Get a Straight Answer

    I am taking an Introduction to Astronomy course and the question of manned space flight came up as we studied atmosphere. Supposedly, upon re-entering the Earth's atmosphere, the re-entry vehicle, say the Command Module from a Saturn V rocket, had to do so within the arc of a precise angle or else it would bounce off of the atmosphere helplessly into space.

    The instructor for this class, a Ph.D. in Physics, seems to think that the Van Allen Belt is the reason for this need for precision, but I believe it has something to with the ionosphere. ... He also told the class that upon launching, all rockets must tend toward the path of least resistance in the Van Allen Belts, which he surmised is near the poles, where, as he put it, "the hole in the donut" is. This to me sounds fishy. As you suggest, plenty of missions, both manned and unmanned have passed through the Van Allen Belts unscathed (or at least at this writing I presume they have.)

    But I ask you, and I beg you for a reply, what is it exactly that causes so much trouble to reentering manned space vehicles to earth? And what part of the atmosphere is it that they must hit at precisely the correct angle in order to avoid bouncing off of it and deflecting back into space?

    I hope you will answer this so that I can at least set my mind and that of the class and the Instructor to rest.

Reply

    The main factor dictating the mode of reentry is the need to dissipate a large amount of kinetic energy, without burning up in the process. Orbital velocity is about 8 km/sec, whereas a typical rifle bullet has about 1 km/sec, so mass for mass, a reentering space vehicle has 64 times more kinetic energy. Dissipating it safely is not easy.

    It is done mostly by forming a shock wave in the rarefied high atmosphere, which heats the air passing through it, heating which extracts energy from the motion. Hot air still hits the vehicle, also the shock wave glows and radiates infra red and visible light, and these heat the vehicle, all reasons why a heat shield is needed--e.g. the tiles on the underside of the shuttle, which faces forward during reentry.

    In order not to overload the heat shield, the descent angle is shallow, so that the shock forms where the atmosphere is not too dense. As the shuttle slows down, it needs (I guess) some lift to keep its trajectory from getting too steep; if the lift were somehow to get too big, it could in principle bounce the shuttle back into space, which would be bad, since it then comes down again far from the planned area, and probably at a bad angle.

    Going through the "hole in the doughnut" is hardly ever done (if at all), because the result is an orbit steeply inclined to the equator and the ecliptic. Most orbits we want are close to those planes. Once you launch into a steeply inclined orbit, it costs far too much to change the inclination. The way manned missions pass the inner radiation belt (that's the intense one) is go through it quickly, it may take just 15-20 minutes to cross.

Subject: QUESTION FOR DR. STERN - HOW A BICYCLES WHEEL KEEP IT UPRIGHT

    Please explain how the spinning wheels of a bicycle help the rider keep the bike upright? My daughter (13 yrs old) and I have tried to research this but seem not to be able explain the complete relationship between centripetal force, angular velocity of the wheel and precession. Thanks in advance of your help.

Reply

    Your question was addressed in
            http://www.phy6.org/stargaze/Snewton3.htm

    I copy that section below:
===============================================
The Bicycle

    A more subtle example is afforded by the bicycle . It is well known that balancing a bicycle standing still is almost impossible, while on a rolling bike it is quite easy. Why?

    Different principles are at work in each case. Suppose you sit on a bike that stands still, and find it is leaning to the left. What do you do? The natural tendency is to lean to the right , to counterbalance the lean with your weight. But in moving the top of your body to the right, by Newton's 3rd law you are actually pushing the bike to lean more to the left. Maybe you should lean to the left and push the bike back? It might work for a fraction of a second, but now you are really out of balance. No way!

    On a rolling bike, balance is kept by a completely different mechanism. By slightly turning the handlebars right or left, you impart some of the rotation of the front wheel ("angular momentum") to rotate the bike around its long axis, the direction in which it rolls. That way the rider can counteract any tendency of the bike to topple to one side or the other, without getting into the vicious circle of action and reaction.
===============================================

You may note that when the bike is rolling slowly, you have to turn the handlebars by quite large angles to keep balance, but when its speed it up, little twists are sufficient. That is because at slow speed, the front wheel has less angular momentum. However, some turning is always essential. About 50 years ago I owned a British "Raleigh" bicycle which had a lock built into its handlebar pivot. When it was locked, the bike would still roll, but only absolutely straight ahead. Neither I nor any of my friends managed to ride it and keep it balanced when the handlebars were locked.

    The design of the front wheel "fork" is quite interesting, turning forward like the letter j. This is an "inverse caster" which makes the ride less stable--but allows a skilled rider nimble moves. If you ever look at a motorcycle, or at the bikes used in a circus on the high wire, they lack this "letter j" feature, the wheel axis is always in line with the handlebar shaft. That helps stability but reduces agility.

    I would like to know if the temperature on the surface of the moon while there is no light (under a dome which you would install on the moon for example) is the zero absolute (0 degres Kelvin)?

Reply

Probably not, though it does get mighty cold on the Moon during lunar night.

    It has to be the night, because during the day it is hard to avoid the Sun's heat. Suppose you build a dome and place your equipment under it, in the shade. The Sun will heat the dome, and the dome will warm up, but after a while it will reach equilibrium (like any object in sunlight)--as much heat is radiated out as infra-red as is received from the Sun as regular sunlight.

    Where is it radiated to? I would guess, about half goes up, half goes down. Thus the shade won't be cool: if the equipment under it cannot get rid of heat, it will gradually become as hot as the roof. Similarly, it can get pretty hot in a closed car standing in the sunlight--even in a closed van, without windows for light to enter!

Hello, We are trying to find out why longitude goes from 0 to 180 and from 180 back to 0, rather than 0 to 360 degrees. Could you possibly answer this question for us. Thank you.

Reply

I don't know. Maybe because in the age of exploration, ships set out from England and other European ports which had longitudes near zero, and moved either east or west. It also turns out that the jump from -180 to + 180 (the international date line) is conveniently set in the middle of a big ocean.

    I don't know, but let me guess. I think astronomers are to blame. When star charts were first made, all stars had to be assigned in some way, so astronomers divided the sky into areas, and naturally, they chose the areas to correspond to constellation. If you look at a star atlas, you will see the sky divided into rectangles and pie shapes, a bit like the map of the states in the western USA. This lets an astronomer state that a nova, or a comet, appeared in (say) Aries, or assign some distant galaxy to the constellation. That too is why, for instance, Section 10 on "Kepler and his Laws" mentions "Tycho's nova in Cassiopeia."

    Astronomers therefore use the word "constellation" to refer to an area in the sky. A new term was now needed for those conspicuous formations involving a few bright stars, and that was"asterism."

    I did not use "asterism" on my web site, though. Most users are not astronomers, and might not even have known Cassiopeia before reading about it. So I stick to common English usage and talk about the "constellation Cassiopeia." In general, given choice of a technical term and plain language, I stick to plain language, which is more user friendly. Amateurs with telescopes and star charts will know what I mean even if I'm not completely technical.

Dear Sir:

        Today is my birthday, and I heard that if I got up early and looked outside, that I would see the stars in the position of when I was born, I'm 48 today. Would you please send me a picture of these positions. I live in northwest suburbs of Chicago. I need an explanation too?

Reply

    I have no star charts available--that is not something I work with. You should really go back to whoever told you what you wrote. Keep in mind two things, however:

(1)         The positions of "proper" stars do not change. Any motion they have relative to us appears so slow (because of the distance) that the eye does not notice. The constellations we see are essentially the same as the ones observed by ancient Greeks. Over the year, the times they rise and set wander over 24 hours, and on your birthday they indeed occupy the same positions (more or less) as they did at the SAME TIME on the day you were born. On any birthday.

(2)         The planets move all the time (and it's the planets which astrologers track). However, none of their motion is commensurate with the motion of the Earth--that is, with the length of the year. After exactly 48 Earth-years, they will not occupy the same positions in the sky.

    Oh to be 48 again! (I'm 72). Congratulations!

    Hi! I am student teaching in a third grade classroom in Tennessee. We just began a unit on space last week. One of my curious little minds asked if the center of the Earth rotated as Earth rotated. I told him that I was unsure, but I would think it would. I've searched and cannot find anything regarding this aspect of Earth. Could you please answer my question and maybe point me to something to back this up for him and the rest of the class? Thank you so much.

Reply

    We now know that the Earth has a core of about half its radius, made mostly of iron  --  molten iron, though in its middle is a smaller solid "inner core." (We know all that by studying the distribution of earthquake waves, reaching us from earthquakes on the opposite side of the Earth. On their way they pass near or through the core, and studying their arrival tells about what they passed through).

    So, does the core rotate at the same rate as the outer part of the Earth-one turn per day-- or not?

    The two MUST rotate together. If there is an inner ball and an outer ball, somewhere there must be a boundary between the two, at which they touch each other.

    (In fact, they do not just touch, but press very hard. Any point on the boundary has above it the weight of 2000 miles or rock, pressing down!)

    If they do not rotate together, one must slide past the other, and they will rub together. There will be friction.

    The faster one will try to get the slower one to speed up.
    The slower one will get the faster one to slow down.
    And this goes on until they move together.

Are you with me, third grade?

Dear David,

    My name is Brendan. I really like to learn about things like space and geography. When I talk about the temperature of the sun, people always ask me, How do you know? So, how do you know the temperature of the sun?

Reply

You ask a very good question. First of all, it shows you think like a scientist. Most people--students in schools, too, even teachers--ask "what is the answer?" How many years is the age of the universe? How far from Earth is the Moon? What is the highest mountain on Earth? Answers like that can be read in books, but once you have them, you really haven't added to your understanding. The meaningful question is usually "How do we know"?

    The temperature of the Sun can be inferred from its color. All hot objects emit light, or radiations similar to light, and the hotter they are, the further their light is to the blue end of the rainbow sequence. A hot pot radiates in infra-red (your eye cannot see it, but your hand can sense it if held close). A red-hot piece of iron glows dark-red, a lightbulb glows yellow (and in a flashlight with a weak battery, orange--the wire in it is not as hot). The very hot star Sirius (brightest non-planet in the sky) emits light near the blue, but since it emits other colors too, to the eye it is bright white.

    The Sun is sort of yellow. Not just that: spectrographs can measure how much light is emitted in each color, and verify that its colors are distributed as you would expect in a hot object. (Light from a rarefied gas would be different, it gets its color not from heat but from the properties of the atoms that emit is; the colors of neon lights for instance--also fluorescent lamps). More than that, the distribution tells what the temperature is. I believe it is 5780 degrees above absolute zero, give or take 10 degrees.

    Glad you asked.

Now about the height of the tallest mountain... see
        http://www.phy6.org/stargaze/Strig1.htm
and the distance of the Moon:
        http://www.phy6.org/stargaze/Shipprc2.htm

        I'm from Mexico, and I'm studying mechanical engineering. I was reading the FAQs and I didn't find an answer suitable for my problem. I seek everywhere I can, and haven't found an equation describing gravity for my needs. I've found one that describes gravity depending on the latitude, but I need one depending on altitude. I know gravity has an accepted value of 9.81 m/s² at sea level, but as I get higher and higher in the atmosphere or as I take my car and reach my city 2000 meter above sea level, which is the gravity there?

Thanks.

Reply

If the Earth were an exact sphere, with density everywhere varying the same way with depth (a good approximation), gravity would only depend on the distance r from the Earth's center, and would decrease with distance like 1/r². That was first proposed by Newton (the story and calculation are in section #20 of "stargazers") and has been confirmed with great accuracy since then. The equatorial bulge of the Earth and variations of valleys and mountains modify this only slightly (and the effects of rotating with the Earth need always be added, too); see section 24a).

    Rising 2 kilometers means about 1/3000 of the Earth's radius r, so gravity at that altitude is weaker by about 1/1500. You will need an accurate instrument to observe the difference. A good pendulum clock should vary by about 1 minute/day from one at sea level at the same altitude (unless the attraction of the mountains reduces the difference, about which I am not sure.)

(Received March 31, 2004)

    Just visited a web site and a mail address of yours appeared. My question: There appears to be a partial eclipse of Venus when you use the backyard telescope. Is this a normal event, or is it connected to the transit-of-Venus expected in June?

Thanks

Reply

    The transit of Venus occurs when the planet comes between us and the Sun, when it appears to be a dark spot crawling across the Sun's disk. At that time, in a little over two months, Venus will present to us only its dark side. As it is now approaching that position, the side we now see is mostly dark and it appears as a crescent.

There are several aspects of the Big Bang theory that puzzle me greatly, none more than the consequences of expansion.

    As explained by current theory, we are not at the centre of the expansion. Since the whole universe is expanding, we would not remain at the origin. Therefore the universe should appear asymmetrical to us. i.e. there should be more of it in one direction than the other. As matter in the universe is emitting light, one side of the view from Earth should appear brighter than the other, all else being equal.

    If we are not at the centre of the universe then as we look further into space and hence back in time, when the universe was smaller, one part of the sky should appear older than all the rest. The further we look and the further back in time we go that area should be smaller still. If we could see right back to the big bang it would be the oldest point in the sky.

    Finally the universe is not expanding at the speed of light so the light emitted from the early universe should have simply rushed past us long ago and become undetectable. How do we see it now?

Reply

Cosmology is not my field, so I limit myself to just a few observations. Your view of the expanding universe has a basic problem, being rooted in concepts of the era before Einstein. Namely, you view as space being infinite and eternal, and regard the universe as a collection of matter and energy expanding in this space, starting from some initial instant which was the big bang, and from some initial location.

    Current thinking is different. The matter and energy of the universe are filling all the space available to them, and have always done so. That space, however, is itself expanding. The big bang is not an explosion in space, like that of a bomb, but an expansion of space itself.

    A common analogy is the surface of an expanding rubber balloon. The rubber is filling all the area available to it, but that area is itself growing, and as it grows, the thickness (in our case, the density of matter in the universe) is gradually decreasing. The way 3-dimensional space is expanding can be described mathematically, but it is not otherwise intuitively graspable, except by that analogy--or perhaps (not sure how well that works) as expansion inside a higher dimensionality.

    Look at it another way. If the universe started 13.5 billion years ago (a good guess), as we look into space, no place we see is more distant that 13.5 billion light years. The volume available is thus limited, though any point sees the same expansion, just as on the balloon.

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