Get a Straight Answer

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For a complete list, including later questions not listed below, click here.
You may also link from here to a listing of questions arranged by topic.

Items covered:

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)

If you have a relevant question of your own, you can send it to
stargaze["at" symbol]phy6.org
Before you do, though, please read the instructions

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

5. 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.

6. 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!

7. 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.

8. 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.

9. 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

₁

₂

Afterwards we have

₁

₂

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

10. 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 Sunís 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.

11. 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.

12. 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

13. 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:

"This storm affects the earth when it is on the western half of the sun, not when it is dead center. This is because the solar wind follows a curved path between the sun and the earth not a straight-line path."
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.

14. 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

15. 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.

16. Mars and Venus

This is a query on Mars. The ferric oxide on the Martian surface contains a lot of oxygen and if heated sufficiently would yield free oxygen. Does the presence of this substance on Mars indicate that at one time there must have been a lot of atmospheric or dissolved (in water?) oxygen available on Mars?

These are my questions about Venus:

1. Why is it that Venus is depicted by Magellan photos as being Red/Orange in colour? If these are radar images how is the "..actual colour of sunlight that reaches the Venusian surface through the thick cloud layer .. " derived? The Magellan images of the surface of Venus are quite bright. Would it really be that bright under such heavy cloud cover?

2. What generates the high velocity (up to 400 km/hr) winds which move from east to west in Venus's upper atmosphere, given that the surface of Venus rotates at a leisurely 6.5 km/hr?

3. What conditions did astronomers expect to find on the surface of Venus before the first probes landed? Thank you for your effort in advance.

Reply

--- Oxygen is not a rare element, Mars, the Moon, etc. have a lot of oxygen, always combined with other elements into stony stuff. The element that is really important to life is hydrogen. The recent signs that the Moon may have hidden water were exciting not because it is hard to find oxygen on the Moon (it must be separated, of course) but because hydrogen is rare. I am not sure about hydrogen on Mars. Jupiter and other cold big planets of course have plenty, and so do their moons, which may be more accessible. Still, they are pretty far away.

--- The colors of radar maps on Venus are probably false colors. The color and intensity of sunlight on the surface should be known, because the Russian landers took photos. However, I don't know it.

--- What drives winds on Venus is not the rotation but the same cause for winds as on Earth--the heat of the Sun. Venus is closer to the Sun, so maybe its atmosphere is more agitated. See the sections in "Stargazers" on sunlight and on the way it creates the Earth's weather

What one observes, in any case, is the wind at the top of the clouds, which might be analogous to the jetstream above Earth, not a good measure of winds at the surface.

I don't know--you must find out by yourself. I know Venus was expected to be "as hot as hell"--but the features of the surface could not be guessed.

17. Where is the boundary between summer and winter?

This may seem like a silly question but is there a line on the earth where on one side is summer and the other winter? Sort of in the same way that you can go back and forth between two days at the International Date Line.

Reply

Yes, there exists such a line and it is called the equator, but the boundary between summer and winter is not as sharp as the one of (say) the international date line.

Right now it is fall here, but spring in Chile and Argentina, and 2-3 months from now it will be mid-winter here and mid-summer in those countries. At our latitudes, these seasons are well defined. However, a broad belt centered on the equator does not have well-defined summer or winter. The Sun's heat fluctuates somewhat as its noontime passage moves north and south. On the equator, for instance, it passes right overhead around the 21st of March or September, 23.5 degrees off to the north on June 21 and the same amount south of "overhead" (zenith) on December 21.

These small changes do not make much of a change in solar heating. The big difference is made by local climate patterns, e.g. seasonal rains like the monsoon. These countries do not have summer and winter the way we do: for instance, when my daughter visited Darwin, Australia, some years ago, she was told that loacally the year had only two seasons, "the wet" and "the dry. "

18. The Ozone Hole

I am a auto mechanic and I have one simple question for you. Scientists say there are 2 holes in the atmosphere, ironically they are around the north and south pole, and they blame these holes on chlorine monoxide or refrigerants i.e. fluorocarbons (CFC) escaping into the ozone. Wouldn't the more likely cause of the holes be the magnetic lines of flux? One more quick question: could there be a way of tapping into that magnetic field as an energy source?

Reply

There do indeed exist two "holes" in the Earth's magnetic field, around the MAGNETIC poles, whose magnetic field lines go very far from Earth and afford an easy connection to the solar wind and to interplanetary plasma phenomena. On those lines we do observe "polar rain", a drizzle of fairly energetic electrons (more energy than those of the ionosphere, less than those of the usual polar aurora) which seem to come from the Sun. Also, when solar activity floods interplanetary space with energetic ions and electrons, that is where they are most likely to come down to Earth.

However, the creation and destruction of the ozone layer does not involve the magnetic field. Instead, its factors are chemistry and sunlight, and the "ozone hole" is around the geographic pole, not the magnetic one.

The ozone layer is maintained as an equilibrium between creation of ozone by ultra-violet sunlight, and its destruction by various natural processes (this is not my field, and I do not know details). During polar winter, the polar cap is dark and ozone is not created, just destroyed (a bit further from the pole, with just a few hours of sunlight and the sun shining at a shallow angle, ozone creation is also reduced). The observation of an "ozone hole" in recent years suggest accelerated destruction, as predicted by Rowland and Molina to come from chlorine in man-made substances.

As for tapping electric currents from space, I don't think it will work, because (1) they are very spread-out, by our standards--how can you tap a current sheet 100-1000 miles wide?; and (2) between us and them lies the atmosphere, a very effective electric insulator--as is well known to power companies, which string their high-voltage cables through air without worrying about the power leaking away.

.

19. What keeps the Sun from blowing up?

Dear NASA, I have a question abour the sun. We all know that the sun is powered by thermonuclear fusion reactions,so why doesn't it explode like an H-bomb?

Reply

What keeps the Sun from blowing up? Gravity.

The Sun is not exactly like an H-bomb--the bomb has fuel that is easier to "burn, " while to Sun "burns" ordinary hydrogen--but there does exist a similarity. Why does a hydrogen bomb explode? Because an enormous amount of energy is released in a short instant and inside a small volume, heating the material to extreme temperature. The material expands forcibly, creating a powerful shock, and that is the explosion.

The Sun releases much more energy every second in its central regions, but those materials are under great pressure, from the weight of all the matter piled up on top of them--the thick outer layers of the Sun, pulled down by a gravity much stronger than anything on Earth. That pressure confines the extremely hot gas in the Sun's core. The heat gradually works its way to the surface, but the Sun does not blow up.

If somehow it could yield to the pressure and expand, the central core would cool down and the nuclear energy release would drop. Then the pressure would decrease again and gravity would reassert itself. Some stars do in fact oscillate, but we should be grateful that the Sun does not belong to that class.

See more in http://www.phy6.org/stargaze/Sun7enrg.htm

20. Those glorious Southern Skies!

Not long ago I visited Chile for the first time and observed the night sky there. At that latitude, the centre of the Milky Way passes overhead, where it makes a grand show...

Why is there such a difference between Southern and Northern Hemisphere? Is it because of the 23 1/2 degree tilt?

Reply

Why such a difference between the southern and northern hemisphere? Because from the point of view of the Earth, all stars are so distant that they appear as if they were attached to a tremendously large sphere, with us in the middle.

At night, standing on the ground, you only see HALF the sphere. If you stood at the NORTH pole, half the sphere would be all you ever saw, appearing to spin around the point right overhead, the zenith. Standing at the SOUTH pole, you would see the other half, spinning around the point overhead, which is on the OPPOSITE end of the sphere from the overhead point at the north pole.

If you lived on the equator, the two poles of the sky would be on opposite sides of the horizon, and as the sphere of the heavens rotated around them, you would in principle see ALL the stars, sooner or later. In practice, those close to the poles will be near the horizon and not easy to see.

Maryland, where I live, is somewhere between the north pole and the equator, so the stars near the north pole are easily seen, and we get to see some stars of the southern hemisphere as well, though not those near the southern pole of the heavens (like the Southern Cross and Alpha Centauri), and many southern stars are only seen here near the horizon.

Similarly, from Chile you won't see the Pole Star, the Big Dipper or Cassiopeia, but the bright stars near the southern pole more than make up for them, and yes, the brightest part of the Milky way is there, too. North of the equator, the best view of the Milky Way is in mid-summer.

The 23 1/2 degree tilt has to do with the way the Sun, Moon and planets appear to move--not with the apparent motion of the distant stars.

.

21. Should we fear big solar outbursts?

I have a question pertaining to your studies in the upcoming year, particularly surrounding the forecast Solar Maximum.

I have heard about the upcoming Solar Maximum starting soon (CNN.com article, Nov. 11, 1999). I have also heard (unofficially), that there could be a very large solar storm near the end of April.

Finally, it is relatively commonly known that there is going to be an unusual alignment of the planets in our solar system at the beginning of May, 2000. Has there been an in-depth study to determine effects of the combination of these phenomena, and the potential impacts on both our solar system, and our planet?

The real question; could this combination of phenomena:

1.) Promote a solar flare, or SME, significantly larger than previously experienced in recorded history?

I have heard of Super flares emitting from G Class Stars, and the theory describes large planets in a close orbit (Jan. 8th Article, Sun-like stars said to emit super flares, CNN). Now, I don't expect this size of phenomena to happen here, but with the unusual planetary alignment, I do believe that this could create larger effects than normal, like a significant Solar-Magnetic Ejection, especially with the excitation of the Solar Phenomena. I'm just curious as to how much larger.

2.) Disrupt the crust of our planet, creating a significant amount of tectonic activity, and if so, by how much?

Now, I know our planets are very far apart, but if the magnetic attractions are larger than normal, and these magnetic attractions promote significant SME activity, this could promote some strange tectonic happenstance, especially with the fragility of our planet and its crust.

3.) Potentially disrupt our magnetic field severely with the combination of solar magnetic and gravitational forces?

I am aware of changes in our earth's history of the magnetic poles, could this happen here with the combination of a large SME and gravitational forces?

No calculations, or in-depth study has occurred, but I have a hunch this should be looked at more closely, and by qualified people.