Get a Straight Answer

    Unbalanced densities create an electric field. A quiet-time electric field does exist in the atmosphere, produced by distant thunderstorms, and the difference over an altitude of (say) 2 meters may be a few hundred volts. That however can only exist because air is a very good insulator. Your body isn't: an electric charge touching it is soon discharged to the ground.

    Your Web site is very useful! I'm 45 years old, with an interest in electricity in space, and still learning.

1.     In the section on The Solar Wind, you mention that "The plasma of the corona is so hot that the Sun's gravity cannot hold it down. Instead, the upper fringes flow away in all directions, in a constant stream of particles known as the solar wind."

    Wouldn't the intense gravitation field of the Sun cause the heavier positively charged particles to fall back to the Sun, resulting in the Solar Wind containing more electrons than protons? Or would any proton falling back, attract an electron along with it?

2.     I read on a number of Web sites that the Solar Wind accelerates away from the Sun. Is there an explanation? Eg. See

        http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1994kofu.symp..401T

Ian

Reply

    Your first question is quite perceptive. A similar question about the Earth's atmosphere was addressed by Pannekoek in 1922 and by Rosseland in 1924 ("Electrical state of a star", Monthly Notices of Roy. Astron. Soc.84, 720-728, 1924). Here is the idea.

    In an atmosphere consisting of a gas of molecular weight M, the density falls off exponentially, at a rate which depends on M, on the temperature of the gas and on the force of gravity. In the Earth's lower atmosphere, the density falls to one half every 5 kilometers or so (or else, by a factor e = 2.71828.. about every 8 kilometers, a distance known as scale height H). High temperatures and low M increase H and spread out the atmosphere, while strong gravity decreases H and makes the atmosphere more compact. Of course, the Earth's atmosphere is really a mixture of gases, but in the first 100 kilometers, enough collisions occur to create a single scale height, some sort of average among the components.

    Above 100 km collisions quickly become rare, so different gases with different M tend to separate, as has been observed. Pannekoek and Rosseland wondered--what about free electrons in the outer layers of a star (or for Earth, in the ionosphere)? Their mass is so small, that their scale height should be much larger. Their layer should extend to great height, while the heavy oxygen ions (the main positive component) should stay well below them.

    They concluded this was not possible, because even a very slight separation of positive and negative charges would create an "ambipolar" electric field, pulling electrons down and O+ ions up. The field would effectively "add weight" to the electrons and would "buoy up" the O+ ions, until effectively each species senses the same downward force, contributed by both gravity and electric forces. With both species having the same "weight", they also will have the same scale height, and the ionosphere would stay electrically neutral, as is observed.

    The corona of the Sun behaves the same way, although the ambipolar field is probably much stronger, because of the high temperature.
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    I am less certain about the second question--the solar wind is really not my field of expertise. In the Earth's lower atmosphere, temperature decreases with height, because heat from sunlight is mainly absorbed by the ground, at the bottom. It is then transported by air flows and by radiation, absorption and re-radiation (by "greenhouse gases") until it gets high enough to be radiated to space, not to return. That height defines the bottom of the stratosphere; other interesting effects occur higher up (e.g. absorption of UV by ozone), but we ignore them now. Because of the upwards flow of heat, the temperature is highest near the ground (where heat comes in) and decreases with height, up to the level where heat is given up.

    Solar physicists expected the same to hold for the Sun, also heated from below, and it is still a mystery, why so much heat is deposited in the corona, making it much hotter than the photosphere below it. Still, once you get to the corona, you would expect the temperature to gradually decrease with height, as one gets away from the source of heat and as the rising gas expands.

    I am living in South Africa. I have a 7year old son who has asked me a very interesting question, which I could not answer for him, with regards to astronomy. The question that he put forward to me was this: "Why when the sun is coming up, does it look so big on the horizon?".
        If you could please help me with an answer, it would really be appreciated.

Reply:

    Your son's question has been asked many times before, usually about the Moon, not the Sun (the Sun is too dazzling to look at, especially when high in the sky). No one is sure of the reason, but it might be an optical illusion--our brain makes us believe objects near the horizon are larger than when they are high in the sky.

    Show your son the web site "Why Does the Moon Look So Big on the Horizon?" by Kathy Wollard, at:
        http://www.word-detective.com/howcome/moonlookbig.html

    As you said any triangle can be formed by 2 right angle triangles.

    How would a person calculate the exact position along the base line for the perpend to the 3 rd point?
(I am programming EXCEL using x,y,z coordinates.)

Reply:

    Let's stick to 2 dimensions (x,y), and say the triangle has a baseline from (x₁,y₁) to (x₂,y₂) and the 3rd point is (x₃,y₃). I hope you meant "every triangle can be DIVIDED into 2 right angle triangles, because two random triangles, with all different sides, cannot be easily combined to one.

    Here is how you do it. Suppose first x₁ = x₂, so the baseline is parallel to the y-direction. Then the point where the triangles join has the same value of x (say x₁) and the line dividing the triangle is parallel to the x-axis, from (x₃, y₃) to (x₁, y₃).

    Otherwise, let the baseline have an equation y = Ax + B .
Then

Now the perpendicular line through (x₃,y₃) has an equation like

    But if the lines are perpendicular, the equations are related, and the relation is C = –1/A (don't ask me why--study coordinate geometry to find out). I assume A is NOT zero. If it is, that means y₁ = y₂, , the base of the triangle is parallel to the x-axis, and you follow the same steps as before, only with the roles of x and y interchanged. Otherwise the line is

and since you know that the 3rd point is on that line

you should be able to derive the value of D. Say now the point on the baseline where both triangles meet is (x₄, y₄). That point is on BOTH lines and therefore satisfies both equations:

    I looked for the Big Dipper so I could find the North Star and show my friends. But I could not see the Big Dipper.

    Is the Big Dipper ever visible from Viet Nam? It looks like Viet Nam is about a thousand miles north of the equator. Thanks for your help. I love your web site

Reply:

    Ho Chi Minh City (aka Saigon) is about 10 degrees north of the equator, so the pole star is about 10 degrees above the horizon. The stars of the Big Dipper circle the pole with radii of 30 to 40 degrees, so they are often below the horizon. (This holds even for much of the continental US, though not for Alaska.)

    Whether you see the constellation depends on the season and on the time of the night. The Big Dipper is on about the same right ascension (same spoke from the pole, if you will) as the constellation of Leo, which is prominent in the evening sky in March and April. At that time your friends should see it in all its glory. Maybe even in December-January, if they look at the sky before sunrise. October is not a good time, however! At such a time it is better to use Cassiopeia for finding the pole star, it is on the opposite side of the pole and should be prominently visible.

(continuing the exchange, in part)

    I urgently await your reply

Reply

    I don't know the answer, however, small holes in the sail should not affect its efficiency. In a sailing ship, the sail gathers the wind, which is a fairly dense gas. In a solar sail, everything happens on the atomic scale, and if a hole exists nearby, it will not divert the flow to itself.

    Solar x-rays are absorbed very quickly in the high levels of the atmosphere. Thus one would not expect any effects in the troposphere (the lowest level where weather is produced). On the other hand, the ionosphere would be very different, and much thinner. The highest layer, the F layer used in long distance communications, might not exist, or anyway its density will be too small to matter much. Other layers probably will suffer too. Radio amateurs may have a tough time.

    For two electromagnetic waves of different frequencies (and wavelengths) and amplitude traveling in the same direction, I have noticed they combine their amplitudes when they superimpose. However, I am also noticing an increase in frequency. Am I correct to think that the two frequencies also combine in some manner so that the resulting wave is the combined frequency of the two waves?

    Or am I perhaps misinterpreting observations in that the increased frequency likely has a different cause not necessarily due to the superimposing? I suppose the situation is similar to a radio wave interacting with a visible light wave.

Reply

    Two waves at SLIGHTLY different frequencies will exhibit beats--a modulation in overall intensity as they slowly get in phase and out of phase again. A freight railroad runs about 2 miles from my home, and I can clearly hear beats when it is pulled by multiple diesel locomotives whose engines run at slightly different speeds. Twin engine airplanes also sometimes sound beats. If your detector smoothes over the rapid variations, it extracts the beat frequency.

    Things get more complicated when wave trains are finite and changing, as in radio waves which carry signals, say music encoded as AM or FM. When such waves are analyzed by frequency (without regard to phase--to where valleys and peaks fall), they always cover a finite "bandwidth" of frequency, around the main one of the carrier signals. Sometimes your radio will receive a mix of two stations, a sign that their bandwidths overlap.

The question continues:

    I appreciated your response. I was using a simple copper wire antenna to receive a consistent frequency radio wave, and I was receiving a frequency slightly higher at periodic times, when I was introducing a different frequency electromagnetic wave.

    Initially I thought that the waves may be combining, but based on your e-mail, could I conclude superposition instead? In other words, the antenna was receiving both frequencies simultaneously, which appeared as one larger frequency on the oscilloscope.

    Is my conclusion correct? Is it possible for the antenna to receive different frequencies from different types of electromagnetic waves simultaneously resulting in a larger frequency reading? I used a radio receiver antenna in place of the oscilloscope antenna, and at the periodic times I could hear choppiness or pulses with the earpiece that coincided with amplitude increases on an oscilloscope attached to the radio receiver. Could the pulses be the beats to which you were referring in that possibly what I was observing was two radio waves of slightly different frequencies? Thus, the oscilloscope attached to the simple copper antenna was giving me the readings of the radio waves and the other electromagnetic wave was not being recorded.

Reply

    Your questions go somewhat beyond my experience. I suggest you look up "The Radio Amateur Handbook" which most amateurs (and some libraries) have, and perhaps discuss them with a knowledgeable amateur.

    In general, superposing signals of one frequency does not generate a higher frequency stable enough to be observed on an oscilloscope. The sidebands I mentioned give irregular wave trains, to which a filter can respond, but not a 'scope.

    However, even if it were possible, I don't think flows of rivers would change. Suppose you did by some miracle manage to slow down the Earth by 10%, or speed it up by that much--what would happen? The equilibrium shape of the Earth would change, as discussed in section 24a of "Stargazers"

Is it possible for Earth to crash into the Sun if it's orbit has stopped? If the orbit did stop how long would this take?

    Yes, if Earth suddenly stopped in its orbit, it would fall into the Sun: but what could make it stop? There is a huge amount of kinetic energy involved, and I for one cannot think of anything forceful enough to stop this motion. Yes, if a planet the size of the Earth and moving in the opposite direction collided with us head-one--it would stop the motion, and the fragments of that collision (both planets would be smashed to little bits) would fall sunward. But nothing remotely resembling such a planet ever been seen, so you can sleep easy.

    By the way, there have been suggestions we should get rid of nuclear waste by rocketing it into the Sun. Not a good idea! Nothing in the solar system is as hard to hit as the Sun itself, even leaving the system completely takes less thrust. To hit the Sun, our spaceship would first have to get free of Earth (velocity needed, 11.3 km/sec) and then add a velocity of 30 km/sec to cancel the orbital velocity around the Sun. That would take enormous thrust, more than any space mission I know.

About the time it takes... use Kepler's 3rd law:

Where T is the orbital period of a planet and a its semi-major axis, half the width of its orbit (k is some number, depending on the units we use). For Earth, T = 365 days, a = 150 million kilometers, a number we indicate by the letter A. So

A very skinny orbital ellipse, one end at Earth and the other at the Sun, has length A, half length A/2. If the period of that ellipse is denoted "t", we get

    Once you arrive at Mars, you will need to match velocities with the planet, using your thrust. You might be able to aim at the atmosphere of Mars and use if for braking your motion.

    The reason I am writing this letter is because I have been having a great deal of trouble calculating the flight path of our rocket. I have learned a great deal from your website, especially about the Hohmann transfer orbit, but I have been unsuccessful in finding an appropriate orbit for so quick a flight.

Reply

    A flight to Mars in two weeks would almost follow a straight line, and would definitely not resemble a Hohmann ellipse (still, two weeks seems rather fast). What you would do is launch a very fast spacecraft to intercept Mars somewhere in its orbit, and on arrival (it will most likely be a one-way trip!) use the thin Martian atmosphere to brake your flight. Without this trick , you will need an enormous reverse thrust to match velocities with Mars. Failing either of these, your spacecraft will whiz by Mars and leave the solar system.

    The question is, in what direction to the Earth's motion should you launch to get away with the smallest initial velocity and initial thrust. If you launch parallel to the Earth's orbit you get the advantage of the Earth's orbital velocity of 30 km/s, but the distance is greater, so you need a greater average velocity to cover it in two weeks. If the Earth's orbit is a circle with radius R1 = 1 AU circle and that of Mars has R2 = 1.52 AU, your distance is approximately 1.14 AU (square root of the difference of squares). If you go out radially, away from the Sun, your distance is only 0.52 AU but you also need cancel those 30 km/s. My guess is, write a short code and calculate the mean velocity at 5 degree increments of the angle of the initial velocity to the Earth orbit, but the result is almost certainly close to radial.

    That gives you the variation of he MEAN velocity with angle. However, the motions are actually in a 1/r potential--the spacecraft starts faster than the mean and ends slower. Modify your code to derive the initial velocity. Then interpolate to find the angle which gives you the trip with smallest initial velocity. All these motions are in straight lines, not ellipses, but for a rough engineering estimate, that's probably good enough.

    The trick, of course, is to design a nuclear engine which is reliable, safe and won't contaminate the environment. If you read my web page on nuclear pace flight, you probably saw that Carlo Rubbia proposed a fast Mars trip like yours--see "Nature", vol 397, page 374, 4 February 1999. I vaguely recall he proposed using an Americium isotope as fuel.