "gravitational particles" reminds me of the theory of George-Louis LeSage (1724-1803), described at

http://en.wikipedia.org/wiki/LeSage_(gravity)

LeSage (like others of his time) tried to explain all Nature in terms of mechanics, and gravity's "action at a distance" was a challenge. (Today we have a broader view of space, believing that "fields" exist which are modifications of space, and which explain such phenomena as electromagnetic waves. I have mentioned some of that is section S-5 of "Stargazers" and in the sections that follow it.)

LeSage's explanation was that space was filled with "ultramundane particles," very fast particles which carried momentum and could easily penetrate matter, a bit like neutrinos. However, occasionally they collided with matter and rebounded elastically, so that two material objects close to each other would be driven towards each other. They would shield each other to some extent and therefore be hit asymmetrically, with more hits on their hemispheres pointing outwards than the ones facing each other.

The trouble with LeSage's theory (among other things) is that it assumes a select frame of reference--the one in which all "gravity particles" are equally distributed in all direction. If an object starts moving in relation to that frame, it encounters more "head-on collisions" than "rear-end collisions" (and even if all those particles move at the speed of light, head-on collisions carry more momentum) and therefore its motion is resisted, as if it moved in a viscous medium.

As for Pluto, it seems now to be not an isolated planet but part of a growing family of distant icy asteroids, the "Kuiper Belt." It isn't even the biggest or the most inclined among them--see the last part of the section on the ecliptic

http://www.phy6.org/stargaze/Secliptc.htm#newplanet .

Until we know more, it's risky to speculate.

245. "Short stay on Mars"

I am an aspiring science fiction author -- I have written a number of short stories and one novel. I try to make the science in my stories as plausible as possible, and hope that you can help me. I saw your page
http://www.phy6.org/stargaze/Smars1.htm
and it almost tells me what I need to know.

For my next story, I need to know at what point of a "quick trip" to Mars (meaning only a month or so on the surface) would the astronauts be at the furthest distance from Earth. i.e. would it be when they land, when they are about to return, or at some point while on Mars?

Thanks for your help.

Reply

You saw the web page Smars1.htm, but did not continue to Smars2 and Smars3. Had you done so, you might have noted a serious problem, making unlikely "a quick trip" of just one month. In order to return to Earth economically, a mission to Mars must wait more than one year (459 days, to be precise) before starting a return trip. Any other timing requires extra thrust, and considering that any mission to Mars already needs enormous rocket power, it seems likely mission planners will try hard to avoid this.

It happens because Earth and Mars have different orbits and velocities. Earth moves faster (30 km/s vs 24 km/s, see end of Smars2.htm), and finishes its smaller orbit in just one year, while Mars takes almost two, For best timing, a spaceship to Mars starts when Earth is slightly behind Mars (positions "1" in the top figure of Smars2.htm), and it gradually catches up with the planet. It loses speed as it overcomes the Sun's gravity, and needs an extra boost on arrival (positions 2) to match the velocity of Mars.

By then, however, Earth has overtaken Mars. A month later, it will be even further ahead! For a spaceship from Mars to catch up with Earth, a lot of extra velocity must be added, and by the time Earth is reached, it will have a huge excess velocity, since the Sun's pull speeds it up. Even the optimal return orbit has about 3 km/s excess velocity--here it will be much larger, and one still must add the extra velocity imparted near Earth by the Earth's pull. All this needs to be removed on re-entry (and if you miss Earth, there may be no second chance).

The best conditions for return exist when Mars and Earth are in the mirror image of the positions at arrival, which happens after 459 days. Halfway through that time Earth and Mars will be diametrically opposite, with the Sun in between. Not only will they be at the greatest separation, but they may find it hard to communicate, though communications via some relay station in orbit around the Sun may still be feasible

The orbit chosen for a mission returning a sample of Mars to Earth will probably be the most economical one. A sample can wait 459 days--not necessarily on the surface, it may also do so in Mars orbit. For human astronauts--unless our propulsion systems get vastly better--I suspect a long stay is needed, in a habitat prepared beforehand by robot missions. Those can dig shelters (against cold nights and solar eruptions), store energy and oxygen, perhaps prepare food-growing modules. In spite of official optimism, I doubt this will happen in the present century.

Sorry--facts of life in space!

246. Weight and Mass

Sir
I am an undergraduate student of applied physics. I got a problem: why earth attracts all the bodies towards itself with same speed, although we know that
F = GmM/ R²
If two bodies have masses m= 10 kg and m = 100 kg, then according to the above equation, the force by earth on 10 kg body must be small compared with the body of mass 100 kg because mass of earth (M) and distance (R) is same for both the bodies. Yet both bodies fall with same acceleration. Why?

Reply

You will find the same question asked in
http://www.phy6.org/stargaze/Smass.htm

and the answer is also given there. Your question was probably asked by Galileo, and the answer, found by Newton about 50 years later, is: all matter has two separate properties, weight and inertia.

Weight

Inertia

An iron ball weighing 100 kg is pushed downwards with a force 10 times larger than an iron ball of 10 kg. However, its inertia, resisting its fall, is also 10 times larger, and the result is that the big ball and the small one fall at the same rate.

However, if you try to roll the two balls on a horizontal road, a motion in which gravity is not involved (except in helping the friction of the road, which we ignore here), you will find it much harder to get the heavier ball moving. It has more inertia!

Physicists also call inertia "mass". Weight is always proportional to mass (by the law of gravity which you wrote down), so to measure the mass of an object, we often measure its weight. However, you cannot do so in the weightless environment of an orbiting space station: to see how it is done there, look in "Stargazers."

247. "The Moon Hoax"

I am looking for some ideas on how to best handle a problem at my son's high school in (named town), Pennsylvania. My son's history teacher showed the class a video on the Moon hoax and asked students to write a paper whether they believed men landed on the moon. I cannot imagine a more limited and biased method of presenting space education as part of a US history class!

My son says his teacher doesn't believe men landed on the Moon. Eric and I used to chase parts at our Air Museum for the old engineer who built Saturn V; that guy would have loved to strangle my son's teacher, not that it would solve the problem. I had Judy Resnik's help early in my career and as a boy [She was an astronaut who perished in the "Challenger" disaster].

Col. Aldrin and his friends at NASA would likely take a very dim view of this. Any other constructive ideas on how to get them pointed in the right direction?

Reply

The "Moon Hoax" story is a legend which no amount of evidence seems able to kill. I have been asked about it quite a few times, and you can find my answer on
http://www.phy6.org/stargaze/StarFAQ4.htm#q68
For a longer discussion (not that length makes a better case) see
http://www.badastronomy.com/bad/tv/foxapollo.html

Considering what you wrote, I wonder how reliable other lessons by that history teacher are. And by the way, the term "Moon Hoax" dates back to 1835--see
http://www.museumofhoaxes.com/moonhoax.html

Judy Resnick was an outstanding person, and her loss was a great blow. She studied at the nearby University of Maryland, and after the Challenger disaster, I listened to one of her teachers tell what she was like. Too bad she's gone.

248. Shuttle re-entry from space

How come when the space shuttle re-enters the atmosphere it does not enter more gradually to avoid heating up a lot? I have wondered about that for a while. Hope you can clear that up for me. Thanks.

Reply

The answer is, such gradual re-entry will not work.

Generally, a craft suspended above the ground (excluding balloons) stays up in one of two ways. Either it has enough velocity (looking at it one way) for the curvature of its fall to match the curvature of the round Earth. Or else, it gets lift from the air its motion encounters, the way airplanes do.

At orbital speed, at least 8 km/sec, the shuttle keeps its height the first way. However, once it enters the atmosphere and slows down, its fall no longer matches the curvature of the Earth, and instead it gets lower and lower. It could have entered more gradually if it could have used the atmosphere to keep its height, the way an airplane does. But at 15-20 times the speed of sound, wings create more resistance than lift, and anyway, presenting a wing edge-forward as an airplane does would concentrate too much heating and pressure on its front.

The shuttle starts re-entry with appreciable altitude and a lot of forward speed, so its fall will take some minutes. The trick is now to lose speed and kinetic energy safely in the time allowed by this fall, and it turns out possible to do so--just barely--without deceleration forces getting too high for the crew. To lose its energy, the shuttle turns its bottom (covered with heat resisting tiles) to face forward, creating a wide shock front in which most of the heating occurs, sparing the shuttle itself. Only at the last stage of its descend does the shuttle actually "fly." By that time, only a small fraction of its energy remains.

249. Energy levels: plus or minus?

The Energy Level Diagram in the NYS Physics Regents Reference Tables lists the ground state energy for Hydrogen as –13.60 eV and the one of infinity as 0.00 eV.

Other tables I have seen reverse these values. I would like to know the reasoning in each case!

Reply

The values –13.6 eV for the ground state is appropriate, though formally of course potential energy is defined within an arbitrary constant, added to ALL values. So if you were to add 13.6 eV everywhere, the ground state energy would be 0, but ay infinity the energy is +13.6 eV, with plus sign.

The situation is similar with gravitational energy--say, planets around the Sun, or satellites around Earth: all bound orbits have negative energy. If you add energy to (say) a satellite), it climbs to a higher orbit with a less negative (total) energy. If the total energy is zero, it is just escaping on a parabolic orbits, and any spacecraft with positive total energy is unbound.

The gravitational case is discussed in
http://www.phy6.org/stargaze/Skepl2nd.htm
The case of quantum energy levels is at
http://www.phy6.org/stargaze/Q3.htm

250. How can such small targets be accurately hit, so far away?

I would like to know how astronautical engineers and spaceflight planners are able to measure points in space that are so far from us, such as the edge of a planet or the moon, with the accuracy needed to calculate and aim the path of a rocket or probe. It seems that there is no device that could measure the exact coordinates of the far off object, then measure precisely that object's edge, determine from them the precise time and angle of the flight path for orbital deceleration, and then steer a rocket exactly to its target!

Reply

I am neither astronautical engineer nor spaceflight planner, so my answer must be hedged with "that is what I believe."

We can determine positions of celestial objects fairly accurately by telescope, A fix within of one second of arc (1") means an error of 0.78 km at a distance of a million kilometers--nearly 3 times the distance of the Moon.

But actual accuracy is even better. We have sent space probes to (say ) Jupiter, and from the delay of radio signals can judge distances quite accurately. The gravity of Jupiter itself is determined by the well-known periods of its satellites. The gravity of the Sun is known (for instance) from the length of the year. The size of planets and the Moon was determined from occultations of stars, but now we have data from probes which have encountered them, took images and even orbited them, all giving more accurate information.

Another helpful factor is that in calculating interplanetary trajectories, we generally only need the center of the target planet (for instance), whose distance from the Sun is determined by the orbital period. When the probe gets close and more information is needed about its relation to the edge of the planet, often additional "mid course firings" of small rockets are used. Horizon sensors or on-board cameras are also useful when the spacecraft is close to the planet and ready to descend.

251. A teacher asks about compiling lesson plans

Hello Mr. Stern, I'm in the College of Education at [name of institution] University. I am studying to be an elementary school teacher and am trying to learn how to write better lesson plans in science. I am writing lesson plans for the fifth grade level at present. The extent to which I might be able to use your lesson plans is unknown to me at this point. For the most part, my search is for information that will allow me to backwards map the State Standards for my methods courses.

Reply

I am not a teacher but a retired physicist, so my advice is not based on experience. But I would say--above all, use common sense! You may start drafting lesson plans by

Defining what you want to achieve through the lesson,
Identifying your source material--text, chapters... maybe more, even web sites, remembering not every kid has a web connection (though public libraries do)
Identifying what examples, stories, diagrams etc. you will use to get your ideas across,
Identify potential stumbling blocks. Where may students become confused, and how do you avoid it? One way is to present an idea (where that is possible) in several ways, or using several different example applications.
Decide how you will make it interesting
Decide how you will test the students' knowledge, and perhaps also:
Identify what is optional and may be omitted, also what more advanced parts are optional but may be given as a special challenge to students who feel the class is too slow.

I do not particularly like state standards--they tend to stress memorizing "facts" and not to learn the reasoning behind them--e.g. the "fact" the Earth orbits the Sun instead of vice versa, rather than the reasoning why we believe it is so (which is actually more interesting). I once wrote a draft standards list of my own and it is on the web:

http://www.phy6.org/outreach/edu/standard.htm

Apart from drafting your lesson plans, go through mine on the web, and maybe others, and then discuss all these with some experienced teachers whom you respect--teachers in K-12, not university people. They should be able to guide you, and especially, tell you if the level is right for 5th grade.

Good luck

Response

Hello again Dr. Stern,

Thank you very much for caring about the success of my endeavors as well as the children I plan to teach. While you claim not to be a teacher, I beg to differ as there are numerous ways to teach and I have learned from both your web site and your personal advice. Your message is so detailed and on the mark that I have to wonder if teachers and physicists ever truly retire.

I too am not particularly a great fan of state standards. It seems to me that if all children leave school with the same answers to the same questions in science, we do little more than march in place. Being a teacher seems to be the easiest occupation in the world, unless of course you are trying to teach children to learn. I take your advice to heart and will put it to thoughtful use.

Your web site is most interesting and useful. Thanks again for the advice and good wishes.

Good luck and good life,

252. Why the Moon has its phases

Going through your site is very interesting and informative. I'll be grateful if you help me know how the moon starts from a crescent and becomes full moon, and then back from full moon to crescent?

Reply

The Moon is ball-shaped, and the Sun shines on it. At any time, half of it is in the sunshine and is bright, and half is in the shadow and dark.

As the Moon goes around the Earth, the amount of the bright side which we see changes, and that determines the "shape" of the Moon. When the Moon is in a direction close to that of the Sun, we see most of the dark side, and only a small crescent that is bright. (Actually, the "dark" part of the Moon is not completely dark at that time. An observer on the Moon at that time will see the Earth almost completely lit-up by sunlight, and the bright "earthshine" brightens up the dark side of the Moon a little.)

When the direction of the Moon and the Sun are at right angles (see picture in
http://www.phy6.org/stargaze/Sarist.htm)
we see half the bright side and half the dark one and get a half-Moon.

And when the Sun and Moon are in opposite directions, we see only the sunlit part and get a "Full Moon." Except, of course, if they are EXACTLY opposite--then the shadow of the Earth falls on the Moon and we get an eclipse of the Moon, a lunar eclipse. It lasts a few hours, until the Moon moves out of the shadow.

253. How can a spacecraft self-rotate?

Mark here, wondering:

You say

An astronaut floating in a space suit cannot shift his position without involving something else, e. g. pushing against his spacecraft. The center of gravity--or "center of mass"--is a fixed point, which cannot be moved without outside help (turning around it, however, is possible).
(http://www-istp.gsfc.nasa.gov/stargaze/Srocket.htm)

Question: How is turning, which I assume is the same as rotating, possible? Thanks!

I've been puzzling this in my head for a year now. Could an astronaut turn around in space, does a gymnast or a diver need a rotational push from the floor or diving board to turn or can they just use the 3 dimensions alternately to create a turn, etc.?

Reply

Dear Mark

It is easiest to solve this problem using a machine as an example. Suppose we have a telescope floating in space, and attached to it are 2 flywheels, capable of rotating around 2 perpendicular axes (fixed relative to the satellite), which we will label (x, y,). The telescope itself points in the third perpendicular direction, labeled (z).

You can visualize this by imagining a sphere with lines of longitude and latitude around the satellite, the polar axis being the y axis, the telescope pointing somewhere in the equator and the x axis perpendicular to it, also in the equator.

Suppose we want to point the telescope in the direction of some latitude and longitude. Spin up the y-wheel: because angular momentum is conserved, the whole telescope will rotate in the opposite direction, and the longitude of the telescope will change. Stop the flywheel when the telescope points at the right longitude: the satellite's rotation then stops, too. Now start up the x-wheel, and the latitude at which the telescope points will change. Stop when you reach the right latitude, and the rotation will stop again.

It is often said that if you drop a cat from a moderate height (not high enough to hurt it), the cat always lands on its feet, even if you drop it holding it upside-down. Films exist that show this indeed is true. The cat has no flywheels, but it can turn the front of its body relative to the rear, and the effect is the same (though it happens so fast that is it hard to analyze).

I guess platform divers and gymnasts have learned the same skill.

254. Stability during a rocket launch

How can a rocket shoot straight into the air without wobbling and toppling over after it becomes airborne?

I don't see any engines firing to the side of the rocket to maintain its course.

Reply

You definitely need a control. Dr. Robert Goddard's 1926 rocket did not have any, since he erroneously assumed placing the rocket engine in front would produce stability. It does not. As a result, the rocket started vertically but soon tipped over and continued horizontally, then crashed.

I am aware of three active control methods used in the past to maintainin an alignment ("attitude") monitored by gyroscopes.

In some rockets, small auxiliary rockets provide thrust to the side, firing downwards but at an angle. Early Atlas rockets had such stabilization. Check some launch pictures--occasionally, they show those small rockets firing!
The main rocket motor can sometimes be swiveled in different directions (around two perpendicular axes, using "gimbals"). I am not sure, but think the shuttle can do so. I vaguely recall pictures of the rocket engine being tilted this way and that before launch, as a final test.
Carbon vanes in the exhaust of the rocket can deflect its jet. I think the early "Redstone" used this, and Dr. Goddard's rockets also had deflecting vanes.

255. Boiling point of water in space

My name is Mike and I reside in the United Kingdom. This is a very short question: at what temperature would water boil at in Outer Space?

It all started with a conversation between a friend of mine and myself. We understood that at higher altitudes water boils at less than 100 degrees C, and I was interested to know that mountain climbers used to use this theory to determine how high they were.

I have searched the Internet for an answer, all currently to no avail, and I was hoping that you could shed some light for us.

Reply

Formally, ANY temperature can serve as "boiling point in space," but to understand how and why, you need know more about "vapor pressure" (vapour pressure in the UK). For instance, you may look up

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/vappre.html

Vapor pressure of a liquid (or even a solid, though its value may be vanishingly small) is a measure of how much its molecules "want" to convert to gas in the surrounding space. For instance, suppose you pump out all the air in a big bulb. The bulb is connected to a reservoir of water, and you let a cupful of water at room temperature, 27 deg C, be sucked into it. What will happen?

The water will boil, and as it does, part will be converted to water vapor. The process will stop when the pressure of the vapor reaches the value of the vapor pressure of liquid water at 27 degrees. At this point, an equilibrium exists--the rate of evaporating molecules is exactly balanced by the rate at which others consense out of the vapor.

Let's say that pressure equals P atmospheres (you can find tables of P against temperature in handbooks). If instead the bulb already contains air at pressure P, the water will be just at its boiling point. Heat it to 40 deg and it will boil, until the steam released adds enough pressure to P for the sum to equal the vapor pressure of water at 40 deg. At 40 degrees the vapor pressure of water is larger than P, because vapor pressure of a liquid increases with temperature.

The rule is, then--water will boil at the temperature where the external pressure equals its vapor pressure. The vapor pressure of water at 100 deg. C is one atmosphere, so at sea level, it boils at 100 C. On a mountaintop where the external pressure is only 0.8 atmosphere, it boils at a lower temperature, the one where the vapor pressure of liquid water only equals 0.8 atmosphere.

And in space, where the external pressure is zero... water boils at any temperature at which it has some vapor pressure. Which in principle means, any temperature above absolute zero.

Suppose an astronaut vents water to the outside. It will always evaporate, but not immediately, because there is also energy to consider. Turning water from liquid to gas requires about 236 calories per gram, more than twice the heat it takes (on the ground) to warm it from near freezing to boiling. That's why sweating on a hot day cools your body: as the water evaporates, it needs extra energy, and by supplying it, your skin gets cooled.

If water is vented to space, part of it immediately boils away, but the rest is cooled to where it freezes, into a sort of instant snow. Ultimately the frozen part also evaporates, since no equilibrium can exist between water (in any form) and a vacuum, but that happens more slowly.

Comets are apparently mostly ice, so they are expected ultimately to evaporate. However, they are so far from the Sun, and therefore so cold, that I would not be surprised if (as long as they stay distant) "ultimately" turns out to be much longer than the age of the universe.

Short question, long answer. If you know any better way, please tell me; but at least, I hope you and your friends have now learned a little more physics, and also have a bit more appreciation for the complexity of nature.

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Last updated 10 May 2005 Edited 1 November 2016

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Items covered:

236. Gases that fill Balloons

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237. Asian tradition on the start of Winter

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238. Why our year starts on January 1

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239. Sticking a hand out a window...

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240. One year of continuous sunlight?

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241. Shielding out radio waves

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242. The way gravity changes with depth

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Continuation of the question

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243. The Sun's Axis

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244. "Gravity particles"?

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245. "Short stay on Mars"

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246. Weight and Mass

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247. "The Moon Hoax"

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248. Shuttle re-entry from space

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249. Energy levels: plus or minus?

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250. How can such small targets be accurately hit, so far away?

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251. A teacher asks about compiling lesson plans

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252. Why the Moon has its phases

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253. How can a spacecraft self-rotate?

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254. Stability during a rocket launch

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255. Boiling point of water in space

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