5 ms·

Landing on the Moon without a Rocket

2 points by gusgordon 14y ago

lutusp 14y ago
There's a problem with this idea -- as you decelerate out of orbit, in order to reduce the craft's velocity so the inflated-bag method will work, the craft's angle with respect to the surface naturally changes and becomes nearly vertical. This is unavoidable. Remember the early Mars missions that used the inflated-bag method? They had to expend a lot of fuel to slow down enough to safely approach the surface, at which point the air-bag method was used (and worked). But this only reduced the amount of fuel required to get to the surface, it didn't significantly change that cost and burden. Just do some numerical orbital simulations -- see what happens when you slowly reduce the velocity and altitude of a craft orbiting the moon so it is able to graze the surface. See what happens when you impact the surface at shallow angles. Hint: the craft's velocity is still extremely high and the air bags have to absorb too much energy -- they will self-destruct. In a hypothetical approach to a perfectly smooth plains area (the moon has plenty of those), in which there are no altitude irregularities and the craft can approach at nearly orbital speed and impact the surface at a very shallow angle, the craft would begin to roll along the surface at such a high velocity that the centripetal forces would tear the craft apart. Speaking as a former NASA Space Shuttle engineer.
- ColinWright 14y ago
  They do specifically say that they're intending to come in at 0.02 degrees or so, which seems to indicate they've considered this. Personally, and not speaking as a former NASA Space Shuttle engineer, it seems unlikely that they'll find ground flat enough for this to work. They mention mountains, but even just craters will provide far, far more variation than 0.02 degrees, so they look like they'll skim the surface nicely, then slam into a wall. But they're not stupid - I wonder what they're not telling us.
  - lutusp 14y ago
    > They do specifically say that they're intending to come in at 0.02 degrees or so, which seems to indicate they've considered this. That won't prevent the craft from contacting the surface at very high speed and beginning to roll. As soon as the roll starts (as soon as linear velocity becomes rotational velocity), the velocity will decline further, which will change the angle of attack and increase the conversion of orbital kinetic energy into centripetal force. Believe me, the craft will disintegrate if it touches the surface at near-orbital speeds, which is implied by the proposed angle of attack. > Personally, and not speaking as a former NASA Space Shuttle engineer, it seems unlikely that they'll find ground flat enough for this to work. Yes, but my reason for stating it like that was to focus on one factor and eliminate others for the moment. > But they're not stupid - I wonder what they're not telling us. I think the author is someone who thinks about these problems, but not with the benefit of experience in orbital analysis or intuition about the kinds of problems that come up.
- gusgordon 14y ago
  With an extremely low average thrust, say, .5 N, the craft would remain in near circular orbit. Maybe you have a different idea of "low thrust", though. The biggest problem seems like it would be the centrifugal force. More simulations would have to be done, but there should be lots of frictional losses, not to mention the kinetic energy lost that goes into generating the rotational energy. It might be possible to overcome this force, especially with a smaller payload.
  - lutusp 14y ago
    > With an extremely low average thrust, say, .5 N, the craft would remain in near circular orbit. Excuse me? To remain in a circular orbit over the airless moon, no thrust is required at all. Read this for a full explanation: http://arachnoid.com/gravitation_equations/orbital_mechanics.html http://arachnoid.com/gravitation_equations/orbital_mechanics... > The biggest problem seems like it would be the centrifugal force. Not centrifugal force (a fictional force) but centripetal force, the force produced by the craft's parts being forced to turn against a natural tendency to move away in a straight line. Which, given the problem's velocity, they will begin to do in short order after tearing themselves to pieces. > It might be possible to overcome this force, especially with a smaller payload. Not in the described scenario -- even a tiny payload would disintegrate upon touching the surface at orbital velocity. Consider that the orbital velocity for a lunar circular orbit is 1.2 km/s or almost 2700 miles per hour -- that would be Mach 3.5 if there were any air available to give that meaning. Given the proposed approach angle, the contact velocity is nearly the same as the circular orbit velocity given above. There's no way that a craft can safely dissipate that kind of kinetic energy by simply rolling along the lunar surface. It's way too fast, too much kinetic energy, the idea isn't practical. Sorry, but physics and mathematics say no.
    - nessus42 14y ago
      > Not centrifugal force (a fictional force) but centripetal force So-called "fictional forces" are used all the time to simplify calculations. Furthermore, if you are going to be pedantic, in GR centrifugal force is just as real as gravitational force. Of course, in GR, gravitational force is also a fictional force. But one makes few friends by constantly pointing this out whenever anyone says "gravitational force".
deleted 14y ago
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2 points by gusgordon 14y ago

There's a problem with this idea -- as you decelerate out of orbit, in order to reduce the craft's velocity so the inflated-bag method will work, the craft's angle with respect to the surface naturally changes and becomes nearly vertical. This is unavoidable. Remember the early Mars missions that used the inflated-bag method? They had to expend a lot of fuel to slow down enough to safely approach the surface, at which point the air-bag method was used (and worked). But this only reduced the amount of fuel required to get to the surface, it didn't significantly change that cost and burden. Just do some numerical orbital simulations -- see what happens when you slowly reduce the velocity and altitude of a craft orbiting the moon so it is able to graze the surface. See what happens when you impact the surface at shallow angles. Hint: the craft's velocity is still extremely high and the air bags have to absorb too much energy -- they will self-destruct. In a hypothetical approach to a perfectly smooth plains area (the moon has plenty of those), in which there are no altitude irregularities and the craft can approach at nearly orbital speed and impact the surface at a very shallow angle, the craft would begin to roll along the surface at such a high velocity that the centripetal forces would tear the craft apart. Speaking as a former NASA Space Shuttle engineer.

ColinWright 14y ago
They do specifically say that they're intending to come in at 0.02 degrees or so, which seems to indicate they've considered this. Personally, and not speaking as a former NASA Space Shuttle engineer, it seems unlikely that they'll find ground flat enough for this to work. They mention mountains, but even just craters will provide far, far more variation than 0.02 degrees, so they look like they'll skim the surface nicely, then slam into a wall. But they're not stupid - I wonder what they're not telling us.
- lutusp 14y ago
  > They do specifically say that they're intending to come in at 0.02 degrees or so, which seems to indicate they've considered this. That won't prevent the craft from contacting the surface at very high speed and beginning to roll. As soon as the roll starts (as soon as linear velocity becomes rotational velocity), the velocity will decline further, which will change the angle of attack and increase the conversion of orbital kinetic energy into centripetal force. Believe me, the craft will disintegrate if it touches the surface at near-orbital speeds, which is implied by the proposed angle of attack. > Personally, and not speaking as a former NASA Space Shuttle engineer, it seems unlikely that they'll find ground flat enough for this to work. Yes, but my reason for stating it like that was to focus on one factor and eliminate others for the moment. > But they're not stupid - I wonder what they're not telling us. I think the author is someone who thinks about these problems, but not with the benefit of experience in orbital analysis or intuition about the kinds of problems that come up.
gusgordon 14y ago
With an extremely low average thrust, say, .5 N, the craft would remain in near circular orbit. Maybe you have a different idea of "low thrust", though. The biggest problem seems like it would be the centrifugal force. More simulations would have to be done, but there should be lots of frictional losses, not to mention the kinetic energy lost that goes into generating the rotational energy. It might be possible to overcome this force, especially with a smaller payload.
- lutusp 14y ago
  > With an extremely low average thrust, say, .5 N, the craft would remain in near circular orbit. Excuse me? To remain in a circular orbit over the airless moon, no thrust is required at all. Read this for a full explanation: http://arachnoid.com/gravitation_equations/orbital_mechanics.html http://arachnoid.com/gravitation_equations/orbital_mechanics... > The biggest problem seems like it would be the centrifugal force. Not centrifugal force (a fictional force) but centripetal force, the force produced by the craft's parts being forced to turn against a natural tendency to move away in a straight line. Which, given the problem's velocity, they will begin to do in short order after tearing themselves to pieces. > It might be possible to overcome this force, especially with a smaller payload. Not in the described scenario -- even a tiny payload would disintegrate upon touching the surface at orbital velocity. Consider that the orbital velocity for a lunar circular orbit is 1.2 km/s or almost 2700 miles per hour -- that would be Mach 3.5 if there were any air available to give that meaning. Given the proposed approach angle, the contact velocity is nearly the same as the circular orbit velocity given above. There's no way that a craft can safely dissipate that kind of kinetic energy by simply rolling along the lunar surface. It's way too fast, too much kinetic energy, the idea isn't practical. Sorry, but physics and mathematics say no.
  - nessus42 14y ago
    > Not centrifugal force (a fictional force) but centripetal force So-called "fictional forces" are used all the time to simplify calculations. Furthermore, if you are going to be pedantic, in GR centrifugal force is just as real as gravitational force. Of course, in GR, gravitational force is also a fictional force. But one makes few friends by constantly pointing this out whenever anyone says "gravitational force".

deleted 14y ago

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