437. Second Moon for Earth?
Dear David P. Stern
I have a question, which might be more fictional than anything else, but I
thought I'd go ahead and ask anyway.
For a while now, I've been very interested in our Solar System. One
question keeps returning in my mind, which is the following: Earth somehow
receives a second moon, and for the fun of it let's get Ganymede into this
email, and what would happen?
What if, 99.999999% unlikely, Ganymede somehow breaks loose from its orbit
of Jupiter and somehow escapes the gravitational pull, strays towards us
and somehow slows down and reaches our planet Earth. (the somehows in this
line are lack of my knowledge on how this should happen.. (I'm not a
Hollywood actor who suddenly knows the answer.. hah!)
What if Earth somehow is able to grab Ganymede within its gravitational
pull and Ganymede becomes our second moon. We now have two, similar sized, moons orbiting our planet. And for
the heck of it, let's say that Ganymede and Luna would oppose each other
and both would have a certain distance from Earth so they won't collide.
What would happen to this change of gravity within our tiny
Earth-Luna-Ganymede system? And more important, what would happen to our
own planet, having two moons pulling on Earth?
Would this be a simple 1+1=2 and whatever we experience now would just be
doubled? Would it keep our Moon from moving away any further or actually
push it away? Would there be drastic climate changes that could harm us?
Reply
Dear Carel
Having more than one moon is nothing special--Earth is the only planet in the solar system which stopped at one moon. On the other hand, Earth's moon is special in having an unusually large mass compared to the planet, more than 1%.
I do not think capturing Ganymede is a possibility. For this to happen, a huge amount of energy needs to be given to it--first, to tear it away from Jupiter, second, to bring it close to Earth, and third, to allow Earth to capture it. An asteroid currently too distant to be seen might just possibly whiz close to Earth, avoid hitting it (that would be a real miracle!) but get captured by some strange combination of motions with the Moon.
It would not be easy. The "Messenger" mission currently on its way to Mercury is trying to enter an orbit around that planet, but to take way its extra speed it needs swing several times in specific encounters with Earth, Venus or Mercury, and even then, it will only have a long, elliptical orbit.
Your question has been asked before--look up
http://www.phy6.org/stargaze/StarFAQ4.htm#q83
The motion of two large moons around the Earth could be complicated. The "Three body problem" in celestial mechanics has few regular solutions, and is hard to predict for the indefinite future (although Earth, Sun and Moon also form such a system). My guess is that the moons will settle into some resonance--say, three orbital periods of nearer one would equal two of the more distant one, like the motion of Pluto (stargaze/Skuiper.htm).
438. Distant companion of our Sun and 2012
Good morning David!
The reason for my question comes from my fascination with the 2012 end-of-times scenario to which so many people have succumbed. Much of it is hogwash but as with any crackpot theory there exists the minuscule possibility of truth. I think there is more credibility in the theory that earth could have a brown dwarf twin with it's own small planetary system than the theory of a planetoid body having a huge elliptical orbit such that it only comes round once every 3600 yrs.
Thanks and have a great day!
Reply
Dear Steve
I certainly do not believe that the Maya astronomers--a stone age culture with no telescopes--observed astronomical processes which modern instruments fail to detect. Many people have written about it, for years, and some of the correspondence is linked from the end of
http://www.phy6.org/stargaze/Scalend.htm
I expect many more letters after November, when a film "2012" is due to be released--also when "Sky and Telescope" will feature a discussion of 2012. Can't help it.
A large astronomical object with a period of 3600 years seems too close--it would probably affect planetary motions in a detectable way. Anyway, known distant small planets are very small. The idea that the Sun may have a dim distant companion star is old: see
http://en.wikipedia.org/wiki/Nemesis_(star)
So far there exists no real evidence
439. Distance between two points on a sphere
Hello sir
I am professional java software engineer.. yesterday I read about you on internet so decided to talk with you...
these days I am exploring the google map services... last week I created an application which take any location name on earth and provide me the correct latitude and longitude as found on Google Earth
for example
Lahore Lati: 33.7181510 Long: 73.0605470
Islamabad Lati: 31.5450500 Long: 74.3406830
I want to get the distance b/w these two cities
I take the haversine farmula to get the result from latitude and longitude... [continues with a computer code]
Reply
I am not familiar with the haversine formula, and also wonder why the Earth's radius (6378 km at the equator, 6371 at the pole) does not appear anywhere in your code. I also did not recognize the computer code language, though one could probably decode it.
Here is how I would have done it--assuming the Earth to be a sphere of radius R, which is OK for moderate accuracy. I prefer working in spherical coordinates (r, theta, phi) or (r, θ, φ ), with φ the longitude but θ equal to 90–latitude, so southern latitude is negative.
(See also
http://www.phy6.org/stargaze/Slatlong.htm).
Then a unit vector along the radius from the Earth's center to a point has components
(Ux, Uy, Uz) = (sin θ cosφ, sin θ sinφ, cos θ)
For any given point, all three components can be derived from the latitude and longitude.
Let (Vx, Vy, Vz) and (Wx, Wy, Wz) be unit vectors to the two points whose distance you seek. Then the angle A between them has a cosine equal to the scalar product (where * denotes multiplication):
cos A = (Vx*Wx) + (Vy*Wy) +(Vz*Wz)
The distance D between the points is simply RA with A in radians. If a quarter of the circumference of Earth is 10,000 kilometers (original definition of the meter) and A is in degrees, then
D = 10000 (A/90)
If D is much smaller than R (as in your example) you probably get reasonable precision by assuming the Earth is locally flat and using the theorem of Pythagoras in rectangular coordinates. Each degree of latitude then has length 10000/90 kilometers and each degree of longitude length (10000/90) cos L, with L the average longitude of the points.
440. Getting sucked in by Gravity
The big mass of the sun has a gravitational pull on us, so is it possible we have just moved a little bit closer to the sun?
Also, basically the same question may be asked about the black hole in the center of the Milky Way. Blacks holes take in anything if I remember right, not even light can't escape. So it should be slowly sucking the Milky Way into it, and in time earth would be sucked in, though that would be a very long way away. Is that possibly right?
Reply
Dear Tyler
None of these will happen, because gravity conserves energy. Let me give an example. The Sun is also surrounded by a huge cloud of slow small icy objects, left over from the time when the solar system formed. Like planets, they move in ellipses (Kepler's first law, a result of the laws of motion).
Some of them do not have enough sideways motion and very slowly fall towards the Sun, in an ellipse that is very narrow and stretched. As they fall, they gain speed, and as they get near the Sun, sunlight evaporates their outer layers, revealing them at comets. However, unless they are aimed exactly at the Sun (very few are) they miss it, swing around and fly away again, losing the speed and kinetic energy they have gained.
The Earth also moves in a nearly stable ellipse--the size of which depends on the energy, which is practically conserved. It is expected to keep its distance from the Sun (if not, the year would get shorter, something sure to be noted). The situation is similar for an object orbiting a black hole, even the one at the center of our galaxy, In the galaxy, energy would be conserved, too (from all we know), even though the motion may not obey Kepler's laws, because of invisible dark matter, which we still do not understand. About that, see
http://www.phy6.org/stargaze/Sun4Adop3.htm
441. The Work of Nikola Tesla
Good evening sir,
I have been enjoying your website but noticed that you haven't mentioned the work of Nikola Tesla at all. Did he not contribute anything to our understanding of the nature of electricity?
Reply
Dear John
Tesla was essentially an engineer who played an important role in introducing alternating currents to general use. See
http://www.phy6.org/Education/FAQs8.html#q123
He contributed nothing important to the understanding the nature of electricity itself, the way Volta, Ampere, Faraday, Maxwell, Hertz and Thompson did (to mention a few).
Later in life, he developed all sorts of ideas for extracting electric power from the atmosphere, and that gave him a reputation of sorts in the area of fringe science. See
http://www.phy6.org/stargaze/StarFAQ4.htm#q72
442. Firing a cannon straight up
David,
Recently several teacher colleagues and I were on a camping trip, and while sitting around the fire found ourselves in a strange debate. The question was posed: If you fired a cannon straight up, would it be possible for the cannonball to return and land back into the barrel? Our debate focused on the effects of wind resistance, etc., as you might suspect, but if such an impediment was removed, would it be possible? Or would the rotation of the earth cause a change in its trajectory on the way back down? Any help would go a long way in preventing such an extreme discussion from recurring. Thanks very much,
Reply
Hello Chris
Few ideas are so crazy that no scientist would look into them, and your experiment was done in 1634, by Marin Mersenne and his friend Petit.
The result was unexpected: the cannonball was never found! Mersenne wrote about it to Descartes, his friend and correspondent, and Descartes suggested the cannonball escaped the Earth's gravity.
I attach an article telling the story (also showing a picture) which appeared in 1954 on p. 76, vol. 22, of the American Journal of Physics. It also cites some older references.
May the Force be with you!
443. Does one see half the sky--or more, or less?
Hello, I am currently going into grade 12 and an interesting question came up between me and some friends at a campfire while staring at the stars. This question could definitely be phrased better for your understanding but this is what it is... If you were on the ocean surface "perfectly flat, largest horizon distance" how much of space is visible from horizon to horizon? Some of us suggested far less than 50%, exactly 50%, and or greater than 50%.
We then created theories and diagrams supporting each percentage. Anyways by the end of the night we concluded that we are technically on the surface or top of a ball looking to either side is an angle SLIGHTLY greater than 180 degrees and as this angle extends out to areas as far as other galaxies we are seeing a larger and larger amount "as the cone's rounded edge circumference increases" If you could please revise this statement or say if it is incorrect. A diagram would be much appreciated!
Reply
Dear Jared
If you lie on the ground and neglect the atmosphere, you see half the sky, 50%. If you stand up you see a bit more, if you climb a building or mountain or look out from an airplane, the higher you go the more you see. You can calculate the extra angle by deriving the distance D to the horizon
http://www.phy6.org/stargaze/Shorizon.htm
Then the tangent of that angle is D/R, with R the Earth radius. The angle does not change with distance of the viewed object.
But in addition there exists another factor, the refraction of light in the atmosphere, which also extends your view. See
http://www.phy6.org/stargaze/StarFAQ6.htm#q99
and
http://www.phy6.org/stargaze/StarFAQ11.htm#q185
Because of refraction, the setting Sun looks a bit flattened. The bottom of the Sun may already be below the horizon (in the "extra" visible part of the sky) but you still see it because of refraction, while the top of the Sun, more distant from the horizon, is refracted by a smaller angle. See table in the first of the letters above.
444. Why are nights dark?
Good evening Sir, I want to ask a question, please answer me immediately if you can. My question is "WHAT IS THE REASON BEHIND THE DARKNESS OF NIGHT ?"
Reply
Dear Toshika
Darkness is the absence of light--it is not a color, but the lack of any light. Almost all our light comes from the Sun, including sunlight reflected by the Moon and planets. Only the light of stars does not come from the Sun--they are very far-away suns, but I guess you do not count their light as a factor. So with the Sun and Moon facing the other side of Earth, the sky has no source of bright light.
People once wondered, "if the universe is infinite, there should be a star in any direction we look" ("Olbers' Paradox"). True, but the universe is not infinite, and in any case it began at a "big bang" about 13.7 billion years ago. So we cannot expect to get any light from distances more that 13.7 billion light years
445. Origin of the Solar System
Can you tell me whose theory it was that the planets started out as plasma
balls ejected by the sun (as opposed to concretions from a nebular disk)?
Is there any evidence that any of the planets are moving away from the sun, as
I presume such a theory must impose?
While I'm at it: Does our sun rotate in any particular direction around the
center of the Milky Way galaxy? Say, as viewed from Polaris?
And:
If we were looking down on the Milky Way as a great disk, what angle would the
Sun's pole and planetary disk subtend? Is that the right word?
Reply
I am not familiar with the theory you describe, but am pretty sure it is incorrect, and not only because it would take enormous energy to eject an Earth-size mass (never mind now Jupiter) from the Sun to the Earth's orbit, or to change its orbit to near circular (rather then hit the Sun on the return trip down the other side of the ellipse).
No, the main reason astronomers discount solar origin of the planetary system is the conservation of angular momentum. All but a few percent of the angular momentum of the solar system reside in the planets: it would be hard to see how this would happen if the planets came from the Sun.
446. Flying to other planets
My name is Mateusz, I have read your article about flight to the mars and I'm wondering if I can use the equations on your site because I'm making a simulation of a spaceship flying to different planets from the earth.
Reply
Dear Mateusz
If your spaceship plans is to imitate the flight to Mars described on "Stargazers," the data for Hohmann ellipse transfer orbits for all major planets are tabulated on p. F-149 to F-153 of my 1982 CRC Handbook of Chemistry and Physics. That handbook is relatively inexpensive (especially if you buy an earlier year's edition!) and should be in most science libraries, though I am not sure that tabulation is still being printed.
However, a mission from Earth to a planet more distant than Mars will almost certainly NOT use a Hohmann ellipse. To do so requires an increase in velocity, and it is usually much more economical to obtain that velocity by a gravity-assist encounter with the Moon and/or another planet, as described in
http://www.phy6.org/stargaze/Stostars.htm
Unfortunately, deriving such orbit requires very precise calculations, and often a "mid-course correction" during the flight (or more than one), because at launch it is very hard to achieve the precise velocity needed.
447. The heat of the Sun's Corona
Hello (from a high school teacher)
I recently found out that the Sun's corona can be hotter than the surface of the Sun. Is this correct - it doesn't seem to make any sense. If so, why? My only guess could be that particles in space are farther between, and therefore heat up more easily, so the actual temperature of the corona wouldn't really "feel" hotter. I'm probably wrong, though.
By the way, I have learned a lot from your website and it has helped me greatly in answering high school students' questions. I might have some more questions in the near future, if that's ok!
Reply
What you encountered by chance has been a major mystery--maybe THE major one--of solar physics for about 60-70 years. No one has found a satisfactory explanation, it seems to contradict the second law of thermodynamics, by which a cooler object cannot supply heat to a hotter one.
It is not related to the separation of particles in space--when you see spectral lines from iron atoms that are missing 13 electrons or so, you cannot but conclude that those atoms are hit by something very powerful (or hot). The X-rays from the corona confirm the heat, and the existence of the solar wind depends on it.
You will find more in
You may send me questions now and then, but please only as last resort!
449. Newton's 3rd Law--In Statics and Dynamics
I teach physics and your Q&A collection is a real treasure for me and my students. Thanks a lot for taking the time to address these questions. I was asked by one of my students (6th grade), a question I could not answer. I hope you can help.
The topic is Newton's 3rd law. I demonstrated the law with various obvious examples (boat, skates, rocket etc.), but when I talked about static systems (e.g. - I press with my finger on the table), I was asked where the opposite force comes from. If a truck stands on the road that means that the road is "pushing" the truck back upwards. Where is this force coming from? If the road has so much force to start with, can't we "harvest" it? Does that mean that static objects has a huge amount of force within them?
Thank you very much for you help.
Reply
Dear Danny
Force is not in short supply in our world. Energy is.
When forces are encountered in nature, it is usually in one of two situations. Either the force encounters an equal and opposite force, and no motion occurs (brick on a table). This is not really Newton's 3rd law: Newton's laws involve DYNAMICS, whereas this is a case of STATICS.
The force may be quite large (chimney standing on its foundation), but it produces no motion, and therefore does not perform mechanical work. It therefore doesn't require energy, and can't provide any.
Or else, the force can produce motion (brick held at arm's length and let go). To be sure, two bodies are involved here too (the brick and the Earth which pulls it down). The difference is that motion is possible, energy is released (or converted), and the brick is accelerated.
By the 3rd law, the Earth moves too, accelerating towards the brick, so that the common center of mass (brick plus Earth) remains the same. Of course, considering the size of Earth, that motion is too small to observe.
I hope that answers your students. I hope they enjoy the Q&A.
Request for clarification
Thanks a lot for this prompt response.
So just to make sure I've got it right (for the educational facts accuracy):
Newton's 3rd law applies only for dynamic cases.
Is this the case?
Many thanks!
Reply
Dear Danny
In a way, yes. If you look at the bottom of the section on Newton's 3rd, you will see a formulation of Newton's laws (I am not completely sure that Mach originated it) which avoids the tricky concepts of "mass" and "force" :
"When two compact objects ("point masses" in phystalk) act on each other, they accelerate in opposite directions, and the ratio of their accelerations is always the same. "
This only makes sense in dynamics. The way the 3rd law is usually formulated, though ("every action has an equal and opposite reaction"), is ambiguous and may be viewed as covering statics as well.