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 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.
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.
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