Hey,

perhaps it`s a stupid n00b question but i still would like to know
lets take the hubble or the ISS for example. they are moving arround the earth in something like 90 minutes? but why? can`t they just stay still in space? i understand that when they are close to earth that those things can simply drop out of the sky when they stay still but surely they can stay still a little further away..

and for hubble it`s a big question for me.. when i look at the stars i try to keep my telescope steady and stable.. on ONE place.. i`m not running arround the house with it.. why does hubble do that? isn`t observing stars verry hard when it`s rotating the earth in 90 minutes?

just something i don`t understand
i hope someone could clear it up for me (please in easy language.. not rocket science )

Thanx,
Mark.

Gravity causes any unsupported object to fall toward the center of the Earth. If the object is coasting horizontally fast enough, the curved surface of the Earth drops away just as fast as the object falls, so it does not crash.

In a low orbit, a speed of about 17,000 mph is needed. At higher elevations the gravity is weaker, so we can get away with slower motion.

This orbital motion does not shake or jiggle Hubble in any way. It is perfectly smooth. Very little force is needed to keep the telescope pointed in a certain direction, and it is accomplished with gyro stabilizers.

The drawback of observing from a low orbit is that the Earth gets in the way for about 40 minutes or so every time around. The future Webb telescope will be placed at a "sweet spot" between here and the Sun, about a million miles out, where it will hover unpowered because of the way Earth and Sun combine their gravitational forces in a way that is too complicated mathematically to explain here. Trust me, it is stable. Of course it will need to be maintenance-free. Servicing missions would be prohibitively expensive.

To expand on Hornblower's comments...
Everything in the universe is in motion. Gravity pulls everything. Earth orbit is the most convenient because it's both close to us, and there are so many possible orbits.
There are specialty "orbits" that do appear still compared to the earth. One is a Geostationery orbit, and others are Lagrange points. The drawback to these is that there is a very limited amount of space.

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I'll have a go at this, but be warned. Pretty much everyone that answers will know more about it than me.

Let's say you are a skateboarder. You go to a new skate park that is known for having deep bowls. One in particular is very deep, with steep sides, but it also has little bowls set in to the sides of the slopes. If you go straight down a slop you can pick up a lot of speed. If you try to move a cross one really fast, you change your direction a little bit, then fly out of it. If you have the speed just right, you will spiral around many times as you slowly descend into it.

Because of the slope, it's not possible to just stand one place. you will always be drawn to the bottom of a bowl. If you are in one of the smaller bowls on the side, you can get far enough away from it that you won't be drawn to it, but you will still be in the big one. The "sweet spot" is that place where the small bowl blends into the bigger one. Right there, the ground is flat, like the very top of a hill. It would be possible to roll off either way.

With this in mind, the big bowl is the Sun's gravity. The smaller one is the Earth's gravity. Even if you get to the point that the gravity of the Earth has no real effect on you, the gravity of the Sun will still be there.

Now, look at satellites. The Hubble/Shuttle/ISS and most others go around about once per 90 minutes. Communications satellites, like the ones used for those little dishes used by DirecTV, Dish Network and others, as well as the big ones used by the networks, cell phones, and governments are all at the exact right altitude to orbit the earth in exactly 1 day. To us it seems that they never move, because we never have to aim the dish again. But since the Earth is spinning the satellites have to go around it.

Pan out to the Moon. It goes around once a month. To us, it seems to go backwards, rising later each day, but it really is moving the same direction the Earth is rotating, just slower. If it ever stopped moving, it would come right in and say Hi. We'd hate that. If something gets even further out the Sun's gravity takes over and the object has to either orbit the Sun, or plunge to a fiery death.

As for the Hubble holding still, it has a motor that keeps it pointed the right way. Many telescopes you can buy for under 1,000 have them too. The ones on earth are set to cancel the rotation of the Earth and keep your view the same. The Hubble has the same basic thing, only it's a lot more accurate. It's taken images that lasted over 100 hours before. This is looking at a spot of the sky about he size of a quarter (2.5 cm) held about 100 feet away (30m). Since the Earth got in the way for a few minutes every 90, that's a pretty impressive bit of tracking.

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Overall, a very nice explanation Hornblower. I'd like to update one minor side-point of what you said.

I'm having some trouble finding the right page, but I'm pretty sure that the JWST will be located about 1.5 million miles further from (not closer to as you indicated) the Sun than the Earth in a 'Halo Orbit'. It is also the case that this orbit is not stable, but is easy to maintain and is stable enough that for the life of the JWST, the stability won't be an issue.

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Isn't the spot between the Earth and Sun already occupied with SOHO? Is the spot big enough to share?

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Already occupied by several Sun-observing satellites, actually. And, being about 250,000 km across, yes it is big enough to share

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SOHO is not actually at that spot, it is in orbit around it--otherwise, it would appear to be directly in front of the Sun when we "looked" at it, and the Sun is not only very bright visually, it is noisy in the radio bands too. Makes reception easier.

Here is a NASA link.
http://www.jwst.nasa.gov/orbit.html
I had the wrong Lagrangian point. L2 makes more sense, because the Earth is also out of the viewing area permanently.

thanx alot for the cllarification of this stuff.
i found it very usefull.

now a little twisted question.. is there any place in this solar system where there is absolutely NO gravity that is affecting you? than i don`t mean the "flat spots".. but just something like a "edge" where the last gravity stops.. so somewhere behind pluto or do we need to look alot further for that.. (behind the asteroid cloud that is behind pluto)

Thanx again,
Mark.

edit::// typos

http://en.wikipedia.org/wiki/Lagrange_point

Way back in the '70s when they (Gerard O'Neill http://en.wikipedia.org/wiki/Gerard_O%27Neill ....and others) started talking about putting cities in space with thousands of colonists, this was where they proposed placing them. It's the points where the gravity wells of the Earth, Moon and Sun somewhat balance.

Actually, that should be "affecting". But I think the answer is, there is no place in the universe where gravity is not affecting you, because gravity has infinite reach (except maybe in a black hole or something like that, but I don't understand that well enough).

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You're right that it's stable on 2 of 3 axes. But it's unstable on the 3rd axis. I'll explain more at the bottom of this post, and the math isn't that complicated.

SOHO is in a lissajous orbit around the Earth / Sun L1 point, about 1.5 million kilometers inbetween Earth and the Sun.

"Still" means 2 different things here. A geostationary satellite appears still with respect to the surface of the Earth. It orbits with the same period as Earth's rotational period. A satellite in a Lagrange point appears still with respect to the Earth's position in a rotating frame of reference that keeps Earth still.

1.5 million kilometers, not miles. A "Halo Orbit" is special case of a "lissajous orbit", like a square is a special case of a rectangle. More on this at the bottom of this post.

It's huge, as Ilya has pointed out.

And that's what makes it huge. Technically, there is only a single point, but since you can orbit this point in Lissajous orbits of varying radii, there's lots of room for as many satellites.


Here's an explanation of the Lagrange 1 & 2 points:

Picture a rotating frame of reference where Earth appears stationary with respect to the sun. An satellite in the L1 and L2 points can move on 3 axes. The tangental axis is parallel to the satellite's orbital velocity vector. The vertical axis is perpendicular to the tangental axis, pointing straight up and down. The radial axis is points towards and away from the sun.




While in the Lagrange 1 or 2 points, this satellite is stable on the tangental axis and on the vertical axis. If perturbed ahead of or behind the Earth / Sun line on this axis, it will be like perturbing a stationary pendulum. It will occilate back and forth about the Earth / Sun line.

If perturbed above or below the Earth / Sun line, it will also behave like a pendulum, oscillating on its vertical axis about the Earth / Sun line.

But if perturbed towards or away from the Sun, there will not be an oscillating motion to rescue it. It will accelerate away from the Lagrange point, never to return.

When you combine the oscillations about the Earth / Sun line on the tangental and vertical axes, the satellite will orbit the Lagrange 1 or 2 point in an oval known as a Lissijous orbit. If the Lissijous orbit is round, then it is called a Halo orbit.

So a satellite in the L1 or L2 points only needs to make correction burns along its radial axis.

So where is this "sweet spot" we call the Lagrange 1 and 2 points?
A little background info:
Imagine a satellite around the sun in a circular orbit. The further it is from the sun, the weaker the pull of the sun's gravity, and the slower the orbital velocity of the satellite. And if the sun were less massive, the satellite would also orbit slower. When a satellite is directly on the Earth / Sun line, the acceleration vectors of the Earth and Sun add together. In the case of the L1 satellite, it results in an acceleration vector towards the sun that is a little weaker than the satellite would otherwise experience at that distance from the sun since the Earth is pulling it away from the sun. So this satellite experiences a gravitational pull identical to what it would experience if it orbited a less massive sun. This means to maintain a circular orbit, it would need to have a slower orbital velocity. The L1 point is the point where its orbital velocity around this seemingly less massive sun gives it a period that matches the period of something at Earth's distance orbiting the normal-mass sun.

To find the approximate distance to this point, you use the Hill Sphere formula: distance = Earth's semi-major axis * cuberoot(Earth's mass / (3 * Sun's mass)) = 1,496,558522.3543201meters or 1,496,558.5223543201 kilometers.

To verify this:
A satellite in the L1 point would be 1496559 kilometers away from the Earth in the direction of the Sun. 149597870691-1496558522.35432 =148101312168.646 meters from the sun.

It would experience an acceleration from the sun's gravity of GM/r^2 = 6.6725985E-11*1.98891691172468E+30/148101312168.646^2 = 6.05053561087579E-03 m/s/s towards the sun.

It would experience an acceleration from the Earth's gravity of 6.6725985E-11*5.97369125232006E+24/1496558522.35432^2 = 1.77971457520933E-04 m/s/s.

This would yield a total acceleration of 6.05053561087579E-03 m/s/s - 1.77971457520933E-04 m/s/s = 5.87256415335486E-03 m/s/s towards the sun.

This is 5.87256415335486E-03 / 5.87256415335486E-03 = 0.970585834219201 or about 97 percent of the acceleration it would experience towards the sun from that distance if it were not in the L1 point. It is as if it is orbiting a star that has 97% the mass of the sun.

The period of this orbit would be P=2*pi*sqrt(a^3/(GM)) = 2*pi*sqrt(148101312168.646^3/(6.6725985E-11* 0.970585834219201*1.98891691172468E+30 ))= 31553106.225464183 seconds or 0.9998 year.

A satellite in the L2 point would be 1496559 kilometers away from the Earth in the direction away from the Sun. 149597870691+1496558522.35432 =151094429213.354 meters from the sun.

It would experience an acceleration from the sun's gravity of GM/r^2 = 6.6725985E-11*1.98891691172468E+30/151094429213.354^2 =5.81319283119933E-03 m/s/s towards the sun.

It would experience an acceleration from the Earth's gravity of 6.6725985E-11*5.97369125232006E+24/1496558522.35432^2 = 1.77971457520933E-04 m/s/s.

This would yield a total acceleration of 5.81319283119933E-03 m/s/s + 1.77971457520933E-04 m/s/s =5.99116428872026E-03 m/s/s towards the sun.

This is 5.99116428872026E-03 / 5.87256415335486E-03 = 1.02019563043814 or about 102% of the acceleration it would experience towards the sun from that distance if it were not in the L2 point. It is as if it is orbiting a star that has 2% more mass than the sun.

The period of this orbit would be P=2*pi*sqrt(a^3/(GM)) = 2*pi*sqrt(151094429213.354^3/(6.6725985E-11* 1.02019563043814*1.98891691172468E+30 ))= 31714290.5276315 seconds or 1.0049 year.

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TOO MUCH NUMBERS i`m sure it looks easy to you but to me it looks hard to understand :P

My explanation has no math or numbers in it (other than the terms L1 and L2). It's only words and diagrams. The numbers come after the explanation and are only there to compute the actual distance, and to verify that I computed it correctly.

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It is hard to understand. At least, it takes a good amount of education and training to understand it--and it's well worth it

Does that mean you understood it ?

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I think I understood it, but my comment was about the math behind the theory, in general.

In your post, I even understood "The radial axis is points towards and away from the sun."

But I'm not sure why you calculated the periods (e.g., 1.0049 year) because clearly the periods of the objects at the Lx points are one year.

But that's why I computed it. As a way of confirming my statement that you can compute the distance with the Hill Sphere formula, and as a way of confirming my statement that the L1 & 2 points are indeed the places where the combined pull of the Sun & Earth mimic a more/less massive sun, causing an object at their distance to have a 1 year period despite being at a distance where the periods would otherwise be significantly faster / slower than 1 year.

The difference between 1.0049 years and 1.0000 years is either because of the figures I used for stuff like sun mass, Earth mass, etc. or because the Hill Shpere formula is only a very good approximation. I suspect the latter.

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Well, you can't expect the same formula to work for both L1 and L2, right? In fact, it's very close to the average of the actual values, which (using your planetary data) seem to be 1.4925E9 meters and 1.5005E9 meters. Their average is 1.4965E9, which is very close to the value 1,496,558,522.35 from the Hill radius formula that you used.

You're right. I looked into it. The Hill Sphere is only an approximation, but a very good one. As the ratio of an object's mass to the Sun's mass approaches zero, the answer generated from the Hill Sphere approaches the true distance to the object's L1 & L2 points, which in the case of the Earth is only off by a few kilometers since the Sun >> Earth.

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4 thousand kilometers, but I guess that's close enough

"A geostationary orbit (GEO) is a geosynchronous orbit directly above the Earth's equator (0° latitude), with orbital eccentricity of zero. From the ground, a geostationary object appears motionless in the sky and is therefore the orbit of most interest to operators of artificial satellites (including communication and television satellites). Due to the constant 0° latitude, satellite locations may differ by longitude only.

The idea of a geosynchronous satellite for communication purposes was first published in 1928 by Herman Potočnik. The geostationary orbit was first popularised by science fiction author Arthur C. Clarke in 1945 as a useful orbit for communications satellites. As a result this is sometimes referred to as the Clarke orbit. Similarly, the Clarke Belt is the part of space approximately 35,786 km above mean sea level in the plane of the equator where near-geostationary orbits may be achieved."
http://en.wikipedia.org/wiki/Geostationary

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Jami Cat
Astronomy is looking up... :)

And I wasn't sure which one was in respect to the OP.
I was hoping to spark a little more thought, clarification, or gain some common ground by the OP.

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Posts discussing Rebel's theory of gravity have been moved to this new thread. All further discussion on that topic must take place there per forum rules.

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