4 ms·

http://what-if.xkcd.com/58/ http://what-if.xkcd.com/58/ This xkcd explains this really well. At orbital height, Earth's gravity's pull is still a full 90% of w

http://what-if.xkcd.com/58/ http://what-if.xkcd.com/58/ This xkcd explains this really well. At orbital height, Earth's gravity's pull is still a full 90% of what we experience on the surface. Orbiting objects "dodge" this fall by going really fucking fast. I think I saw a comparison once that it takes about 15~20X more energy to attain orbit speed as it does to attain orbital height.

deleted 13y ago
[deleted]
barking 13y ago
So is this centrifugal force counteracting gravity? So then gravity would be like your hand's grip preventing the bucket from flying away when you swing a bucket of water? If this is the case then I think I'd understand the way planets don't fall into the sun better. The way I've heard it described before as something like 'forever falling into the sun' confusing.
- jacquesm 13y ago
  It's actually a very gradual spiraling inward. This is because the momentum imparted initially is slowly dissipated (by impacts & interaction with the gravitational field of the sun), an extremely small factor. The question is whether the sun will go nova before or after we (or our sterilized remains) finally arrive :) That's not the only effect, all of this is counteracted by the decrease in mass of the sun and the tidal interaction between the sun and its planets (which slows down the rotation of the sun), both of these should push the earth outward from the sun. I'm not sure how it all works out (whether it is a net positive or a net negative) but it can't be much or we'd have drifted out of the zone where life is sustainable long ago.
  - barking 13y ago
    Thanks. I can understand the mass of the sun having an effect but why would the speed of rotation of the sun have an effect on earth's orbit?
    - jacquesm 13y ago
      I think that's this effect: http://en.wikipedia.org/wiki/Tidal_acceleration http://en.wikipedia.org/wiki/Tidal_acceleration Speaking about the rotation of a gaseous body is a bit tricky anyway, not all parts of the sun rotate equally fast. (inside/outside, equatorial/polar).
      - barking 13y ago
        That's a very interesting link, the diagram of the earth moon system explains it quite well, I think. I just googled and found this article that reports the same effect between the earth and the sun http://www.newscientist.com/article/dn17228-why-is-the-earth-moving-away-from-the-sun.html#.UjbUUX_JLIc http://www.newscientist.com/article/dn17228-why-is-the-earth...
  - deleted 13y ago
    [deleted]
  - jlgreco 13y ago
    Tidal forces between the Earth and its Moon in fact result in the Moon (very slowly) getting further and further away from the earth.
    - jacquesm 13y ago
      Yep. The actual experimental verification of this makes for very interesting reading. see: http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment and lots of links referenced from there. Amazing precision, I still find it quite incredible that such a distance could be reliably measured to such precision and that meaningful conclusions could be drawn from those measurements.
- jonmrodriguez 13y ago
  Yes, I think you're correct
  - barking 13y ago
    Yay me!
- jlgreco 13y ago
  > So is this centrifugal force counteracting gravity? I think that is probably a relatively reasonable way of considering it, but to be clear, centrifugal force isn't actually something that exists. It wouldn't be drawn in a free body diagram unless the diagram was drawn with a non-inertial reference frame, but even then it is only drawn so that equations meant for inertial reference frames will still work. (http://en.wikipedia.org/wiki/Centrifugal_force_(rotating_reference_frame) http://en.wikipedia.org/wiki/Centrifugal_force_(rotating_ref...) I think the best way to get an intuitive understanding of how orbit works is with Newton's cannonball thought experiment, and perhaps playing around with a simulation of it.
  - barking 13y ago
    Thanks I looked it up on wikipedia and they have a nice simulation there. http://en.wikipedia.org/wiki/Newton%27s_cannonball http://en.wikipedia.org/wiki/Newton%27s_cannonball I didn't understand the elliptical orbit at first and was writing a supplementary question here but as i did so the following explanation occurred to me. As it spirals outward it's no longer travelling at right angles to the force of gravity so now gravity exerts some force parallel and opposite to its direction of travel and starts to slow it down. It slows down at a rate greater than the required orbital speed does and eventually it's travelling at less than orbital speed and starts to travelling closer to earth again and now gravity starts to act to increase its speed because the ball is again not travelling perpendicular to the force of gravity but this time gravity's pull increases its speed. Now the ball's speed increases faster than the required orbital speed does until it exceeds orbital speed again. And so ad infinatum!
ColinWright 13y ago
Extremely simple sums tells you orbital kinetic energy is about 10 times gravitational potential energy. Orbital velocity is about 8 km/s [0], so KE/kg is 1/2 v^2 which is about 32 * 10^6. At 320 km altitude the PE/kg is about g.h = 10.320.1000 = 32 * 10^5. So getting up to the right height is about one tenth of the energy needed. [0] You can compute this using Pythagoras and the fact that the Earth's radius is about 6.4 * 10^6 meters.
- Gravityloss 13y ago
  One can also look at it this way, though it does not explain the energies' relation to each other: The total gravitational potential energy you start from, from Earth's surface, is far from zero, as you are already at over 6000 km height. The 300 km height is a small blip on top of that. A 5 % change. The velocity you start with is only the spinning of the earth (lower velocity closer to poles, vector sum too because you must launch to an inclined orbit from there) which is about 0.4 km/s. So you need a 20x change. (Coincidentally reciprocal of 5%.)
  - ColinWright 13y ago
    But that's really misleading, as the energy involved in increases in velocity go as a square, whereas increases in height are linear. In fact, if you go to 600 km height then you double the PE change, but don't much change the KE requirement. I'm not sure your comment would really help someone who doesn't already know what's going on, but I'd appreciate replies from people who are trying to understand more about this. Maybe I'm just the wrong audience.
    - lutusp 13y ago
      The easy way to solve this class of problem is to compute two quantities -- one, the object's energy at rest on earth's surface, and two, the object's energy when it gets into orbit. The first is mostly gravitational potential energy (GPE), the second is a mixture of GPE and kinetic energy (KE). They're both easy to compute. Finally, to figure out how much energy is required to go from the launch pad to orbit, simply subtract one energy number from the other, remembering that the total energy is the sum of KE and GPE. A complicating factor is the fact that a rocket's mass decreases as it ascends (as the fuel is burned), but this isn't a deal-breaker -- all one needs to do is profile the ascent, the increase in velocity alongside the decrease in mass, using a numerical integral. That's not difficult at all (and it has to be a numerical integral). More here: http://arachnoid.com/gravitation_equations/orbital_mechanics.html http://arachnoid.com/gravitation_equations/orbital_mechanics...
6ren 13y ago
Another way to visualize this is if you had a perfectly flat planet with no atmosphere, you could orbit at head height. You could sort of do this on the moon, but mountains.
- jlgreco 13y ago
  > You could sort of do this on the moon, but mountains. Here is am image of the moon, taken from orbit around it, a mere 28.4 miles above the surface: http://neverworld.net/lunar/lo2_h162_3a-m.jpg http://neverworld.net/lunar/lo2_h162_3a-m.jpg
tocomment 13y ago
Along this line of thinking I was wondering how much energy it would take to reach the height of one of the lagrangian points? And once we get there would it be possible to stay there and resist the pull of earth? If so that would be cheap space access right there.
- tocomment 13y ago
  Seriously a downvote? Why not explain why it's wrong?
  - metageek 13y ago
    IIRC, it takes just as much energy to reach L4 (or L5) as to reach the Moon--they're in the same orbit. If you've got settlers on the Moon, or beyond the Earth-Moon system, L4 is probably a good place to build a commercial and industrial outpost; but, for your only space installation, something lower down is much cheaper.
    - tocomment 13y ago
      I'm talking about earth sun Lagrange points. Those wouldn't be orbiting the moon.

deleted 13y ago

[deleted]

barking 13y ago

So is this centrifugal force counteracting gravity? So then gravity would be like your hand's grip preventing the bucket from flying away when you swing a bucket of water? If this is the case then I think I'd understand the way planets don't fall into the sun better. The way I've heard it described before as something like 'forever falling into the sun' confusing.

jacquesm 13y ago
It's actually a very gradual spiraling inward. This is because the momentum imparted initially is slowly dissipated (by impacts & interaction with the gravitational field of the sun), an extremely small factor. The question is whether the sun will go nova before or after we (or our sterilized remains) finally arrive :) That's not the only effect, all of this is counteracted by the decrease in mass of the sun and the tidal interaction between the sun and its planets (which slows down the rotation of the sun), both of these should push the earth outward from the sun. I'm not sure how it all works out (whether it is a net positive or a net negative) but it can't be much or we'd have drifted out of the zone where life is sustainable long ago.
- barking 13y ago
  Thanks. I can understand the mass of the sun having an effect but why would the speed of rotation of the sun have an effect on earth's orbit?
  - jacquesm 13y ago
    I think that's this effect: http://en.wikipedia.org/wiki/Tidal_acceleration http://en.wikipedia.org/wiki/Tidal_acceleration Speaking about the rotation of a gaseous body is a bit tricky anyway, not all parts of the sun rotate equally fast. (inside/outside, equatorial/polar).
    - barking 13y ago
      That's a very interesting link, the diagram of the earth moon system explains it quite well, I think. I just googled and found this article that reports the same effect between the earth and the sun http://www.newscientist.com/article/dn17228-why-is-the-earth-moving-away-from-the-sun.html#.UjbUUX_JLIc http://www.newscientist.com/article/dn17228-why-is-the-earth...
- deleted 13y ago
  [deleted]
- jlgreco 13y ago
  Tidal forces between the Earth and its Moon in fact result in the Moon (very slowly) getting further and further away from the earth.
  - jacquesm 13y ago
    Yep. The actual experimental verification of this makes for very interesting reading. see: http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment and lots of links referenced from there. Amazing precision, I still find it quite incredible that such a distance could be reliably measured to such precision and that meaningful conclusions could be drawn from those measurements.
jonmrodriguez 13y ago
Yes, I think you're correct
- barking 13y ago
  Yay me!
jlgreco 13y ago
> So is this centrifugal force counteracting gravity? I think that is probably a relatively reasonable way of considering it, but to be clear, centrifugal force isn't actually something that exists. It wouldn't be drawn in a free body diagram unless the diagram was drawn with a non-inertial reference frame, but even then it is only drawn so that equations meant for inertial reference frames will still work. (http://en.wikipedia.org/wiki/Centrifugal_force_(rotating_reference_frame) http://en.wikipedia.org/wiki/Centrifugal_force_(rotating_ref...) I think the best way to get an intuitive understanding of how orbit works is with Newton's cannonball thought experiment, and perhaps playing around with a simulation of it.
- barking 13y ago
  Thanks I looked it up on wikipedia and they have a nice simulation there. http://en.wikipedia.org/wiki/Newton%27s_cannonball http://en.wikipedia.org/wiki/Newton%27s_cannonball I didn't understand the elliptical orbit at first and was writing a supplementary question here but as i did so the following explanation occurred to me. As it spirals outward it's no longer travelling at right angles to the force of gravity so now gravity exerts some force parallel and opposite to its direction of travel and starts to slow it down. It slows down at a rate greater than the required orbital speed does and eventually it's travelling at less than orbital speed and starts to travelling closer to earth again and now gravity starts to act to increase its speed because the ball is again not travelling perpendicular to the force of gravity but this time gravity's pull increases its speed. Now the ball's speed increases faster than the required orbital speed does until it exceeds orbital speed again. And so ad infinatum!

ColinWright 13y ago

Extremely simple sums tells you orbital kinetic energy is about 10 times gravitational potential energy. Orbital velocity is about 8 km/s [0], so KE/kg is 1/2 v^2 which is about 32 * 10^6. At 320 km altitude the PE/kg is about g.h = 10.320.1000 = 32 * 10^5. So getting up to the right height is about one tenth of the energy needed. [0] You can compute this using Pythagoras and the fact that the Earth's radius is about 6.4 * 10^6 meters.

Gravityloss 13y ago
One can also look at it this way, though it does not explain the energies' relation to each other: The total gravitational potential energy you start from, from Earth's surface, is far from zero, as you are already at over 6000 km height. The 300 km height is a small blip on top of that. A 5 % change. The velocity you start with is only the spinning of the earth (lower velocity closer to poles, vector sum too because you must launch to an inclined orbit from there) which is about 0.4 km/s. So you need a 20x change. (Coincidentally reciprocal of 5%.)
- ColinWright 13y ago
  But that's really misleading, as the energy involved in increases in velocity go as a square, whereas increases in height are linear. In fact, if you go to 600 km height then you double the PE change, but don't much change the KE requirement. I'm not sure your comment would really help someone who doesn't already know what's going on, but I'd appreciate replies from people who are trying to understand more about this. Maybe I'm just the wrong audience.
  - lutusp 13y ago
    The easy way to solve this class of problem is to compute two quantities -- one, the object's energy at rest on earth's surface, and two, the object's energy when it gets into orbit. The first is mostly gravitational potential energy (GPE), the second is a mixture of GPE and kinetic energy (KE). They're both easy to compute. Finally, to figure out how much energy is required to go from the launch pad to orbit, simply subtract one energy number from the other, remembering that the total energy is the sum of KE and GPE. A complicating factor is the fact that a rocket's mass decreases as it ascends (as the fuel is burned), but this isn't a deal-breaker -- all one needs to do is profile the ascent, the increase in velocity alongside the decrease in mass, using a numerical integral. That's not difficult at all (and it has to be a numerical integral). More here: http://arachnoid.com/gravitation_equations/orbital_mechanics.html http://arachnoid.com/gravitation_equations/orbital_mechanics...

6ren 13y ago

Another way to visualize this is if you had a perfectly flat planet with no atmosphere, you could orbit at head height. You could sort of do this on the moon, but mountains.

jlgreco 13y ago
> You could sort of do this on the moon, but mountains. Here is am image of the moon, taken from orbit around it, a mere 28.4 miles above the surface: http://neverworld.net/lunar/lo2_h162_3a-m.jpg http://neverworld.net/lunar/lo2_h162_3a-m.jpg

tocomment 13y ago

Along this line of thinking I was wondering how much energy it would take to reach the height of one of the lagrangian points? And once we get there would it be possible to stay there and resist the pull of earth? If so that would be cheap space access right there.

tocomment 13y ago
Seriously a downvote? Why not explain why it's wrong?
- metageek 13y ago
  IIRC, it takes just as much energy to reach L4 (or L5) as to reach the Moon--they're in the same orbit. If you've got settlers on the Moon, or beyond the Earth-Moon system, L4 is probably a good place to build a commercial and industrial outpost; but, for your only space installation, something lower down is much cheaper.
  - tocomment 13y ago
    I'm talking about earth sun Lagrange points. Those wouldn't be orbiting the moon.