Get a Straight Answer

Thank you.

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Near the ground the atmospheric density drops to 1/2 every 5 kilometers (8 kilometers = 5 miles, very nearly), so at 10 km where jets fly, density is down to 1/4. This continues more or less up to 100 km (the halving distance varies a bit, with temperature), where collisions between molecules become relatively rare. Higher up the oxygen and nitrogen each decrease at its own rate.

The space shuttle flies at 300-400 kilometers, but even there enough air remains to seriously limit orbital lifetime. Also, enough of that air is ionized--electrons ripped off molecules by the Sun's extreme ultra violet, leaving behind positive "ions"--to reflect radio signals. Satellites orbit at 600-1000 km and up, and that, too, is where the first signs of the radiation belt can be observed, particularly off the Atlantic coast of Brazil, where the magnetic field is relatively weak.

Somewhere between here and there, you enter "space. "

(1) My students had a couple of questions that I thought were interesting. I told them I'd ask ya'll.

1. What would you weigh is you were at the exact center of the earth?
   
2. What would you weigh is you were 3 meters from the center of the earth?

Please include supporting evidence for your answers

(2)

I am confused about Newton's discussion of the force on a particle within a sphere in the Principia. In one place, he says that the force would be zero, since the attraction of all the particles in the sphere would cancel each other out.

Just a little further on, he says that the force would be directly proportional to the particle's distance from the center of the sphere.

Can you clarify these two seemingly conflicting statements?

Thank you for any light that you can shed.

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    Yours is an old question, first tackled by Newton, as the "Hollow Earth Paradox." If the Earth where a hollow sphere (inner and outer surfaces spherical) and someone dug a hole that reached the hollow interior, and then stepped into it--what would that person experience?

    Newton's answer--there would be no gravity inside the hollow. Any object thrown into it--say, a stone--would continue in a straight line with constant velocity (ignoring air resistance).

    Newton's argument was roughly as follows. Take an object at a point in space anywhere in the cavity and draw from it a double cone (like a teepee, extending to both sides). Each side of it will cut part of the sphere, and the gravity of the two parts will tend to pull the object in opposite directions (make a drawing and you will see).

    Newton showed that the pulls of both part cancel each other: one part may be closer, but then it will also be smaller. Since all directions can be covered by a collection of such cones, the total force is zero. Today we get this result much more quickly by the theory of the potential, but that takes three-dimensional calculus, which Newton did not have.

    Now: Imagine you are somehow in the middle of the solid Earth--by some magic, not crushed by the rocks, suffocated or incinerated. In your mind you can divide all matter on earth into two parts: a smaller sphere containing everything that is CLOSER than you to the Earth's center, and a hollow sphere containing everything that is MORE DISTANT.

    By Newton, the hollow sphere exerts no pull, while the interior sphere, like the Earth, pulls as if all its mass were concentrated in the middle (that's another thing easily shown from potential theory). If you are halfway to the center, and the density everywhere is the same (actually, matter gets compressed towards the middle) then only 1/8 of the Earth mass is pulling you, but at half the distance, the pull is 4 times stronger ("inverse squares law"), so the final result is 1/2 of the gravity on the surface. At 1/N times the radius, the pull of gravity is just 1/N the pull at the surface.

    As you get deeper and deeper, the inside sphere gets smaller and its pull is weaker, so gravity too weakens. At the center, it is zero. At 3 meters from the center, it is the pull of a 3-meter sphere of rock, experienced on its surface--the pull of a tiny asteroid.

    Please note--that is just the pull of gravity on YOU. The rocks above you are also all pulled down, all the way to the surface of the Earth, and their weight is likely to crush you before you get very far. There may perhaps exist a cave a mile deep, but if so, none is much deeper, because there is too much weight piled on top.

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    The cheap and simple way is to build a shelter--especially since the dangerous events are the ones of solar outbursts, which are rare and last a day at most. You can calculate the shielding, but 20-50 cm rock should do a pretty good job (remember gravity is weaker, too, they will weigh less than on Earth).

    Have you read Ben Bova's "Mars"? It's fanciful science fiction, but his physics seems OK. The Mars astronauts are hit by a solar outburst halfway to Mars and wait it out, huddled in a special shielded area of their spaceship.

    The other night, I was watching a program about the Apollo missions and how they might be a hoax staged by NASA. Many interesting points were raised and by apparently learned men. The point concerning me most was a claim that the craft used to travel to the moon, and the LEM that took them to its surface, would not have provided enough protection from the high levels of radiation in the VanAllen Belt and cosmic radiation traveling through the solar system. The man postulated it would take roughly three feet of lead to shield against such radiation. Can you tell me what material protection the crew members of the Apollo missions had or what measures were taken to protect them?

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    That program on TV created a flurry of inquiries--I think yours is 5th. ... The cosmic radiation in space might indeed take 3 feet of lead to reduce it significantly--which is about as much protection as the atmosphere gives us.

(By the way, a widespread misconception of the public is that lead makes a better shield than other materials. It does for x-rays, which is why dentists and doctors use lead aprons, etc. However, for high- energy particles of space, all substances are more or less equivalent, for the same amount of grams per square cm.)

Keep in mind, though , cosmic radiation is a very weak source--it carries about as much energy as starlight. So the number of rads one gets from it is not high.

I am a writer, so my knowledge is always far behind the "truths" I write. But I was reading your article "Orbits", on the web, while researching a story I am writing, and liked the voice of your writing. So I thought I might ask you some questions. I hope you have the time.

I have searched extensively (without much luck) for information on the issues of reentry. I understand a layman's idea of how you get something into orbit and get it to stay there. I have also read about the demise of the Compton Observatory. Yet when I read about reentry, the issues of angle and speed are not clear.

If for instance a craft were out there going at 17500 mph and I just slowed it down to 60 mph--what would happen? Or if I just stopped it? Would it come crashing down like a lump and burn up along the way. Why does the shuttle enter the atmosphere at such speed and not just pop in with a large parachute? Perhaps it could use rockets to slow the gravity pull while it descends to an altitude where a parachute or simple gliding could be initiated without all that heat.

My problem in my story is that I need for a smallish satellite or vehicle (whatever) to be able to be descending into the atmosphere from orbit (in or out of control), fairly slowly (>400mph or thereabouts-- preferably, much less) between 60000 feet and the ground. Do I have a prayer - other than a Hollywood solution?

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You wrote:

"If for instance a craft were out there going at 17500 mph and I just slowed it down to 60 mph--what would happen? Or if I just stopped it?"

You can't do so, no more than you can with your car: if it is going at 60 mph you cannot just slow it down to 10 mph, or to a stop, unless you apply the brakes. If you do that, the brake pads rub against the drums or disks and convert the energy of the car into heat.

The speed of an orbiting object is enormous, and it too must be dissipated as heat: re-entering at a shallow angle lengthens the time spent in re-entry, so the heat is generated more gradually and can be radiated away without the spacecraft getting too hot. A small satellite can survive it, if suitably protected or if lucky enough: in the 1960s, USAF airplanes retrieved film packages from spy satellites, ejected in well-protected (but relatively small) capsules. When such satellites reach the denser atmosphere, they slow down--perhaps to 150 mph or less. The film capsules would then deploy a parachute, and as they floated down, a twin-boom transport airplane, with rear door open, would trail a line with a hook, snag them and pull them in. If they were to hit the ground (in principle--actually, rerieval was usually above water) they would of course get banged up somewhat, but their instruments etc. could perhaps survive.

Further message (shortened)

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The nearest thing I can recall is an airliner flying over Damascus, Syria, during one of the tense periods between Syria and Israel. The Syrians fired a missile at it, and by some miracle it stuck in a wing and did not explode. The airplane landed safely, but it was a close call.

I went through the site at: http://www.phy6.org/Education/wlagran.html and was just wondering if there is any information on L3? As I read the article I may have missed its mention. If not could you tell me if there is any information you could point me towards?

Thanks Very Kindly

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The Lagrangian point L3 is not mentioned because it is only of theoretical interest and has no practical application. If only Earth and Sun existed, L3 would be on the Earth-Sun line, but on the opposite side of the Sun. At that distance, a satellite at exactly 1 AU (Astronomical Unit = Sun-Earth distance) would experience a bit more gravity than the Earth, being pulled not just by the Sun but also by the Earth. Its orbital period would therefore be a tiny bit less than a year. L3 is at a slightly larger distance, where the greater distance from the Sun reduces the Sun's pull just enough to make the satellite's orbital period exactly one year, so that it would keep a fixed position relative to the Earth.

Practically, the feeble pull of the Earth at this distance is negligible compared to other effects, such as the attraction of Jupiter. Besides, it would be hard to communicate to such a spacecraft, since it would be permanently behind the Sun!

Look also at "From Stargazers to Starships", home page at http://www.phy6.org/stargaze/Sintro.htm. Sections 34 there calculate the Lagrangian points.

The reason I stumbled across your page is that I work on astronomical, science fiction and fantasy art and illustration and was looking for a formula that would allow me to calculate the distance to the horizon of any body once one knows the radius of that body. I've been working up a piece illustrating a scene taking place on Ceres and it would be nice to have the horizon more or less where it ought to be.

Any suggestions where I might find such a tool?

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There is a great story by Arthur Clarke, "Hide and Seek," in which a man in a spacesuit successfully evades a spaceship on a moon of Mars, where the horizon is quite close. Have you read it?

Hi - I am looking for the way of calculating orbits of two planets passing each other.

For example:

• Planet A - orbits every 30 years
• Planet B - orbits every 12 years

Trying to answer this for a grade school student - trying to assist.

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Your example does not sound too practical--it's like asking when does Jupiter overtake Neptune, even at the closest approach they are still very far apart. So let me reformulate

• Earth orbits the Sun in 1 year, the frequency is 1 rev/year.
• Mars orbits the Sun in 1.8822 years, same direction
          Its frequency is 1/1.8822 = 0.5313 rev/year

Draw the two orbits, one inside the other. Then assume you put the whole set-up on a turntable, rotating with a frequency –1, that is that is, 1 revolution per year but in the opposite sense. Forces in the rotating frame would be quite different, but we ignore that--the motions alone are what counts here.

Frequencies add up, rotation periods don't. The orbital frequency of Earth in the rotating frame is 1 – 1 = 0 revolutions/year, that is, it does not move. That of Mars is 0.5313 – 1 = –0.4687 rev per hear, in the opposite direction, which make sense--the Earth moves faster, Mars viewed from it moves backwards. Its period T' will be, in the frame where the Earth sits still

That's you answer. You can use 12 and 30 years, if you wish, the same way.

"...the most economical way of achieving that mission...may well be sending the spacecraft towards Jupiter....It would then make a tight loop around the planet, overtaking it in such a way that practically all its orbital velocity around the Sun is lost, and then fall toward the Sun. "

"Falling toward the sun"...I'm having difficulty visualizing this. Is it similar to the slingshot effect used to get Apollo 13 home? Would that be a grossly incorrect description? If so, could you offer a short, simple lay person description? I'm writing for a power industry internal newsletter and cannot get overly technical.

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"A little knowledge is a dangerous thing. " I would guess that you reached my Pelton wheel site using a search engine, and did not note it was unit #35a--the last unit, in fact--of a big educational site about spaceflight, astronomy and physics. You can reach the home page from the "back to index" button on the bottom, and after you do that, go to the preceding unit #35 where gravity-assist orbital maneuvers are discussed. Even though the spacecraft and moving planet (or moving moon) never touch, the result is similar to an elastic collision. Just as Sammy Sosa's home runs are a combined product of the speed imparted by the pitcher and that of the bat, so it was for Voyager 2 and Jupiter, for instance.

It works both ways: head-on collisions gain energy, overtaking ones give it up. Usually in space we want an extra boost, but in a turbine we want the water to lose as much as possible, because by the conservation of energy, what is lost from the water jet is gained by the water wheel. To hit the Sun with a spacecraft, it's not enough to escape the Earth's gravity: do that and you are still circling the Sun, like the Earth, with the same velocity, 30 km/sec. To give up as much is just as hard as gaining an equal amount--which is more than what is needed to escape the solar system altogether.

I am not sure change of velocity was a consideration with Apollo 13, as much as hitting the Earth edge on, so that the spacecraft would be braked by the high atmosphere. Hit any lower and it would hit the Earth (and burn up before doing so), hit any higher and you miss the Earth, go out to space, and long before you are back for another try you have run out of oxygen.

Could you please assist me with an estimation of the energy absorbed from the sun at the arctic (polar) regions? I would also like to know, if possible, the amount of energy (in Joules or calories) that a polar bear needs for a day. Thanks in anticipation.

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The polar bear is unaware
Of cold that cuts me through
Why not? He has a coat of fur!
I wish I had one too!                         Hilaire Belloc

Your request is one of the strangest I have ever received--do my correspondents feel that NASA knows everything? What is this information intended for?

About the polar bear: I will make a wild guess, 60,000 calories/day. But I really do not know. However a book exists which probably contains the answer, "The Fats of Life" by Caroline Pond. I saw its review in the "New Scientist" supplement of 6 May 1999, the review talks about the bear's metabolism but does not count calories.

The amount of sunlight at the pole depends on season. At the pole, in midwinter it is of course zero, in midsummer (clear sky) I would guess about 0.3-0.4 Kw/sec sq. meter, of which I suspect most is reflected by the ice.

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What a question! Some are, most are not--but the answer depends on what you plan to do with the information.

The Sun covers about half a degree of the sky. So rays that come from opposite edges of the Sun have directions which differ by half a degree and are not parallel.

Rays which reach your left and right eye from a distant star, on the other hand, are very close to parallel--they may meet somewhere at the star, at the same point (and then they converge) or separated points (and then they probably diverge), but the eye and even the the telescope cannot resolve such details.

Rays from the same point on the Sun are pretty much like those from a star. But again, "same point" is hard to pin down--even sunspots may be thousands of kilometers wide.

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The reason I needed confirmation or refutation is simple. My hubbie has been observing the shadow cast by the sun on our south facing balcony railing at a specific time of day. One of his statements showed that he felt the sun's rays radiate ! I disagreed both with his conclusions and his statement.

You're a star for answering so clearly and quickly and leaving both of us room to be right. Is psychology your second area of specialization ?

I was struck by the visual similarity between a Pelton bucket and the split "clamshell" type of thrust reverser on jet aircraft. By an impulse analysis I estimate that the rearward thrust is about 157% of the forward. Is this about right?

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I did not understand all you wrote, but please remember the zeroth law of thermodynamics, "there ain't such a thing as a free lunch. "

I don't believe a clamshell thrust reverser can generate more thrust forwards than backwards! There is a fine question of energy, which goes as the square of the velocity--but thrust is momentum transferred, and shooting a mass m with velocity v forward or backwards gives momentum + mv or - mv. In a Pelton wheel, the buckets move relative to the jet, but in a jet-engine clamshell they don't--so they can deflect, but not add any energy.

Sorry to be taking up your time, you are without a doubt very busy with answering many questions, but I am a 7th grader in View Ridge doing a report on "distance from the earth to the sun at 1 day from all 12 months." After searching for 3 hours I could not find this data and I thought maybe you could direct me on how and where to find this data or maybe you could send the data to me if it is easily accessible to you. Well thanks for your time and reading this message.

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I am not at all sure what your teacher is asking, or why--or what you are supposed to learn from it all. The mean distance of Earth from the Sun is the astronomical unit (AU), equal according to my handbook to 149,599,000 kilometers.

From the nautical almanac; the Sun's distance on day N of the year in AU is

A formula exists for finding N based on the way astronomers count days, but I suspect counting from January 1 is OK, because then R is smallest around January 3, which is indeed well known. Being in 7th grade, do you know what a cosine is? If not, look up in the mathematical section of "Stargazers".

The question is:

Suppose the total mass of the mission payload is M_p, and the total mass of the launch vehicle plus propellant is M_t. How does M_t scale with respect to M_p? If

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There does exist a certain flexibility in K, however. You do have to escape the Earth's gravity, at least to a circular parking orbit, and this sets a lower limit to K--it depends on your fuel, and the staging of the rocket, but it is going to be around 20 or more, usually more.

To fly from there to Mars, all sorts of tricks can be employed--ion engines, solar sails, etc.--and the duration of the trip is also a variable. Finally, the spaceship is to stop at Mars, not fly past it, and that requires either extra thrust or some delicate aerobraking. I think the article goes into all this, but the bottom line is--it isn't easy, and would take both ingenuity and luck to be pulled off successfully.

    I found your web page course "From Stargazers to Starships" and thought you might know what I am looking for specifically.

Thank you for any suggestions.

Dan

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    Tycho Brahe had no telescope, you know, but he built some very accurate quadrants--instruments resembling giant protractors, with sights to line up with stars. With these he tracked the positions of stars very accurately, taking into account factors such as refraction of light in the atmosphere. Kepler's laws were essentially based on the observations of Mars alone. A 4-inch telescope can be aimed much more accurately (the eye has a resolution around a minute of arc, at most), but this does not help much in precise tracking, unless you can determine with equal precision the direction in which it is pointing. That probably needs a strong foundation and accurate and strong mounting.

    That, however, need not be not your problem. You can always "fake your data" and take positions of Mars from the ephemeris table of the astronomical (or nautical?) almanac. Amateur astronomers in your area will have it, also university libraries--you might even find it on the web, I have not looked.

The question is, however, what then? To check the position of Mars in the sky, you need track both Earth and Mars. I don't know if motion of the Earth in a circle around the Sun (the idea of Copernicus) is good enough, but you can try it for a start. In fact, try the same for Mars, too, using the proper orbital period for each: you will get retrograde motion, though positions will not agree exactly.

    To check the uneven motion of Mars is harder. You must draw its ellipse (not accurately--only the calculation needs to be accurate) and mark on it a large number of points, say 36. Using the equation of the ellipse (and you will also need the eccentricity of that ellipse-something Kepler had to find out the hard way), calculate for each point its distance from the Sun (a computer will help). Also calculate the area covered by each 10-degree pie slice, approximating each slice by a triangle, and then from the law of areas you get the time needed to cover that angle, in some arbitrary units. Those units can be found by noting that all the times must add up to one orbital period. Of course, you must also know in which direction apogee is, the greatest distance from the Sun.

    If you now mark 72 positions of the Earth on its orbital circle over 2 years (about one Mars year), each will have its date. For the same date, you can estimate the position of Mars from your 36 points, each of which also has a date, and get the direction to Mars. Oh well--you may need 3-dimensional geometry, because the positions in the Almanac are probably not in ecliptic coordinates (Mars and Earth both move near the plane of the ecliptic) but in declination and right ascension. You can see it will be a big job.

    I do not know how Kepler's calculated all this (it took him years). The position of Mars in the sky (as you can see above) involves BOTH Kepler's first and second laws. My GUESS (just a guess, I may be wrong) is that he assumed the Earth moved in a circle, and that Mars moved in a circle too (he knew the periods of both from long-term observations). He then assumed (like Copernicus) that both moved at constant rates around those circles, and found (as you might--see above) that the positions did not fit. He then MAY have assumed that the position of the Sun was some distance from the center of the circle of Mars, and found in that case that the law of areas could explain most irregularities. FINALLY, he had to give some logical reason for the displacement of the Sun from the center of the orbit, and the most elegant idea he could find was that the orbit was an ellipse and the Sun was at one focus.

    I vaguely recall that the book "The Sleepwalkers" by Arthur Koestler describes those years in Kepler's life, and maybe you will find some clues there. Be warned, though--that is a work of literature by a writer not familiar with math. Maybe you can then write me what you found.

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    First of all--the moon looks pretty much the same from anywhere on Earth, at least its light-shadow phase-Richmond or Los Angeles, there is no big difference. The direction in which the light-dark division points may differ, because while the moon seen from Australia and here is pretty much the same, the "up" direction in which our heads point is not.

    To find the phases of the moon, you might consult an almanac. Or else, you might look up the Jewish date, using a web site such as