Orbital mechanics: Accelrating takes you into a slower orbit?

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I'm confused by something very simple and stupid. When the mass of a satellite is tiny compared to the body being orbited, we can approximate orbital velocity as sqrt{(GM)/r}

Now its obvious that as r gets larger, the orbital velocity drops. Higher orbits are slower orbits. A geosynchronous satellite orbits at around 11,038kph. But the ISS orbits at about 27,735kph.

All of this makes sense, but here I get to the part I'm being stupid on. To get into a higher orbit, you must accelerate. To get into a lower orbit you must decelerate. How does it happen that you ACCELERATE and end up going SLOWER?

Is this explained by the fact that you accelerate into an elliptical orbit and then must adjust that orbit later (presumably be decelerating?) to make it circular at a higher orbit?

Sorry, its just been too long since I took College physics and I've forgotten IMPORTANT stuff! :)
 

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  • #2
AlephZero
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If you want to get from a circular orbit to another circular orbit with a bigger radius, you have to do some work against the gravitational potential energy field.

You have two basic choices about what to do first, accelerate or decelerate. One of them does positive work on the satellite, the other negative!

Also, you have to accelerate again when you reach the apogee of the elliptical orbit, to get the right speed to convert the ellipse into a circle.
 
  • #3
DaveC426913
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Kilarin:
Accelerating will put you in an elliptical orbit, with perigee at your original altitude and apogee at your new altitude.

But this elliptical orbit, even though it has more potential energy than your smaller, circular one, is slower at its apogee as it climbs. So, you've added energy, but you are going slower at the highest point.

Think about riding a bike up a mountain. You put lots of energy into it. By the time you get to the top, you've gained lots of potential energy, even though you might be barely moving.

In both cases, You've converted your kinetic energy into potential energy.
 
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  • #4
Simon Bridge
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"acceleration" is a special term in physics and refers to rate of change in momentum.
Work out the total momentum in each orbit and, from there, you will be able to tell the average acceleration required to go between them.

It is easier to understand in terms of energy.
The higher orbits require work against gravity to achieve, therefore you have to expend energy.
Your question is, effectively, "how can you accelerate and end up with a slower tangential speed" - the answer is that the extra energy goes into the rest of the orbit.
 
  • #5
rcgldr
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One method used to change from one nearly circular orbital path to another involves acceleration twice, the first to get into an elliptical path, the other at the apogee of the elliptical path to get into a circlar path again. Wiki article:

http://en.wikipedia.org/wiki/Hohmann_transfer_orbit
 
  • #6
DaveC426913
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Is this explained by the fact that you accelerate into an elliptical orbit and then must adjust that orbit later (presumably be decelerating?) to make it circular at a higher orbit?
No. The second burn is also an acceleration. This one you do at apogee.

The trick to changing orbits is when you do your burns.
 

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  • #7
Drakkith
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By acceleration, do we mean that we input energy to climb out of the gravitational well a bit using the thrust from a rocket? Which would mean that you aren't actually gaining speed, but merely countering the effects of gravity?

Similar to speeding up in a car to 80 mph and then keeping a steady velocity up a long hill. You haven't increased your velocity, but you've used energy to gain potential energy. Is that right?
 
  • #8
rcgldr
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By acceleration, do we mean that we input energy to climb out of the gravitational well a bit using the thrust from a rocket? Which would mean that you aren't actually gaining speed, but merely countering the effects of gravity?
At the time thrust is used, the rocket is accelerating to a speed that takes it out of a circular orbit into an eccentrical elliptical orbit. Once at the apogee (peak distance), to prevent returning back "down" the elliptical path, the rocket is accelerated again to achieve a new circular orbit. The wiki link from above explains this, along with some math:

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

It's almost always more efficient to use thrust to increase speed with velocity nearly perpendicular to the pull of gravity, (versus to directly oppose gravity), then let the speed carry the rocket away. The apollo missions used the same method for "translunar injection":

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

flight path angle here is angle above "perpendicular" to gravity:
http://history.nasa.gov/SP-4029/Apollo_18-24_Translunar_Injection.htm

http://www.braeunig.us/apollo/apollo11-TLI.htm

Similar to speeding up in a car to 80 mph and then keeping a steady velocity up a long hill. You haven't increased your velocity, but you've used energy to gain potential energy. Is that right?
More like speeding up to 80 mph, then coasting while going up the hill, where the car gains potential energy, but ends up moving very slow by the time it reaches the top of the hill.
 
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  • #9
DaveC426913
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By acceleration, do we mean that we input energy to climb out of the gravitational well a bit using the thrust from a rocket? Which would mean that you aren't actually gaining speed, but merely countering the effects of gravity?

Similar to speeding up in a car to 80 mph and then keeping a steady velocity up a long hill. You haven't increased your velocity, but you've used energy to gain potential energy. Is that right?
Do you speed up. But the burn only lasts so long. Once the burn ends, you are coasting and slowing under the pull of gravity. Since you haven't reached apogee, you continue to decelerate until you do. At that point, your speed will have dropped below what is was in the lower orbit.

Again, think about coasting along a bike path.
You coast at 20mph.
You decide to take a left turn up a hill.
You accelerate to 30mph, turn upward and coast.
Your speed immediately starts dropping as you climb.
By the time you get 50ft up the hill, your speed has dropped to 10mph.
You turn back downward, but stay coasting.
You accelerate as you come down.
By the time you hit the bottom, you are doing, not 30 but 40mph.

Net result:
You put energy into the system, giving yourself kinetic energy.
You convert that kinetic energy into potential energy - meaning your kinetic energy drops.
The total energy in the system is greater than when you started - even though there are points on your path where you kinetic energy is quite low.
That is the crux of the answer to the OP's question. (Why, do you accelerate to ends up going slower?)

This is where the bike analogy breaks down: To hold with the orbit model, the bike analogy needs to be altered thus: You cannot steer your bike. To stay at the top of the hill you'd need to start pedaling again (adding more energy to the system to maintain your position on the hill).
 
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  • #10
Drakkith
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Ah ok, I see now. You use the thrust to increase your speed, not against gravity, but perpindicular to it. At least in that particular orbital transfer linked above. That makes much more sense now.
 
  • #11
DaveC426913
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Ah ok, I see now. You use the thrust to increase your speed, not against gravity, but perpindicular to it. At least in that particular orbital transfer linked above. That makes much more sense now.
Yes. Your thrust axis is tangential to - and in the direction of - your orbital path.
 
  • #12
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Thank you very much folks, that explains it. I KNEW it was something simple. When you go up in orbit, you are, of COURSE, decelerated by gravity. DUH! You've put energy into the system and you now have more potential energy, but a lower velocity. Makes PERFECT sense. The bicycle analogy was great.

Thank you VERY much.
 
  • #13
rcgldr
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Again, think about coasting along a bike path.
You coast at 20mph. You decide to take a left turn up a hill. You accelerate to 30mph, turn upward and coast. Your speed immediately starts dropping as you climb. By the time you get 50ft up the hill, your speed has dropped to 10mph. You turn back downward, but stay coasting. You accelerate as you come down. By the time you hit the bottom, you are doing, not 30 but 40mph.
Assuming no losses such as drag or rolling resistance, you would need a speed of 38.6742 mph to climb up a 50 ft hill. You stated that the rider accelerated to 30 mph before turning and coasting up the hill. 30 mph would would allow the rider to then coast and climb up a 30 foot hill (30.0864 ft), after which the rider could turn around and coast back up to the orignal 30 mph (assuming no losses). What is the source of the extra 10 mph in your example? Perhaps you meant to state that the rider increased his speed by 10 mph once at the top of the hill with a second impulse of acceleration?

Yes. Your thrust axis is tangential to - and in the direction of - your orbital path.
Note the orbital path isn't circular once thrust is applied. In the case of the table of numbers for the apollo missions listed above, the flight path angle was a bit over 7 degrees "outwards" from being perpendicular to gravity, but I assume the ships orientation and the thrust are kept perpendicular, as if the ship had a bit over 7 degrees "inwards" angle of attack versus it's path during acceleration.
 
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  • #14
Drakkith
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Rcgldr, accelerating up to 30 mph just means that the person will be able to coast up a larger hill than 20mph would allow, or the same hill but have more speed at the top.
 
  • #15
rcgldr
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Rcgldr, accelerating up to 30 mph just means that the person will be able to coast up a larger hill than 20mph would allow, or the same hill but have more speed at the top.
My issue was what is the source of the extra 10 mph at the end of the return trip back down the hill (the 40 mph)? As I just asked, was there a second acceleration at the top of the hill to produce an additional 10 mph?
 
  • #16
Drakkith
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My issue was what is the source of the extra 10 mph at the end of the return trip back down the hill (the 40 mph)? As I just asked, was there a second acceleration at the top of the hill to produce an additional 10 mph?
Ahh, I see the issue. No, there was no 2nd acceleration. I assume Dave just didn't do a thorough calculation.
 
  • #18
DaveC426913
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Assuming no losses such as drag or rolling resistance, you would need a speed of 38.6742 mph to climb up a 50 ft hill. You stated that the rider accelerated to 30 mph before turning and coasting up the hill. 30 mph would would allow the rider to then coast and climb up a 30 foot hill (30.0864 ft), after which the rider could turn around and coast back up to the orignal 30 mph (assuming no losses). What is the source of the extra 10 mph in your example? Perhaps you meant to state that the rider increased his speed by 10 mph once at the top of the hill with a second impulse of acceleration?
You do realize that this is a deliberately loose analogy to do nothing more than show how accelerating can be followed by a slower velocity, yes?

There are no calculations, only round numbers.

The source of the extra 10 mph is the original acceleration.

If a cyclist moving at 20mph turned up the hill without accelerating, he'd get a ways up the hill, then come back down and return to 20mph.

If a cyclist moving at 20mph turned up the hill with accelerating, he'd get a ways up the hill, then come back down and return to 30mph (or whatever).

The hill conserves his energy by converting back and forth between kinetic and potential. His original acceleration to the higher velocity is also conserved.

But the whole point we're trying to show here is that, halfway up the hill, when he turns around, his velocity is much lower - despite the fact that he accelerated to get here. That is the crux of the OP's question.

Note the orbital path isn't circular once thrust is applied. In the case of the table of numbers for the apollo missions listed above, the flight path angle was a bit over 7 degrees "outwards" from being perpendicular to gravity, but I assume the ships orientation and the thrust are kept perpendicular, as if the ship had a bit over 7 degrees "inwards" angle of attack versus it's path during acceleration.
Again. The point isn't tangential versus 7 degrees off; the point is tangential versus perpendicular. We're looking for qualitative differences, not quantitative differences.
 
  • #19
rcgldr
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Again, think about coasting along a bike path. You coast at 20mph. You decide to take a left turn up a hill. You accelerate to 30mph, turn upward and coast. Your speed immediately starts dropping as you climb. By the time you get 50ft up the hill, your speed has dropped to 10mph. You turn back downward, but stay coasting. You accelerate as you come down. By the time you hit the bottom, you are doing, not 30 but 40mph.
What is the source of the extra 10 mph in your example? Perhaps you meant to state that the rider increased his speed by 10 mph once at the top of the hill with a second impulse of acceleration?
You do realize that this is a deliberately loose analogy to do nothing more than show how accelerating can be followed by a slower velocity, yes? If a cyclist moving at 20mph turned up the hill with accelerating, he'd get a ways up the hill, then come back down and return to 30mph (or whatever).
My concern was how the bicyclist ended up at 40 mph on return instead of the 30 mph he accelerated to before going up the hill.

Note the orbital path isn't circular once thrust is applied. In the case of the table of numbers for the apollo missions listed above, the flight path angle was a bit over 7 degrees "outwards" from being perpendicular to gravity, but I assume the ships orientation and the thrust are kept perpendicular, as if the ship had a bit over 7 degrees "inwards" angle of attack versus it's path during acceleration.
Again. The point isn't tangential versus 7 degrees off; the point is tangential versus perpendicular. We're looking for qualitative differences, not quantitative differences.
I wasn't disagreeing with you here. My assumption that thrust was always perpendicular (or at least nearly so) to the instantaneous direction of gravity, and the rate of increase in velocity accounted for the 7 degree "outward" direction of the path. However I'm not sure of this and was hoping someone reading this thread would know if this was the case.

In the case where the change in velocity is instantaneous, the path instantanly changes from circular to elliptical. You need at least a second change in velocity to end up with a circular path again, which is what I was getting at.
 
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  • #20
DaveC426913
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My concern was how the bicyclist ended up at 40 mph on return instead of the 30 mph he accelerated to before going up the hill.
Ah. That is odd. He would have come back down doing 30. Apologies.

I was messing about with the numbers, partly because it is not all that good an analogy. (On a bike, your altitude is not inextricably linked to your velocity, as it is in an orbit, so it's a bit awkward to show how the initial acceleration would have immediately started you up the hill).



I wasn't disagreeing with you here. My assumption that thrust was always perpendicular (or at least nearly so) to the instantaneous direction of gravity, and the rate of increase in velocity accounted for the 7 degree "outward" direction of the path. However I'm not sure of this and was hoping someone reading this thread would know if this was the case.

In the case where the change in velocity is instantaneous, the path instantanly changes from circular to elliptical. You need at least a second change in velocity to end up with a circular path again, which is what I was getting at.
Ah. I thought you were just being nitpicky about the angle of the thrust force.

Thing I'm finding is that the OP's question really gets answered before even getting to the second burn. While in the elliptical orbit, the craft has undergone acceleration to reach a higher velocity, yet has ended up at a lower velocity - intuitively paradoxical until you realize the kinetic versus potential energy trade-off. This - to me - was the crux of the OP's confusion.
 

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