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Post by wvengineer on May 12, 2015 23:47:44 GMT
It's been quite a while since I took a class in orbital mechanics, so I am having some trouble think about this one.
Setup
You are on a ship in orbit around earth. The orbit is mid to high so that the there is no air drag (like there is on the ISS). For easy math, lets say the orbit is a true circular orbit around the equator. While on an EVA, you throw a ball (or other object) directly to earth. What will happens to it?
I am debating between the following:
A: The earthward vector will cause it to slowly spiral down to earth.
-or-
B: The Earthward Delta V will change the orbit of the ball into one similar the ship, but different. The new orbit would be slightly elliptical in shape and at one or more points will intersect with your current orbit.
What are your thoughts?
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Post by the light works on May 13, 2015 0:35:33 GMT
It's been quite a while since I took a class in orbital mechanics, so I am having some trouble think about this one. Setup You are on a ship in orbit around earth. The orbit is mid to high so that the there is no air drag (like there is on the ISS). For easy math, lets say the orbit is a true circular orbit around the equator. While on an EVA, you throw a ball (or other object) directly to earth. What will happens to it? I am debating between the following: A: The earthward vector will cause it to slowly spiral down to earth. -or- B: The Earthward Delta V will change the orbit of the ball into one similar the ship, but different. The new orbit would be slightly elliptical in shape and at one or more points will intersect with your current orbit. What are your thoughts? can we change that to "launch" instead of throw? the key question in my mind is, does a lower orbit have to have higher velocity or lower velocity to maintain a stable orbit? (and yes, that is a question) |
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Post by Cybermortis on May 13, 2015 22:38:33 GMT
Leaving out the effects of drag at lower altitudes.
The further away you are from the center of whatever you are orbiting the higher the energy needed to maintain that orbit. Imagine swinging a chain around, then stopping and allowing it to wrap around your arm or hand. When it is being swung the end will be moving fairly slowly, but as it wraps around your hand and the chain shortens the speed (or rather number of rotations) increases.
Since we are talking about orbital mechanics this would mean that the ball would start to orbit faster and faster around the Earth in a spiral.
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Post by the light works on May 13, 2015 22:48:09 GMT
Leaving out the effects of drag at lower altitudes. The further away you are from the center of whatever you are orbiting the higher the energy needed to maintain that orbit. Imagine swinging a chain around, then stopping and allowing it to wrap around your arm or hand. When it is being swung the end will be moving fairly slowly, but as it wraps around your hand and the chain shortens the speed (or rather number of rotations) increases. Since we are talking about orbital mechanics this would mean that the ball would start to orbit faster and faster around the Earth in a spiral. I'm not sure the chain model translates. to further clarify my question, assuming both orbital positions are above atmospheric drag: if you have one object in a stable orbit at a given elevation, and you wish to set an object in a lower orbit, will that lower object have to have a higher velocity or a lower velocity to maintain a stable orbit? as I see it, by launching the ball towards earth, you are sending it to a lower orbit with the same velocity. that will either result in it having too high a velocity for its orbit, and the orbit becoming eccentric until it returns to a stable orbital path for its velocity, or having too low a velocity for its orbit and losing altitude until it enters the atmosphere. since most of our spacecraft trigger re-entry by decelerating, I would guess the former is true, but without knowing the relationship between altitude and velocity, I could very well be wrong. |
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Post by GTCGreg on May 13, 2015 23:27:55 GMT
If you were to throw the ball straight toward the Earth, it's orbital velocity would be too high to stay in that lower orbit. It would actually "bounce" back to an orbit slightly higher than your original orbit and then fall back slightly below the original orbit. It would continue this oscillation below and then above the original orbit. This would create an elliptical orbit for the ball with it's average the same as the original orbit before it was thrown.
If you wanted the ball to travel in a lower orbit than you are, you should throw it backwards to the direction you are traveling, not down toward the earth. Throwing it forward would cause it to go to a slightly higher orbit. Throwing it up (away from the Earth) would have the same effect as throwing it down toward the Earth, causing its orbit to become elliptical.
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Post by the light works on May 13, 2015 23:57:12 GMT
If you were to throw the ball straight toward the Earth, it's orbital velocity would be too high to stay in that lower orbit. It would actually "bounce" back to an orbit slightly higher than your original orbit and then fall back slightly below the original orbit. It would continue this oscillation below and then above the original orbit. This would create an elliptical orbit for the ball with it's average the same as the original orbit before it was thrown. If you wanted the ball to travel in a lower orbit than you are, you should throw it backwards to the direction you are traveling, not down toward the earth. Throwing it forward would cause it to go to a slightly higher orbit. Throwing it up (away from the Earth) would have the same effect as throwing it down toward the Earth, causing its orbit to become elliptical. follow up question: would the eccentric orbit essentially result in it orbiting the platform you threw it from? ahead of it as it traversed from lower to higher and slightly behind as it traversed from higher to lower? - I'm thinking of the difference in length of the orbit between the lower orbit and the higher one. |
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Post by wvengineer on May 14, 2015 1:45:39 GMT
The orbit would be around the major gravity well in the area. In this case, the Earth. The gravity from the ship, even though it is much closer is miniscule in comparison.
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Post by the light works on May 14, 2015 2:13:44 GMT
The orbit would be around the major gravity well in the area. In this case, the Earth. The gravity from the ship, even though it is much closer is miniscule in comparison. however, if I am understanding the mechanics correctly, the ball goes faster in relation to the platform when it is in the lower orbit and slower when it is in the higher orbit - the looping around the platform is not due to the platform's gravity, but just the mechanics of its trajectory. |
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Post by GTCGreg on May 14, 2015 2:38:12 GMT
If you were to throw the ball straight toward the Earth, it's orbital velocity would be too high to stay in that lower orbit. It would actually "bounce" back to an orbit slightly higher than your original orbit and then fall back slightly below the original orbit. It would continue this oscillation below and then above the original orbit. This would create an elliptical orbit for the ball with it's average the same as the original orbit before it was thrown. If you wanted the ball to travel in a lower orbit than you are, you should throw it backwards to the direction you are traveling, not down toward the earth. Throwing it forward would cause it to go to a slightly higher orbit. Throwing it up (away from the Earth) would have the same effect as throwing it down toward the Earth, causing its orbit to become elliptical. follow up question: would the eccentric orbit essentially result in it orbiting the platform you threw it from? ahead of it as it traversed from lower to higher and slightly behind as it traversed from higher to lower? - I'm thinking of the difference in length of the orbit between the lower orbit and the higher one. Or will it hit you on its way by. |
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Post by the light works on May 14, 2015 2:56:05 GMT
follow up question: would the eccentric orbit essentially result in it orbiting the platform you threw it from? ahead of it as it traversed from lower to higher and slightly behind as it traversed from higher to lower? - I'm thinking of the difference in length of the orbit between the lower orbit and the higher one. Or will it hit you on its way by. yes. |
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Post by silverdragon on May 14, 2015 5:48:38 GMT
This is a "depends" question in so many ways.
We have all seen meteorites, or witnessed the passing of "shooting star" meteorites as they burn up in earth's atmosphere....
Some make it to earth, some dont.
How fast they are travelling is important, the angle of approach is important, the make-up of the object....
"A Ball", of what?...
A Football would perhaps just explode?...
A Ball of solid ice would melt.
Maybe a ball of pure iron would burn a little, maybe a lot of it would get to earth.
"perfect trajectory", if you throw it right, with enough force, it getting to come in at high speed.
So how will it land.
Getting that perfect trajectory would depend on how you throw it and with what force, yes you would have to throw it somewhat backwards, but also allow for the earths movement around the sun and in space, make allowances for passing of the moon, and a whole lot more....?...
I think this is why the use computers to make that kind of maths for orbiting and landing space craft....
However, I am now wondering and trying to work out, if it was a ball of pure metal iron for example that would survive the burn up through the atmosphere....
If you just threw it directly towards the centre of earth from your own point of view on an orbital craft, just how much would you miss by.
And how much does that depend on in which direction the orbit is....
If the orbit is going forwards on the Earths passage through space, compared to backwards, does that alter which direction the ball would miss?....
I suspect yes.
I am watching this thread with acute curiosity....
Mainly because much of the maths I am using comes from Monty Python.....
So travelling at 40 thousand miles per hour, does a ball kicked out of the back of the spacecraft drop down straight, same as they did with the 30mph launch on the back of that truck.
I am guessing that if you were in geo-stationary orbit, and threw directly towards the earth, by the time the ball got there, the earth would not be there?....
BTW, for anyone who knows that song, and I bet thats all of you, how many red it without singing it in their head?...
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Post by the light works on May 14, 2015 14:10:49 GMT
I think you are adding factors that are not in the original question. the scenario, as I understand it, is that the launch platform is orbiting significantly above the outer boundary of the atmosphere - so as to make atmospheric friction irrelevant to the model.
material of the ball is irrelevant, because the question is not about surviving reentry.
the ball is launched directly towards the center of the earth, in order to release it with the same velocity as the platform but in a smaller orbit.
so the question is, does the ball develop a stable orbit inside the orbit of the platform, does it de-orbit, or does it develop an unstable orbit wobbling between inside the platform's orbit and outside the platform's orbit?
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Post by GTCGreg on May 14, 2015 14:37:09 GMT
I think you are adding factors that are not in the original question. the scenario, as I understand it, is that the launch platform is orbiting significantly above the outer boundary of the atmosphere - so as to make atmospheric friction irrelevant to the model. material of the ball is irrelevant, because the question is not about surviving reentry. the ball is launched directly towards the center of the earth, in order to release it with the same velocity as the platform but in a smaller orbit. so the question is, does the ball develop a stable orbit inside the orbit of the platform, does it de-orbit, or does it develop an unstable orbit wobbling between inside the platform's orbit and outside the platform's orbit? it develops an orbit wobbling between inside the platform's orbit and outside the platform's orbit. But it's not unstable, just elliptical. Many satellites are put in an elliptical orbit on purpose. [/quote] |
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Post by the light works on May 14, 2015 14:39:52 GMT
okay, not a decaying orbit, but not necessarily a perfect ellipse, either.
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Post by GTCGreg on May 14, 2015 14:48:21 GMT
okay, not a decaying orbit, but not necessarily a perfect ellipse, either. Why not a perfect ellipse? I would think it would be about as good an elliptical orbit as your going to get. |
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Post by the light works on May 14, 2015 14:54:04 GMT
okay, not a decaying orbit, but not necessarily a perfect ellipse, either. Why not a perfect ellipse? I would think it would be about as good an elliptical orbit as your going to get. because the wavelength of the wobble would have to be exactly right to get a perfect ellipse. |
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Post by GTCGreg on May 14, 2015 15:01:34 GMT
Why not a perfect ellipse? I would think it would be about as good an elliptical orbit as your going to get. because the wavelength of the wobble would have to be exactly right to get a perfect ellipse. I'm no expert on orbital science, in fact, just thinking about all this gives me a headache, but I believe a perfect ellipse would form naturally. The only thing that may make it more complex is if the ball's mass distribution was off center and it was slowly spinning. |
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Post by the light works on May 14, 2015 15:17:27 GMT
because the wavelength of the wobble would have to be exactly right to get a perfect ellipse. I'm no expert on orbital science, in fact, just thinking about all this gives me a headache, but I believe a perfect ellipse would form naturally. The only thing that may make it more complex is if the ball's mass distribution was off center and it was slowly spinning. I see no reason in the laws of physics why the period of the wobble would be influenced to change to exactly 2 full waves per orbit. now granted, I am not a physicist, but it seems to me if it were so, the moon would only orbit the earth twice per year. |
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Post by GTCGreg on May 14, 2015 15:37:58 GMT
I'm no expert on orbital science, in fact, just thinking about all this gives me a headache, but I believe a perfect ellipse would form naturally. The only thing that may make it more complex is if the ball's mass distribution was off center and it was slowly spinning. I see no reason in the laws of physics why the period of the wobble would be influenced to change to exactly 2 full waves per orbit. now granted, I am not a physicist, but it seems to me if it were so, the moon would only orbit the earth twice per year. I don't know why the elliptical orbit would sync up with the rotation around the earth but from what I've read, it does. I don't have the necessary knowledge to debate that. I could very well be wrong. I'm sure the relative position of the Sun and Moon would also have some effect, but not that great of one. |
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Post by the light works on May 14, 2015 15:53:45 GMT
I see no reason in the laws of physics why the period of the wobble would be influenced to change to exactly 2 full waves per orbit. now granted, I am not a physicist, but it seems to me if it were so, the moon would only orbit the earth twice per year. I don't know why the elliptical orbit would sync up with the rotation around the earth but from what I've read, it does. I don't have the necessary knowledge to debate that. I could very well be wrong. I'm sure the relative position of the Sun and Moon would also have some effect, but not that great of one. I think the period of the oscillation would be more dependent on how hard the ball was launched than anything else. a harder launch would have a longer period (and deeper amplitude) than a weaker launch. |
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