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About asteroids hitting Earth
A friend asked if I could find anything on an asteroid heading for earth & a laser that supposedly is in space that will eliminate the asteroid before it hits earth. Is there any such thing, or is he reading too many sci-fi books?

Reply
About asteroids heading for Earth: the best account I know is a section "The Shoemaker Comets" in the book "First Light" by Richard Preston. As for lasers capable of destroying one, they are (at least right now) pure sci-fi.
The above book makes an interesting point: it would be very hard to spot an asteroid heading for Earth. Astreroids are usually detected in photographs of the sky (via a telescope) by the fact they move across the line of sight, leaving a streak rather than a spot. If they move across the line of sight, they are not going to hit Earth; if they are heading straight for Earth, they leave no streak and attract no notice.
The swirling of water in a draining tub
I am a grandmother, soon to be 55 years old, who often gets into heated arguments in her maillists! My question, to avoid the argument by having facts, is: The direction of draining water in the tub, sink and toilet is said to be the opposite in Australia. Someone said the Coriolis Effect governs this, and it is a myth. Someone said the CE has nothing to do with it, and it is a myth. Then again someone said it is NOT a myth. I am on the fence with this one, as I cannot argue something I do not have one idea about!! Can you help me? Thanks in advance!

Reply
Your first respondent was right: the Coriolis effect governs it, and it is a myth. The Coriolis effect can govern the swirling of fluid flows, and where it does, the swirling is opposite in opposite hemispheres. However, it is only appreciable on a very large scale. Hurricanes obey it: tornadoes, which are much smallers, do not, and neither do kitchen sinks, which are much smaller still.
For details and explanations, look up on the world-wide web here.
Dispensing water at zero-g.
I am a student currently studying for a degree in engineering. As part of this degree we have been given the task of designing a water heater and dispenser for use in zero gravity. It has been suggested to use a bladder in a pressurized container heated using microwave or Radio Frequency technology. The heater must heat approximately 100 ml to 80 degrees celsius, and the entire system can use no more than 12 volts. How would you suggest a layout for this type of system might look? Any information that you could provide would be most appreciated.

Reply
Our research group is concerned with plasmas and magnetic fields in the rarefied medium between here and the Sun. We have no expertise at all in zero-g and space-station hardware.
This does not stop me from speculating about your request, of course. The key word is "dispensing": what do you mean by that? You cannot just have a tap and let out the hot water--it will form globs that drift away in zero-g and ultimately contaminate your circuitry or mess up your living quarters.
So you need three elements--a container where the water is heated, a tank from which the water is obtained and a container for the hot tea or whatever you want to make with the water.
The first and third may well be plastic bladders, whose volume can adjust. The reservoir would have an outer bladder with water only and and inner bladder filled with air, and as the astronaut with a squeeze-bulb pumps air into the inner bladder, water is squeezed out.
The heating vessel--you could use RF heating, but I suspect it will be somewhat heavy, will need stepping up the voltage from 12 volt and also will have to be shielded from radiating. In an environment where every superfluous gram costs a great deal, wouldn't a simple resistive heating element--with a thermostat, of course--be simpler? It could be in a cylindrical container with a spring loaded piston which is initially at the bottom. When refilling it from the reservoir the water enters from below and pushes the piston up, against a ratchet, and when the astronaut wants a drink, he or she releases the ratchet and the spring loaded piston pushes the water out, into the third container. The trick is to never mix any air with the water--once they are together, they are hard to separate.
This opinion comes to you with no warranty by NASA or anyone. Have fun! Now, let me go back to serious work...
Robert Goddard and World War II.
I just came upon your website "Stargazers to Starships". Great site! I am researching a school project on Goddard and found the information here useful. I have a few questions, perhaps you can help me.
- When the U.S. was spending money on the a-bomb during WWII, do you think this prevented the government from providing Goddard with money to research rockets for the military?
- The German's developed the V-2 rocket, and Goddard believed they copied his design. I read that Germany sent spies to observe Goddard. Did Goddard know he was being watched?
Reply
To answer your questions:
- During WW-II, the government also spent money on rockets. The A-bomb project did not cut into this. Goddard was part of that effort, but the biggest rocketry effort was probably at Caltech, with Theodore Von Karman and Frank Malina. See the section in "Stargazers" on the evolution of the rocket.
  Part of the problem was that Goddard preferred to work alone, while the Caltech people brought in bright students and had much better engineering support.
- I never heard about the Germans spying on Goddard, and it seems very unlikely. They too had much better engineering support and took Goddard's ideas--DeLaval nozzle, liquid fuel, using the fuel to cool the engine, steering vanes in the exhaust etc.--and developed them beyond what Goddard himself was able to do.
  A similar thing happened in WW-I. The Wright brothers invented the airplane in 1903, but the Europeans took their work and expanded it greatly, so that the German, British French and even Russian airplanes in that war were far superior to the ones America produced. After America entered the war, its pilots all flew British and French machines.

Asymmetry of the Moon's orbit.
Subj: Moon's perigee/apogee
I have just read your article in "Stargazers", but what I am trying to dertermine is the observed variations in the time between events of perigee and apogee. For example it may be 14 days from perigee to apogee and say 12 days from apogee to perigee. Then at a further time the periods can be reversed. I am seeking an explanation of this dynamic variation.

Reply
I looked up the ephemeris tables of the Moon, and you are right: counting only the times between minimum and maximum distance from Earth, those distances ARE variable, more than one would expect for, say, an Earth satellite in a long elliptical orbit.
All I can give you now is a guess. The motion of the Moon is really a 3-body process, influenced by the Sun as well, with further perturbations perhaps due to Jupiter etc. The orbit is close to a circle, which means that a pull of a few 1000 km this way or that can shift the time of largest and smallest distance by a great amount, in contrast to what it would do to a high-eccentricity orbit.
The literature comments "The orbit of the Moon is complicated" and I think your question illustrates that complexity. If you look at page D-46 of the US Astronomical Almanac, for instance, you will see that even the "low precision formulae" for lunar motion are alarmingly long, and better approximations (found for instance in "Astronomical Algorithms" by Jean Meeus) are even longer.
So the bottom line (as they say) is that Kepler's laws still hold, but actual motions may be complicated by additional factors.
Measuring distance from the Sun.
I hope this isn't too dumb a question, but when a planet's distance from the sun is given, should that be assumed to be from the center of the sun to the center of the planet, or is it a measure of the surface of the sun to the surface of the planet?
-- and if it's surface to surface, then what is considered the 'surface' of a gas giant?

Reply
Your question isn't dumb, and it has a simple answer: from the center of the Sun. A spherical mass--Sun, Earth, red giant or whatever-- pulls objects outside it with the same force as it would, if all its mass were concentrated in its middle. As far as gravity is concerned, the position of the surface makes no difference.
By the way--the Earth does not orbit the center of the Sun. If the solar system contained only it and the Sun, the two would orbit their common center of gravity. Of course, the Sun being much more massive, that point is very close to the center of the Sun.
With more planets, the system orbits around the common center of gravity, which I suspect is close to the center of gravity of its heavyweights--Sun, Jupiter and Saturn. Viewed from some other solar system, far away, this would make the Sun's position wobble a bit, in response to the motions of the planets. In recent years, astronomers have observed such subtle wobbles in the motions of quite a few nearby stars, and concluded that like the Sun, they had planets, too--big planets, like Jupiter. It is still too hard to detect the effects of lightweights such as Earth, but progress is being made.
Keep up your interest!
Who owns the Moon?
Dear Gentlemen,
If you be so kind as to reply, please tell me, is it true that the Moon has a formal proprietor and who is this man?
Thank you in advance for your kindness.
Reply
I do not know who told you differently, but the moon belongs to all of us together, even you, even I. When Neil Armstrong stepped onto the moon he said "We came in peace in the name of all of mankind" and that still holds true.
Acceleration of a Rocket
I've looked your site and have taken some information but I need more for my project. In my project I want to search on the G force on the rockets at launch.

Reply
I really do not know. The g-forces on a rocket vary with the design. Manned rockets stay under about 5g, unmanned scientific satellites may be launched at up to 10-12g, small sounding rockets with strongly built instruments sometimes reach 30g, and missiles can also accelerate very rapidly. The greatest acceleration is usually not at launch but just before burn-out, because the thrust of the motor changes little (or not at all), while the mass goes down as fuel is burned.
Rebounding Ping-Pong Balls (re. section #35)
Can you send or suggest any more references to support the 20 miles ping pong ball in and the 60 miles back out? I am having trouble with my colleagues who say it should be 40 miles back.

Reply
I hope you have a nice bet riding on this matter, because in that case you win. The correct velocity is indeed 60 mph. The way I gave it in "Stargazers" was meant to make it intuitively easy, but a rigorous calculation gives the same result.
In what follows we agree that velocities from right to left are positive, from left to right are negative.

Initially you have

A ball of mass M₁ moving with velocity – V₁, against...
A paddle of mass M₂, moving with velocity V₂

Afterwards we have

A ball of mass M₁ moving at velocity +W₁
A paddle of mass M₂ moving with velocity W₂

We assume the paddle is much more massive--M₂ >> M₁ (actually, it is mostly the mass of the hand behind the paddle), so that V₂ and W₂ are almost the same (=the impact does not slow the paddle by any great amount).

Conservation of momentum:

M₂V₂ – M₁V₁ = M₂W₂ + M₁W₁ (1)

Conservation of energy (we assume the encounter is perfectly elastic-- approximate for the ping-pong ball, very well observed by gravity-assist maneuvers of spacecraft around planets):

M₂V₂²/2 + M₁V₁²/2 = M₂W₂²/2 + M₁W₁²/2

multiply by 2:

M₂V₂² + M₁V₁² = M₂W₂² + M₁W₁² (2)

In both numbered equations we collect all M₂ terms on the left and all M₁ terms on the right:

M₂(V₂ - W₂) = M₁(W₁ + V₁) (3)

M₂[V₂² – W₂²] = M₁[W₁² – V₁²] (4)

By a well known factoring identity, for any two numbers A and B

A² – B² = (A + B)(A – B)

so (4) becomes

M₂(V₂ – W₂)(V₂ + W₂) = M₁(W₁ – V₁)(W₁ + V₁) (5)

If we divide equals by equals, what remains is still a valid equality. So let the left side of (5) be divided by the left of (3), the the right side of (5) by the right of (3):

V₂ + W₂ = W₁ – V₁ (6)

Add V₁ to both sides

V₁ + V₂ + W₂ = W₁

V₁, V₂ and W₂ are each 20 mph. Therefore, the rebound velocity W₁ equals 60 mph. QED

Rebounding ping-pong balls and gravity assist
Hello!
I've enjoyed browsing your fine web site, From Stargazers to Starships, and I figured I would take a moment to let you know that. I was particularly intrigued by the chapter "Project HARP and the Martlet. " There is one possible error I found in the site. In Sections 35 and 35a, on planetary swing-bys and the so-called "slingshot effect, " you state that the maximum velocity increase is imparted to a spacecraft when it approaches a planet head-on, or retrograde to its orbit. My reading indicates that the opposite is true. My sources are the following web sites:
JPL's Basics of Space Flight
http://www.jpl.nasa.gov/basics/bsf4-1.htm#gravity
Scientific American
http://www.sciam.com/askexpert/astronomy/astronomy10.html
I expect the discrepancy stems from the fact that the model used in From Stargazers to Starships is based on the ping-pong paddle example. The key difference is that the force exerted by a ping-pong paddle on a ball is repulsive, whereas gravity is attractive. Thus the numbers are the same but the sign is reversed.
By the way, I did find the analogy to the Pelton turbine very interesting. Thank you again for a very informative web site!

Reply
I believe that the ping-pong analogy is still valid, because it can be reduced to simple arguments of the conservation of momentum and energy, which should hold equally in a planetary-assist maneuver. Some other correspondent questioned this result, and as a result, you can find that calculation in item #9 of the question-and-answer section of "Stargazers," linked at the end of the home page [the item preceding this one].
What seems to confuse the issue is the following. A spacecraft would get its biggest boost if it approached head-on, made a hairpin turn around the rear of the moving planet and returned along a path 180 degrees from its first approach (that would be the ping-pong analogy). Viewing the encounter from far north, if we put the moving planet at the center of a clock dial with the Sun�s direction at 12 o'clock, we would see the planet moving towards 3 o'clock, so our satellite has to approach from that direction and return to it again.
When the Voyager and Pioneer spacecraft approached Jupiter and Saturn, however, they were coming from the Earth, which is roughly in the same direction as the Sun; in any case, their initial orbital velocity, which was essentially that of the Earth, which moves in the same direction as other planets', did not allow a head-on approach. Instead, they entered around 12 o'clock on the dial. They still rounded the night side and exited around 3 o'clock, which gave them an apprecible boost, though perhaps not the biggest one possible.
I have some old issues of "Science" on these events and in the one of the Pioneer 10 fly by, for instance (page 304, 25 January 1974), the satellite enters at 1 o'clock and leaves a bit after 3 o'clock. For the Voyager 1 fly-by of Saturn (p. 160, April 10, 1981), entrance is around 11:30 and exit around 4:30 on the same dial.
You are right, of course, in that the force on the ping-pong ball is repulsive while the planet's gravity attracts the spacecraft. However, the strongest attraction occurs when the spacecraft is at its closest approach, on the night side, and its direction then is along the velocity of the planet, the same direction as the force exerted in the ping-pong analogy.
With all this, I am grateful for your message. It again shows that at least some users go into the details of "Stargazers". Quite a few errors were caught only thanks to people like yourself who checked out such details.
Why don't we feel the Sun's gravity pull?
Dear Dr. Stern:
I have asked several teachers and many other people the following question but have not received any respectable answer:
The Earth is 93 million miles from the sun. Other planets, and even much denser planets I might add, are much further yet from the sun. The obviously strong gravitational attraction of the sun holds all of these planets in orbits around the sun. If gravity could be simply defined as a force that attracts matter, and the sun's gravitational pull is sufficient to hold the Earth in orbit, what keeps it from pulling me off the Earth? In fact, the gravitational pull of the sun is so weak at this distance that It can't even produce enough pull to raise a hair on my head. So how can it hold the Earth and several even denser planets (even further out) in orbit?
So--if the gravitational force of the sun is powerful enough to hold the Earth in orbit, then how could the Earth's gravitational force be powerful enough to hold me down, counter-acting the gravitational force of the sun? Please unconfuse me!

Reply
Dear student
Two effects are at work, each of which would be quite sufficient:
(1) The force of gravity goes down with distance squared. For example, since the Moon is about 60 times further from the center of the Earth that you or anyone who is standing on the surface, the pull of the Earth on each pound or kilogram of the Moon is 60 x 60 = 3600 times weaker than the pull on the same mass on the surface.
So: the Sun is indeed more massive, but also much more distant. As a result, its pull on each kilogram or pound at the Earth's distance is only about 0.06% of the Earth's pull near the surface.
(2) Being on the orbiting Earth, your body already responds to the Sun's gravity, by sharing the Earth's velocity of 30 km/s around the Sun. Therefore there is nothing left over from the Sun's pull to make you move any more.
In a similar way, an astronaut in orbit feels weightless, because the Earth's gravity is already fully employed in keeping up the orbital motion. The astronaut is not beyond the reach of Earth's gravity: if it were so, the spacecraft would fly away never to return, rather than stay in orbit. It is just that--like a stone in free fall--gravity is already doing to the astronaut all it can. It also does so on the spaceship the astronaut rides in, leaving no extra force pulling the astronaut down to the floor, or in any direction.
How hot are red, white and blue (etc.) stars?
Hi, My name is Donny and I have a question that I cannot seem to find an answer to.... How hot, exactly, is a blue star, a red star, a white star, and other color of star?

Reply
Your question has an answer, but you have also to learn a bit about what color is. Look at the following web site
http://www.phy6.org/stargaze/Sun4spec.htm
The stars for which statements about temperature are made are glowing dense bodies of gas, so for them the "black body spectrum" is relevant. In that spectrum, the curve of intensity against wavelength (color) typically rises to a peak and then drops.
The total area under the curve tells how bright the light is: the hotter the emitter, the higher the curve and the brighter the light. You know this from experience: a flashlight with a weak battery glows weakly in orange, a flashlight with a good battery glows bright yellow, and if you connect a 3-volt battery to a 1.5 volt lightbulb, you get a very bright, very white flash, and then darkness, because you have heated the wire inside the lightbulb so much that it melted, and you have just lost your lightbulb.
And in that sequence, you also see the color move along the rainbow: orange with a little heating (a feeble red when the battery is almost dead) yellow under normal operation, white when it's too hot. The color you see is where the peak is--and if it is blue, you see white, because all other colors are also emitted, and white light is what the eye then sees.
Stars are like that too. We can say the Sun's photosphere radiates pretty much like a black body at 5780 degrees absolute or about 5500 degrees centigrade, by the way its colors are distributed. The color tells how hot it is, and I think "blue" here means white-blue; such a star would be at about 10,000 degrees.
David
How does the solar wind move?
I am confused about the solar wind and don't want to mislead my students. On a web site about magnetic storms I read the following:
Is the solar wind influenced by the magnetic field of the sun so it has a curved path to the earth? Or is this too much of a simplification? What really happens?

Reply
Physics and astronomy get complicated at times. Will the interplanetary magnetic field curve the path of the solar wind? Without peeking at the observations, one can only say "it depends," and what it depends on is the ratio between the density of particle energy (density n times average of 0.5 mv^2) and the magnetic field energy density (B^2/2 mu-zero). This ratio is often called "beta" in plasma physics, and it's an important quantity in experiments aimed at confining a plasma for nuclear fusion. If beta is much less than 1, the magnetic field is the dominant factor and particles meekly follow its field lines, making containment easy. Practical fusion however requires a greater beta, and if beta exceeds 1, the plasma starts pushing the magnetic field around. The way it does so is by subtly segregating its charges, to create a charge density and hence an electric field, and electric fields can allow a plasma to move whichever way it wants.
Suppose the magnetic field is constant and equals B₀ in the z direction, and the plasma is moving along the x axis. Then an electric field E₀= -vB₀ in the y direction will allow it to do so, canceling the magnetic force on any electric charge q, equal to qvB₀ along -y. (It also works out with spiraling particles).
The same happens with the solar wind, where beta may be 5 or more. As a result, the solar wind moves radially out, though it gets buffetted a bit, and it's not clear by what.
Now what about the MAGNETIC field? There is a rule (for plasmas with high beta, satisfying the "MHD condition"), that "particles that initially share a field line, continue doing so indefinitely" (there exist some extra "fine print" conditions, but we ignore them here).
What follows below is the original answer sent to the questioner. Later this was converted to a graphical exercise, Section S-6a Interplanetary Magnetic Field Lines, linked to section S-6. You can either link there or continue below (or both), as you choose.
Take a sheet of paper, put on it a small circle--that is the Sun viewed from far north of it, or rather, it is a circle in the corona, some level above the Sun, where the solar wind begins. On this scale, let's say the solar wind moves one inch (1") per day (or if you wish, 2 cm). Draw from the center 6 or 7 radial rays 13.3 degrees apart. Mark as "P" the point where the first ray--the one furthest clockwise--cuts the circle. We look at 6 ions located at P, and therefore presumably on the same field line--let's number them 1, 2...6. We have advance information that 1 will be released into the solar wind today, 2, tomorrow, 3 the day after, and so on. Mark P with 1--that is where ion no. 1 is today.
Next day, P is on the second ray. Point 1 has moved 1" outward, radially, and Point 2 is at the base of the new ray, ready to go. Next day: Point 1 is now 2" out on the first ray, point 2 is 1" out on the 2nd, point 3 at the base of the 3rd, ready to move. And so on.
Five days later, 1 is 5" out on the first ray, 2 is 4" out on the 2nd 3 is 3" out on the 3rd, etc., and 6 is at the base of the 6th ray. However, all these points started on the same field line, so they are still strung out along one line. CONNECT THE DOTS marking the outermost ions on the 6th day and you have a spiral line of the interplanetary field: if the ions started on the same line, they must still be on one.
The solar wind in all this has moved radially. But now and then the sun releases bursts of high energy particles, say from flares. The energy of these particles may be high enough to endanger astronauts in interplanetary space--but their density is very low, so their beta is also low. THEY therefore are guided by the magnetic field lines (rather than deforming them to their own flow), and therefore they move spirally.
The solar wind takes about 5 days to cover 1 AU. Therefore, if the Earth is to receive particles guided by an interplanetary field line when it is on the first ray, the emission has to be at the base of ray 6--that is, near the western limb. The high-energy particles take only an hour or so to arrive, depending on their energy of course.
The shape of the orbit of Mars
Dear NASA,
I have read your articles about stargazers and I believed this is one of the most interesting subjects in astronomy. Here is a question which came up when I was reading 'Planetary evolution', could you please help, thanks. Mars moves in an elliptical orbit around the Sun, what is the relative distance of the Sun to this ellipse? Would it be at one end of the major axis of the ellipse?

Reply
The eccentricity of the Mars orbit is 0.09337, the semi-major axis of the orbit is A = 1.524 AU (1 AU is the mean Sun-Earth distance, about 150,000,000 km; AU stands for astronomical unit) and distances of perigee (closest approach) and apogee (most distant) are B = 1.381 AU and C = 1.666 AU (letters are just notation for here).
These are the numbers. What do they mean? I remember seeing long ago a German physics text from the 1920s drawing the orbit of Mars. One side of the line was a circle, one side was the orbit, and the varying thickness of the line showed the difference between the two. It was hard to see that difference!
Let us calculate the length and width of the ellipse. The length (through the two foci--the line on which the Sun is located) is 2A = B + C = 3.048 or 3.047 AU. The displacement of the center from either focus is D = (C-B)/2 = 0.1425 AU and the width is 2G where

G² = A² – D² G = 1.51732 AU 2G = 3.035 AU I do not think you or I would be able to distinguish an oval with dimensions (3.048, 3.035) from a true circle! The position of the Sun at one focus is however notably asymmetrical, about 10% of the distance from the center to the edge.
David
What if the Earth's axis were tilted 90° to the ecliptic?
I was recently looking at the webpage "Seasons of the Year" and I read about what would happen if the earth's axis were perpendicular to the ecliptic. I was just wondering if you could give me some insight on what would happen if the ecliptic was inclined at a 90-degree angle with respect to the celestial equator? Would this mean that earth's orbit would travel along this "new ecliptic" while the north and south poles are travelling along this "new ecliptic"?

Reply
The hypothetical case you describe does in fact exist: for some unknown reason, the spin axis of the planet Uranus is almost exactly in the ecliptic.
That means that at some time one pole (let me call it the north pole, even though that "north" direction is almost perpendicular to the northward direction from Earth) points at the Sun. Then the northern hemisphere is in constant light and the other one in constant darkness. Half an orbit later--42 years or so--the roles are reversed. And halfway between those times, the planetary rotation axis is perpendicular to the Sun's direction, making day and night alternate in a way similar to what the Earth experiences at equinox.
I leave it as an exercise to you to figure out whether Uranus ever receives sunlight the way Earth does at solstice.

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Author and Curator: Dr. David P. Stern
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Last updated 9-17-2004

Get a Straight Answer

Please note!

Items covered:

About asteroids hitting Earth

Reply

The swirling of water in a draining tub

Reply

Dispensing water at zero-g.

Reply

Robert Goddard and World War II.

Reply

Asymmetry of the Moon's orbit.

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Measuring distance from the Sun.

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Who owns the Moon?

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Acceleration of a Rocket

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Rebounding Ping-Pong Balls (re. section #35)

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Rebounding ping-pong balls and gravity assist

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Why don't we feel the Sun's gravity pull?

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How hot are red, white and blue (etc.) stars?

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How does the solar wind move?

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The shape of the orbit of Mars

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What if the Earth's axis were tilted 90° to the ecliptic?

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