Expert: Courtney Seligman - 9/5/2010

Question
QUESTION: Hi and thanks for taking the time to answer my question(s),

I am currently watching the program 'Universe' on the Tv and i am always interested in working out the scientific theory of the content of the episode in layman's terms in my own mind.  My question is in relation to why the moon does not fall to the earth due to the earths superior gravitational pull.  I am aware that the earth is effectivly dragging the moon towards the planet due to the gravitational pull and also that the moon is spinning to the side of the planet which creates the orbit. My question is why is this enough to keep the moon in a steady orbit? and not just creating a spiral effect which would eventually result in a collision. Reading between the lines i am asuming that the moon has a conflicting gravitational pull (which is not nearly enough to alter the earths orbit around the sun) but is enough to stop it from being 'pulled' towards earth completely as it attempts unsuccessfully to drag the earth towards it, sort of like a gravtational 'tug of war' between the two 'planets' where the moon has just enough strength to stay in orbit around the Earth.  Is my analogy correct or anywhere near the mark? your explanation would be greatly appreciated, also is this relationship similar to the earth's orbit around the sun?

Thanks a lot:) Ian

ANSWER: The Moon's pull does have an effect on the Earth -- it makes the Earth move around the center of mass of the Earth-Moon system, while our pull on the Moon makes the Moon do the same thing. For a fairly detailed (albeit slightly incomplete) discussion of how this works, see my page on "Gravitational Interactions of the Earth and Moon: Barycentric Motion
, at http://cseligman.com/text/moons/earthmoongravity.htm

However, although that should explain their mutual effects in some ways, it won't address your basis question -- why doesn't the Moon fall down? That has to do with general orbital motion, which is a combination of two factors:

(1) If a body is moving through space, without any force acting on it, it will move in a straight line, with constant speed (this is one way of paraphrasing Newton's First Law of Motion, the Law of Intertia). So if the Moon had no force on it, it would move in a straight line, instead of orbiting the Earth, and soon be far out in space.

(2) If a body has a force acting on it, its motion will change at rate which depends upon the ratio of the force to its mass, and in the direction of the force (again, this is a paraphrase of one of Newton's Laws; in this case, his Second Law, the Force Law). In other words, when the Earth pulls on the Moon, the straight-line motion the Moon would have is changed, in a particular way.

The result of these two laws is that the Moon does not follow a straight-line path, but one which gradually curves, according to the effect that the Earth's gravitational force has on it. That effect will cause the Moon to fall away from its straight line path, toward a path which is closer to the Earth, and more in the direction of the Earth. In addition, if the Moon is headed upward, the force of the Earth will slow it down; whereas if the Moon is headed downward, the force of the Earth will speed it up.

At every moment, as the Moon moves, it has the tendency, from the First Law of Motion, to move in a straight line, in whatever direction it happens to be moving. But in the next moment, it will move in a slightly different direction, which is a little more toward the Earth, and a little closer to the direction to the Earth than the original motion. This naturally leads to the (and your) conclusion that if this were to continue, the Moon would inevitably spiral inwards, and end up hitting the Earth. However, it turns out this is wrong, and its actual path is (ignoring other bodies, such as the Sun) an ellipse which repeats over and over again, without ever getting any closer to or further from the Earth, than the time before.

To understand this, suppose that we drop something. It falls straight down, at a rate determined by gravity.

Now throw the object horizontally to the side, and see what happens. It still falls straight down, at the rate determined by gravity, because that is unaffected by the motion of the object. But it ALSO moves horizontally, so its path is not straight down, but straight to the side (by an amount which depends upon its speed) and straight down, at the same time. In other words, it has two simultaneous motions, sideways and downwards. This produces a curved path, which is usually described as a parabola, but is actually an ellipse with a focus at the center of the Earth (for small bodies) or at the center of mass of the Earth and the other object (for larger bodies, like the Moon).

Now imagine throwing the object to the side, faster and faster. No matter how fast you throw it, if you look at its motion during a very short interval of time (so that we can ignore the curvature of the Earth, beneath the falling object), it will fall by exactly the same amount. This is true for all speeds, up to and including the speed of light. If you imagine shining a beam of light horizontally, and wait for it to travel 10 feet (which takes about a billionth of a second), it will fall exactly the same distance as anything else would in that amount of time. Admittedly, since the amount of time is so small, you might not be able to see that it "fell"; but it does fall, exactly the same as anything else, and it is equally true that any one object would fall the same distance, due to gravity, as any other object, under the circumstances described.

Now, let's throw something fairly fast, and watch it move sideways (due to its motion) and downwards (due to gravity). It follows a curved path, which ends with it at the same height it would have had, if it had fallen straight down. BUT... and this is the critical point... this is only true, relative to the horizontal defined by the surface of the Earth at the place where it started off.

When you drop something straight down, it falls toward the same surface that it started above. The distance that you see it fall is exactly the same as the distance that it does fall. But if you throw something to the side, while it falls, it also moves around the Earth; and since the surface of the Earth is a sphere, the surface that it lands on is not on the same horizontal plane as the surface it started above, but is tilted downward and to the side, according to how far it moved to the side, while it fell.

For slow sideways motions, we can ignore the complications introduced by the fact that the Earth isn't flat, but a sphere. If you throw something twenty feet to the side, the surface there is essentially at the same horizontal angle and height as where it started off. But suppose you throw it very fast -- say, 18 thousand miles an hour to the side.

In this case, during one second, the object will fall 16 feet (that's how far an object falls in the first second, from a standing start, due to gravity), and move 5 miles to the side (18 thousand miles per hour divided by 3600 seconds per hour). Now, 5 miles off to the side, how does the surface compare to the surface where we started? The answer is, it is 16 feet below the original horizontal, and tilted slightly downward, by exactly the same amount as the downward tilt of the falling body's motion. (This works out the same because I knew the answer in advance, and chose 18 thousand miles an hour as the speed for that purpose).

This is the secret of orbital motion. Suppose you watch the International Space Station passing overhead. After a while, you will see it drop below your horizon. When you see that happen, it has actually "fallen" from its original height, relative to your horizon, to the plane of your horizon. But you know it hasn't actually fallen down -- it's still up in orbit. You just can't see it, because it's gone beyond the curve of the Earth's horizon. So why hasn't it fallen down, as it appears to have done, even though the Earth's gravity was pulling on it? Because it's sideways speed is 18 thousand miles per hour, the "orbital velocity" of objects near the Earth. And as a result, as its path slowly curves, "falling" toward your horizon, the Earth underneath it -- not the Earth where it passed above you, but the Earth off to the side, below its orbital motion -- is "falling" out from under it.

In some ways, I've belabored the point. But hopefully, by doing so, you can see that any object in orbit is always falling toward the object it is orbiting. It just doesn't fall down, because as it goes around the object, the surface of the object "curves away" from it, relative to the horizontal surface that we might imagine as the starting point.

Now, the situation I've just described presumes that the sideways motion is the same as the "orbital speed". If the sideways motion isn't the same as the orbital speed, things are a little different. Just as, when you just drop something, it falls straight down, objects moving sideways at less than the orbital speed will fall toward the surface below them. If they are moving very slowly sideways, they'll fall at essentially the same rate as objects which are not moving sideways at all; but if they move at nearly the same speed as the orbital speed, they will hardly seem to fall downward, at all. It's a matter of degrees, so to speak. Very little sideways motion produces very little effect; but large sideways motions make the downward rate of fall seem smaller and smaller, until at the orbital speed, the rate of fall seems to be zero (the object is "in orbit").

The same thing happens, but in reverse, if the speed is faster than the orbital speed. Instead of appearing to fall toward the surface, the object seems to "rise", relative to its original height. At speeds very close to the orbital speed, this effect is small; but at substantially faster speeds, the orbiting object seems to move off, almost as if it weren't being pulled down, at all. In fact, if the sideways speed exceeds the square root of two times the orbital speed (in other words, is a little over 40% faster), the object will go off into space, and never return.

For an object like the Moon, part of the motion (at perigee and apogee, the nearest and furthest parts of the orbit) is apparently horizontal. The rest of the motion is downwards (while moving from apogee to perigee) or upwards (while moving from perigee to apogee). During the downward motion, the object goes faster and faster, until it is going so fast that it seems to stop falling downward, and start moving toward a more horizontal motion, and then an upward motion. During the upward motion, the object goes slower and slower, until is is going so slow that it resumes what we would consider a normal "falling" movement. The increase of speed during the downward movement is equal to the decrease of speed during the upward movement, and the change from downward to upward motion as the object goes fast is equal to the change from upward to downward motion, as it goes slower. So in the end, the motion just repeats itself, over and over again, without ever changing (again, as noted above, if we ignore other bodies, such as the Sun),.

This is probably a far longer and more complicated answer than you expected, but since I don't have a webpage dealing with basic orbital motion, I couldn't make it too short, without just waving my hands and hoping you would accept an answer, without any real explanation. I could hardly blame you if you would like me to go over any part of it in a simpler way, or if there is something I have forgotten to cover, that you would like me to discuss. In the meantime, I hope that between the page on barycentric motion and this "page" on basic orbital motion, you will have a much better understanding of the situation. But if not, please feel free to contact me again.

---------- FOLLOW-UP ----------

QUESTION: Thank you for the detailed explanation although i am not fully understanding of the mathematics of your answer I think i understand the visual aspect as i like to see diagrams and create visual images in my head to understand theory's and i think you explanation was enough to help me visualize what happens.  I think i understand that objects in space with a gravitational pull will have an orbital speed (eg Earth/18000 mph) and that if an object is traveling slower than that speed it would be gradually dragged towards Earth and if as you explained if the object was travelling faster than the orbital speed it would appear to rise off into space as the gravitational pull couldn't compensate for the speed of the object.  Your explanation reminded me of Newton's analargy when he said if a cannon ball was fired from a point on earth at exactly the right speed despite gravity pulling on the cannon ball to drag it to the ground it would travel all the way around and eventually hit you on the back of the head.  I am also aware that the moon has an eliptical orbit which explains your answer that between perigee and apogee the path is almost horizontal,,, however because of the downward and upward motion,,,is the perigee sort of the part of the orbit where its shaped like the base of an egg? and the apogee where the moon has moved downward to the pointy tip of the egg like orbit? and am i right in assuming that the orbit is slightly tipped downward towards the apogee which would cause the downward motion sort of like a rollercoaster affect? Sorry i just hate it when i cant understand something. Also what determined that the moon was actually traveling at the Earth's orbital speed? did it happen by chance? or was it determined when the solar system was created or did the relationship between the central mass of the 'Earth and Moon' system have a bearing on the speed the moon was travelling? thank you so much for your explanations and time:)

Ian

ANSWER: First, I agree that my discussion would have been much easier to follow with diagrams. When I gave the lecture my answer was based on, I used a lot of diagrams, and keenly felt the disadvantage of not being able to use them in my reply to your question. I'm sorry that as a result, it was not as clear as it should have been.

As for your followup, it is obvious that you understand what I covered up to where you discuss Newton's cannonball, and are just making sure you are correct. After that, things are a little off, so I'll try to explain. Fortunately, some of what you are asking is covered (with diagrams!) on my website, so I'll refer to the appropriate pages, as I cover that. In particular, you should find it useful to refer to Ellipses and Other Conic Sections (http://cseligman.com/text/history/ellipses.htm), Kepler's First Law (http://cseligman.com/text/kepler1.htm), and Kepler's Second Law (http://cseligman.com/text/kepler2.htm). I would recommend opening those pages in a separate window from this answer, so you can refer to them while you read this.

I should point out that the diagrams of ellipses are generally very elongated, to make their non-circular nature clear; but as shown in the comparison of ellipses of different eccentricities (about halfway down the page on Kepler's First Law), you need a pretty big eccentricity for an ellipse to look significantly elongated. Even the ellipse of 40% eccentricity, although obviously "wider" than "high", looks more nearly circular than the very elongated ellipses used to illustrate apogee and perigee, on the rest of the page. For a comparison with real orbits, you might take a look at Planetary Orbits (http://cseligman.com/text/sky/orbits.htm). The third image from the top shows the orbits of Neptune and Pluto, and how Pluto is actually closer to the Sun than Neptune at times, because of its relatively large eccentricity (25%, more than any other planet). But as shown in the diagram immediately below that, the elliptical orbit of Pluto is almost identical to a circle of the same size and center; so although we will discuss ellipses as though they are very elongated, in reality, most orbits are more nearly circular than not.

Now, as shown in the diagram at the top of the page about Kepler's First Law, peri-something and ap-something represent the two "pointy" ends of the ellipse. When the orbit is for a planet around the Sun, the ends are called perihelion and aphelion. When the orbit describes the motion of an object around the Earth, the ends are called perigee and apogee. In each case, the prefix peri- means "close to" and represents the closest place in the orbit to the object you are going around, and the prefix ap- or apo- means "far from" and represents the furthest place in the orbit. The suffix -helion refers to the Sun (Helios, in Greek mythology) and the suffix -gee refers to the Earth (Gaea, in Greek mythology). So for the Moon, apogee is the "end" of the orbit where the Moon is furthest from the Earth, and perigee is the "end" where it is closest to the Earth. At both those locations, the Moon is moving exactly sideways relative to the direction to the Earth, and can be thought of as moving horizontally, given the fact that the Earth and its gravity would define "down".

Now, at apogee, the Moon is moving horizontally, but too slowly to stay at that distance (that is, it is going slower than the orbital speed at that distance), so it "falls" toward the Earth, just as any object does, if it is moving slower than the orbital speed (technically, it's called orbital velocity, but that includes its direction, and we're only discussing how fast it is going, so "speed" is OK).

From apogee to perigee, the Moon is steadily moving downward, getting closer and closer to the Earth. But at the same time, it is going faster and faster, because the Earth's gravity is also downward, and when you pull or push something in the direction it is going, it moves faster. So the downward pull of gravity makes the Moon go faster and faster, as it gets closer and closer. You can see this in the page on Kepler's Second Law, which shows that objects move faster near perihelion, and slower near aphelion.

During the first part of its downward motion, the Moon is going, as stated above, slower than its orbital speed at that distance. But as it gets closer to the Earth, its speed increases more rapidly than the orbital speed. At first, this doesn't make much difference, and the Moon heads more and more downward, as well as faster; but by the time the Moon is getting fairly close to the Earth (in the diagrams shown, about 70% of the way from apogee to perigee), it is going at the orbital speed for that distance. (I hate to digress, but it is useful to briefly explain something here, about orbital speeds.)

Digression (sigh): In the paragraph above, I refer to "its orbital speed at that distance". It is important to note that as you move further from the Earth, the Earth's gravity is weaker, and things don't have to go as fast, to keep from falling down. Close to the surface, the orbital speed is 18 thousand miles an hour; but at the orbit of the Moon, which is 60 times further from the Earth, the orbital speed is nearly 8 times slower, and is only a little over 2000 miles an hour. Because of this, when the Moon is near apogee, it doesn't have to move very fast, to stay at that distance. When it is near perigee, because it is closer, it has to move faster, to stay at that distance. It probably seems peculiar, but although the Moon doesn't have to move as fast to stay at apogee, it is moving too slowly to do so. And as you will see in a few moments, although it has to move faster to stay at perigee, it moves TOO FAST to do so. (End of digression.)

Now to return to the discussion of two paragraphs ago. At apogee, the Moon doesn't need to go very fast, to stay at that distance; but despite that, it is going to slow to do so, so it starts to fall inwards (and downwards). By falling inwards, which is in the direction of gravity, its speed increases; and this increase is at a faster rate than the increase in orbital speed at different distances, so eventually, instead of going slower than the orbital speed, it will be going faster than the orbital speed. I wish I had a diagram to show this, but perhaps you can draw one, like this: Use the X-axis (from left to right) to represent the distance from the Earth, with perigee on the left, and apogee on the right. Use the Y-axis (from top to bottom) to represent speed, with higher speeds at the top and lower ones at the bottom. Now draw a diagonal (it should be a curve, but a straight line will do for the concept), representing the orbital speed at different distances. It should be lower on the right (and greater distances), and higher on the left (at closer distances). Then draw a second diagonal, representing the ACTUAL speed at different distances. This diagonal should be steeper, so that it is below (slower) than the orbital speed diagonal at apogee, and above (faster) than the orbital speed diagonal at perigee.

The graph you have just drawn shows how the speed of the Moon changes as it approaches the Earth. At apogee, it is moving slowly -- in fact, so slowly that even though the orbital speed at apogee is relatively low, the actual speed is lower yet, and the Moon is not going fast enough to stay at that distance, so it falls toward the Earth.

As the Moon approaches the Earth, it speeds up. At some point, it is going at the local orbital speed. IF AT THAT TIME IT WAS MOVING HORIZONTALLY, it would stay in a circular orbit at that distance. But it is NOT moving horizontally, because it has been falling downward, ever since it was at apogee (refer to the pages on orbital motions, and note that as the orbit goes from apogee to perigee, the general direction of motion is sideways and downward).

Since the move is moving downward, even though it is going at the orbital speed, it cannot maintain a circular orbit. Instead, it moves closer and closer to the Earth. In the process, it picks up more and more speed, until it is going considerably faster than the orbital speed.

Now here is the tricky part. We can easily imagine how about 70 or 80% of the orbit works. Suppose we start at a point between perigee and apogee where the Moon is heading upward, away from the Earth. It is headed upward, but gravity is pulling it downward, which is backward. When you push or pull backward on something, it goes slower. So as the Moon moves away from the Earth, it slows down. Eventually (at apogee), it slows down so much that it cannot rise any more, and begins to fall. It then falls downward, in the way that I've described above. So if you ignore the bottom part of the orbit, things are easy to understand. Just as when you throw something up, it goes up for a while then falls back down, as the Moon heads upward away from perigee toward apogee it slows down, then falls back to perigee.

The hard thing to understand is, if the Moon was falling downward all the way from apogee to perigee, why doesn't it keep falling downward, and spiral into the Earth, after reaching perigee? (I'm thinking here, of your original question.) The reason is, it is going too fast.

Recall the argument from my original answer (unfortunately, still without any diagrams; so perhaps you could try to draw diagrams to represent what is going on). If you drop something, it falls straight down. If you throw it to the side, it falls down (the same amount, in a given amount of time) and moves to the side some distance, which depends upon its speed. To visualize this, draw two horizontal lines. The bottom one represents the horizontal plane at the surface of the Earth where you dropped the object. The upper one represents the horizontal plane the object was on, at the moment you dropped it. Now draw a straight vertical downward line from the top line to the bottom one, at the left end of each line. That represents how things fall, if just dropped. Now draw another vertical a little ways off to the right, and draw a curve which starts at the left end of the upper line and is nearly horizontal at that point, and gradually curves downward as it moves toward the right, so that the curve ends where the second downward line reaches the lower horizontal line. If done correctly, the curve will be a smooth curve, which is a part of a parabola. The nearly horizontal part represents the way an object thrown sideways moves at the start of its fall. At that time, gravity hasn't had time to make it fall very fast, so it is moving sideways more than it is falling down. The more vertical part, where the parabola meets the vertical line, represents the way the object is moving after some time has passed, and gravity has forced it to fall at the same rate it would have fallen, if it had just been dropped. It is not moving straight down, because it is falling AND moving to the side; but it is moving more nearly downward than at the start, because it is falling faster than at the start.

Repeat this drawing another two or three times, each time placing the vertical line further to the right, and drawing as smoothly falling a curve as you can from the start (at the far left of the top horizontal line) to the end (at the bottom of the vertical line). Each curve should be a little more horizontal, because of the greater horizontal speed involved in going further to the side, during the same time. (In case this isn't clear, the following paragraph explains the picture.)

The series of curves and vertical lines you have just drawn represent the motion of objects thrown to the side at different speeds. If you just drop something (the leftmost vertical line), it falls straight down. If you throw it to the side, it follows a curved path (the one connecting the start, to the first vertical line). If you throw it faster to the side, it follows a more horizontal curved path (each successive vertical line, further to the right, requires a more horizontal curve to show the motion of the falling body). Note that in every case, the object falls the same distance (the distance between the two horizontal lines) in the same time. Only the sideways motion is different.

Now add one more curve to your diagram. Starting at the far left of the LOWER horizontal, which represents the horizontal plane of the Earth's surface at the place you dropped the object, draw a curve which is as near a circle as possible, and curves downward to the right about the same amount as the next to the last curve you already drew. That is, if you drew four curves, make the circular curve below the horizontal line more or less parallel the third curve.

Finally, for this part of the discussion, imagine that the circle you just drew is the curved surface of the Earth. For the object that was dropped, and fell straight down, the landing point is directly below the starting point, and the distance that it fell took it from above the surface of the Earth, to the surface of the Earth. BUT FOR ALL THE OTHER CURVES, although the object "fell" the same distance, as the Earth's surface "curved out from under" it, the object does not fall all the way to the surface, but ends up above the surface. For the slower moving objects, the distance that the Earth's surface curves out from under them isn't very much, so they seem to have fallen almost as far as if they weren't moving sideways at all. But for the faster moving objects, the Earth's surface curves away from them almost as much as they fell, so they don't appear to have fallen as much.

We now consider the last two curves -- the one that is more or less parallel to the curve of the Earth's surface, and the one that goes even further to the right than that curve. The one more or less parallel to the surface represents a circular orbit around the Earth. The object is falling toward the Earth, but the surface is falling out from under it at the same rate that it is falling, so it appears not to be falling at all. In the case of near-Earth orbits, if something is moving at 18 thousand miles an hour to the side, it moves 5 miles to the side and 16 feet downward, in the first second after it is "dropped" and "thrown sideways". But as it happens, the curvature of the Earth is such that if you move 5 miles to the side, the surface curves 16 feet downward, relative to a horizontal plane, so although the object has fallen 16 feet, it isn't any closer to the Earth than when it started; and although it is headed downward, relative to its original horizontal motion, so is the surface of the Earth, so the object still appears to be moving horizontally.

In other words, in a circular orbit, the object continually falls toward the Earth, but the Earth's surface falls out from under it, both in distance and direction, in the same way the object is falling, so it doesn't appear to be falling at all.

This is a much more complicated explanation that Newton's mountain and cannonball, but it does have a useful purpose. Namely, let's consider the very last curve, which goes even further to the side than the one that is parallel to the surface of the Earth. You will find that although the object has fallen the same as any of the others, the fact that it is going further to the side means that the Earth's surface falls out from under it, more than it falls. As a result, it is actually further from the Earth, and headed "upward" relative to the surface below it. It seems odd, but it is true, that when you go faster sideways than the orbital speed, even though you are falling as much as ever, you actually seem to rise.

Returning to the Moon falling toward the Earth. During the first part of its path, as it falls it heads more and more downward, but as a result of the pull of gravity, goes faster and faster. By the time it is most (but not all) of the way toward perigee, it is going so fast that even though the orbital speed is higher closer to the Earth, the Moon is moving faster than the orbital speed. As a result, it begins to "rise" relative to the path it was following. Instead of falling downward at a steeper and steeper angle, it begins to pull out of its dive. It is still falling faster and faster, but its increased speed, instead of making it fall "toward" the Earth more, makes it harder for the Earth to "hold onto" it, so it gradually loses its downward motion, and by the time it reaches perigee, it is moving horizontally, exactly the same as it was at apogee.

But whereas at apogee, the Moon was moving horizontally and too slowly to maintain its distance, at perigee, it is moving horizontally and too FAST to maintain its distance. So it continues to "rise", relative to the surface of the Earth, even though it is being pulled down harder than ever by gravity, and is actually falling away from the path it would follow in the absence of gravity, faster than ever. Its too-fast speed causes it to move away from the Earth, back toward apogee. Of course, as it rises, the Earth's downward pull becomes a backward pull, so the Moon slows. But at least at first, it is headed upward too fast for gravity to stop it from rising. So it goes up and up, slower and slower, until it stops going up, at apogee.

The point of the long discussion is that the Moon doesn't exactly slide all the way down to perigee. It is pulled downward, faster and faster, throughout that downward movement; but the fast that it is going faster means that instead of going downward more and more, it starts to pull out of its downward motion partway down, and is moving horizontally again, by the time it reaches perigee. Then, since it is going faster than the orbital speed at that distance (the diagram with two diagonals), it rises, all the way from perigee to apogee, but with everything exactly backward. Instead of speeding up, it slows down. Instead of falling downward, then pulling out of its dive, it rises, then gradually falls to a horizontal direction (at apogee), then continues to fall, as it comes back to the first (downward) part of its path.

I might note that it is inevitable that the upward and downward motions be identical, since the increase of speed due to gravity on the way down is the same as the decrease of speed due to gravity on the way up. However, it is not inevitable that the motion correspond to an ellipse. That only occurs if the way in which the force of gravity changes with distance happens to follow one of two laws. In one, which applies to springs, the force increases uniformly with distance. That leads to elliptical paths centered on the object that is being orbited. In the second, which applies to gravity, the force decreases as the inverse square of the distance. That leads to an elliptical path with the object that is being orbited at the focus (not the center) of the orbit. NO OTHER FORMULA FOR THE FORCE RESULTS IN A REPEATING ORBIT OF ANY SHAPE, let alone an ellipse. It seems quite marvelous that the laws of nature are just what they need to be, to make things work as they do. (Some people would say that's because if they weren't, we wouldn't be here to discuss it. Others just wonder why the laws work so well.)

I hope that clears up the confusion about how the orbits work. Asking you to draw the diagrams involved feels like cheating, but since I can't post a drawing here, and don't want to make you wait while I create a new web page, it seemed the best way to answer the first part of your question (starting with sliding down, then back up again).

Now as to the last part of your question, I'm not exactly sure what you mean by "the Moon traveling at the Earth orbital speed". I think you mean, how did the Moon manage to have the speed required to orbit the Earth? If so, the answer is, it started off that way, 4.5 billion years ago (give or take a few million years), although as it happens, in a much smaller, much faster orbit (tidal slowing, which is a completely different subject, has caused the Moon's orbit to gradually grow over the aeons).

Now, it might seem strange that the Moon started off with a speed capable of keeping it in orbit (note that I did not say with THE speed), but it's actually very easy for that to happen. Right now, the Moon has a particular speed at apogee, which happens to allow it to fall inward to perigee, picking up enough speed so that once there, it rises right back to where it started off (at apogee), with the same position, the same speed, and the same direction of motion, so everything just repeats, the same as before, over and over and over. But suppose that instead of having the speed it does at apogee, it were going 1% slower. Would that cause it to fall into the Earth? No. It wouldn't. All that would happen is, since it is going slower, it would have to fall further, in order to pick up enough speed to pull out of its downward motion. As a result, when it got to the other end of the orbit (perigee), it would be about 1% closer than in its current orbit, and moving about 1% faster than it moves at perigee now. The extra 1% speed would mean that even though it would be closer to the Earth than at its current perigee, it would have plenty of speed to "rise" toward apogee, and return to exactly where it started off. In fact, it wouldn't make any difference if it were to start of 5%, 10%, or even 70% slower than it does at its current apogee. All that would do is move its perigee closer and closer to the Earth, and make its speed at perigee faster and faster, as well. In the process, its orbit would become more elongated, so that instead of being a nearly circular orbit, it would be more obviously eccentric. But as long as perigee was well away from the body of the Earth, it would maintain its elliptical orbit, over and over, the same as now. (This discussion ignores the effect of the Sun, which tends to force the Moon toward a more circular orbit, as a result of the mutual gravitational interactions of the Earth, Moon and Sun; but that "n-body" problem is far more complex, and way beyond anything that could be answered in a forum like this.)

To summarize:

In that part of the orbit which starts with the Moon headed upward, you can just imagine it rising and falling the same as anything thrown upward from the surface of the Earth. But in the part where the downward motion reverses and turns into an upward one, you have to realize that where the actual speed is greater than the orbital speed, even though the Moon is "falling", its too-fast sideways motion makes it "rise" relative to the circular path which represents the surface of the Earth, or any constant distance from that surface.

The Moon has been going around the Earth in a similar (but slightly different) way for the best part of 4.5 billion years. Thanks to various factors, it happens to have a nearly circular orbit. But even if its speed at apogee was only a fraction of the orbital velocity, it would still have a stable elliptical orbit around the Earth, due to the interplay of motion and force discussed in explaining the previous paragraph.

I realize that this answer is even longer and more complicated than the first one, but hopefully by referring to the diagrams on my website, and helping you visualize the diagrams that aren't there, it makes more sense. But again, if you need additional clarification, do not hesitate to ask.

(P.S. For some reason, the spell-checker on AllExperts will not work on my computer. I've tried to be careful about my spelling, but apologize for any errors I failed to notice.)

---------- FOLLOW-UP ----------

QUESTION: Thank you so very much again for helping me understand the relationship between orbital velocity and gravity, I'm pretty sure i understand the concept of varying orbital velocities at apogee and perigee which made it easier to understand why a 70 percent decrease in velocity at apogee would not effect the Moon's orbit of the Earth, just one more question if you would be so kind, just how HUGE would the Moon be in the nights sky at Perigee if there was a 70 percent decrease in orbital velocity and what would be the distance from the Earth? I promise i will leave you be after this last question.  I just have a thirst for knowledge and i am too poor at mathematics to work the answer out for myself.

Thanks again so much for the diagrams and detailed explanations:)

Ian

Answer
It's been more than a decade since anyone asked me to do such a calculation, and with a 70% reduction in velocity at apogee, the perigee moves so far inward that even a small error in the calculations can substantially change the result. As a result, I did the calculation two ways, one of which is supposed to give a correct answer, and the other a limit to the value. Both gave about the same result, so I am reasonably confident that I am not giving you misinformation.

I estimate that the lunar perigee would be about 6600 miles from the center of mass of the Earth-Moon system, which would put the center of the Moon about 5600 miles above the surface of the Earth at the place where the Moon is overhead, and closest to the surface of the Earth. That is about 40 times closer than its usual distance of around 240 thousand miles, so it would be about 20 degrees across in the sky. The main problem with this answer is that if the distance is off by even a thousand miles, it would cause a relative change in distance of nearly 20%, and make the Moon closer to 15 or 25 degrees in size. But no matter how you look at it, HUGE would indeed be the appropriate way to state it.

I might also note that if such a thing could come to pass, the tides the Earth and Moon exert on each other would be increased by more than sixty thousand times. Tides raised on the Earth by the Moon would be nearly 20 miles high, and tides raised on the Moon by the Earth would be nearly 400 miles high. This would cause gigantic earthquakes on the Earth, and probably fracture and possibly tear the Moon to pieces; so it's a good thing that such a thing can't possibly happen.