Listed below are questions submitted by users of "From Stargazers to Starships" and the answers given to them. This is just a selection--of the many questions that arrive, only a few are listed. The ones included below are either of the sort that keeps coming up again and again, or else the answers make a special point, often going into details which might interest many users.
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.
(a) Why are Satellites Launched Eastward?
What is a "Sun Synchronous" orbit?
(b) Why are satellites launched from near the equator?
(a) Distance to the Big Dipper
(b) Big Dipper star names
The effect of the Color of Light on the Output of Solar Cells
Does any location get the same number of sunshine hours per year?
Why isn't Longitude measured from 0° to 360°? "Constellation" or "Asterism"?
What if a Neutron Star hit the Sun?
Why did the Moon appear Red?
(1) Why don't its particles separate by weight?
(2) What accelerates the solar wind?Why does the rising Sun look so big?
Superconductors work, universe expands--with no energy input. Why?
Does precession affect the time of summer? (2 questions)
(a) Temperature in space
(b) Exposure to the space environment: Freeze or burn?
291a.Angle between ecliptic and Milky Way
291b.(Continuation)--About the year 2012
Is humanity changing the climate, or is it the Sun and the Earth's magnetism?
If no stars were seen--could Earth's orbital motion be discovered?
Can the Sun interfere with the visibility of the Big Dipper?
396A Posssibility of Asteroid Hitting Earth (1)
396B Posssibility of Asteroid Hitting Earth (2)
Si tiene una pregunta propia relevante, puede enviarla a
stargaze[símbolo "arroba"]phy6.org
Sin embargo, antes de hacerlo, lea las instrucciones
1. Sobre los asteroides que golpean la Tierra
Un amigo me preguntó si podía encontrar algo en un asteroide que se dirigía a la tierra y un láser que supuestamente está en el espacio que eliminaría el asteroide antes de que golpee la tierra. ¿Existe tal cosa, o está leyendo demasiados libros de ciencia ficción?
Respuesta
Acerca de los asteroides que se dirigen a la Tierra: el mejor relato que conozco es una sección "Los cometas Shoemaker" en el libro "First Light" de Richard Preston. En cuanto a los láseres capaces de destruir uno, son (al menos ahora) pura ciencia ficción.
El libro anterior plantea un punto interesante: sería muy difícil detectar un asteroide que se dirige a la Tierra. Los astreroides generalmente se detectan en fotografías del cielo (a través de un telescopio) por el hecho de que se mueven a través de la línea de visión, dejando una raya en lugar de una mancha. Si cruzan la línea de visión, no chocarán con la Tierra; si se dirigen directamente a la Tierra, no dejan rayas ni llaman la atención.
2. El remolino de agua en un tubo de drenaje
¡Soy una abuela, que pronto cumplirá 55 años, que a menudo tiene discusiones acaloradas en sus listas de correo! Mi pregunta, para evitar la discusión al tener hechos, es: Se dice que la dirección del drenaje del agua en la bañera, el lavabo y el inodoro es la opuesta en Australia. Alguien dijo que el Efecto Coriolis gobierna esto, y es un mito. Alguien dijo que el CE no tiene nada que ver con eso, y es un mito. Por otra parte, alguien dijo que NO es un mito. ¡Estoy indeciso con este, ya que no puedo discutir algo de lo que no tengo ni idea! ¿Me puedes ayudar? ¡Gracias por adelantado!
Respuesta
Su primer respondedor tenía razón: el efecto Coriolis lo gobierna, y es un mito. El efecto de Coriolis puede gobernar el torbellino de los flujos de fluidos y, cuando lo hace, el torbellino es opuesto en hemisferios opuestos. Sin embargo, sólo es apreciable a gran escala. Los huracanes le obedecen: los tornados, que son mucho más pequeños, no, y tampoco los fregaderos, que son mucho más pequeños aún.
Para obtener detalles y explicaciones, busque en la red mundial aquí.
3. Dispensación de agua a cero g.
Soy un estudiante que actualmente estudia para obtener un título en ingeniería. Como parte de este grado nos hemos dado a la tarea de diseñar un calentador de agua y dispensador para uso en gravedad cero. Se ha sugerido utilizar una vejiga en un recipiente calentado a presión utilizando tecnología de microondas o radiofrecuencia. El calentador debe calentar aproximadamente 100 ml a 80 grados centígrados y todo el sistema no puede usar más de 12 voltios. ¿Cómo sugeriría que sería un diseño para este tipo de sistema? Cualquier información que pueda proporcionar sería muy apreciada.
Respuesta
Nuestro grupo de investigación se ocupa de los plasmas y los campos magnéticos en el medio enrarecido entre aquí y el Sol. No tenemos ningún tipo de experiencia en hardware de gravedad cero y estaciones espaciales.
Esto no me impide especular sobre su solicitud, por supuesto. La palabra clave es "dispensación": ¿qué quiere decir con eso? No puede simplemente tener un grifo y dejar salir el agua caliente; se formarán grumos que se alejarán en gravedad cero y, en última instancia, contaminarán sus circuitos o estropearán su vivienda.
Entonces necesitas tres elementos: un recipiente donde se calienta el agua, un tanque del que se obtiene el agua y un recipiente para el té caliente o lo que quieras hacer con el agua.
El primero y el tercero bien pueden ser vejigas de plástico, cuyo volumen se puede ajustar. El depósito tendría una vejiga exterior con agua solamente y una vejiga interna llena de aire, y cuando el astronauta con una perilla bombea aire en la vejiga interna, el agua es exprimida.
El recipiente de calentamiento: podría usar calentamiento por RF, pero sospecho que será algo pesado, necesitará aumentar el voltaje de 12 voltios y también deberá protegerse de la radiación. En un ambiente donde cada gramo superfluo cuesta mucho, ¿no sería más simple un simple elemento calefactor resistivo, con un termostato, por supuesto? Podría estar en un contenedor cilíndrico con un pistón accionado por resorte que inicialmente está en la parte inferior. Al rellenarlo desde el depósito, el agua entra desde abajo y empuja el pistón hacia arriba, contra un trinquete, y cuando el astronauta quiere beber, suelta el trinquete y el pistón accionado por resorte empuja el agua hacia el tercer recipiente. El truco es nunca mezclar aire con agua, una vez que están juntos, son difíciles de separar.
Esta opinión le llega sin garantía de la NASA ni de nadie. ¡Divertirse! Ahora, déjame volver al trabajo serio...
4. Robert Goddard y la Segunda Guerra Mundial.
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?
Respuesta
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.
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.
Respuesta
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?
Respuesta
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.
Respuesta
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.
Respuesta
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.
Respuesta
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 V2Afterwards we have
A ball of mass M1 moving at velocity +W1 A paddle of mass M2 moving with velocity W2We 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
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!
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.
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!
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.
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?
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
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?
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.
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?
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
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"?
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.