3 ms·

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 bi

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".
  - lutusp 14y ago
    > So-called "fictional forces" are used all the time to simplify calculations. Consider the context -- someone wants to soft-land on the moon at 2700 miles per hour. Under those circumstances, I suggest avoiding any other gratuitous fictions until more has been learned. > But one makes few friends by constantly pointing this out whenever anyone says "gravitational force". Yes, unless you're instructing a student. Someone who plans to slide into home base at 1.23 km/s is a student and should receive the pedantic version.
    - nessus42 14y ago
      > Consider the context In this context, I'm sure that determining the "centrifugal force" at each point on the inside of the ball to calculate the pressure on the outside wall, and comparing that to the tensile strength of the outside wall, should work just fine. At least for that one aspect of the problem.
      - lutusp 14y ago
        > In this context, I'm sure that determining the "centrifugal force" at each point on the inside of the ball to calculate the pressure on the outside wall ... That's a centripetal force -- the unified wall's resistance to its component parts' tendency to move away in a straight line. Source: http://en.wikipedia.org/wiki/Centripetal_force http://en.wikipedia.org/wiki/Centripetal_force Quote: "Centripetal force (from Latin centrum "center" and petere "to seek") is a force that makes a body follow a curved path: its direction is always orthogonal to the velocity of the body, toward the fixed point of the instantaneous center of curvature of the path. Centripetal force is generally the cause of circular motion." Perhaps in this case, argument for argument's sake isn't constructive. > At least for that one aspect of the problem. As is often the case when quoting fictitious forces, it can only confuse the student. Maybe as a side effect, it might make him think he can land an intact spacecraft on the moon by sliding to a stop.
        nessus42 14y ago
        > That's a centripetal force It's also centrifugal force, if you are working in the rotating reference frame of the ball, which in the case of calculating the internal pressure, is probably a bit easier to do. > it can only confuse the student. I'm not confused.
- gusgordon 14y ago
  I meant .5 N average thrust in the direction opposite the velocity vector. Energy would slowly be lost and the craft would remain in something very simar to circular orbit, ie. it would hit the surface at a near zero angle.
  - lutusp 14y ago
    > Energy would slowly be lost and the craft would remain in something very simar to circular orbit, ie. it would hit the surface at a near zero angle. Yes, but at 1.2 km/s -- way, way too fast for anything but complete destruction of the craft. The craft approaches the surface at an angle near zero. The craft's lower extent begins to rub against the surface and (assuming a perfectly flat terrain) begins to dissipate energy through friction, and by "rolling" -- by converting linear velocity into rotational velocity. The lost energy causes the angle of attack to increase and push the craft into the surface more forcefully. This increases heating and rotational velocity. Within seconds, the combination of high rotational velocity and frictional heating causes the craft to disintegrate, to fly apart. 1.2 km/s is just too high a velocity with which to approach the moon's surface.
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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".
- lutusp 14y ago
  > So-called "fictional forces" are used all the time to simplify calculations. Consider the context -- someone wants to soft-land on the moon at 2700 miles per hour. Under those circumstances, I suggest avoiding any other gratuitous fictions until more has been learned. > But one makes few friends by constantly pointing this out whenever anyone says "gravitational force". Yes, unless you're instructing a student. Someone who plans to slide into home base at 1.23 km/s is a student and should receive the pedantic version.
  - nessus42 14y ago
    > Consider the context In this context, I'm sure that determining the "centrifugal force" at each point on the inside of the ball to calculate the pressure on the outside wall, and comparing that to the tensile strength of the outside wall, should work just fine. At least for that one aspect of the problem.
    - lutusp 14y ago
      > In this context, I'm sure that determining the "centrifugal force" at each point on the inside of the ball to calculate the pressure on the outside wall ... That's a centripetal force -- the unified wall's resistance to its component parts' tendency to move away in a straight line. Source: http://en.wikipedia.org/wiki/Centripetal_force http://en.wikipedia.org/wiki/Centripetal_force Quote: "Centripetal force (from Latin centrum "center" and petere "to seek") is a force that makes a body follow a curved path: its direction is always orthogonal to the velocity of the body, toward the fixed point of the instantaneous center of curvature of the path. Centripetal force is generally the cause of circular motion." Perhaps in this case, argument for argument's sake isn't constructive. > At least for that one aspect of the problem. As is often the case when quoting fictitious forces, it can only confuse the student. Maybe as a side effect, it might make him think he can land an intact spacecraft on the moon by sliding to a stop.
      - nessus42 14y ago
        > That's a centripetal force It's also centrifugal force, if you are working in the rotating reference frame of the ball, which in the case of calculating the internal pressure, is probably a bit easier to do. > it can only confuse the student. I'm not confused.
gusgordon 14y ago
I meant .5 N average thrust in the direction opposite the velocity vector. Energy would slowly be lost and the craft would remain in something very simar to circular orbit, ie. it would hit the surface at a near zero angle.
- lutusp 14y ago
  > Energy would slowly be lost and the craft would remain in something very simar to circular orbit, ie. it would hit the surface at a near zero angle. Yes, but at 1.2 km/s -- way, way too fast for anything but complete destruction of the craft. The craft approaches the surface at an angle near zero. The craft's lower extent begins to rub against the surface and (assuming a perfectly flat terrain) begins to dissipate energy through friction, and by "rolling" -- by converting linear velocity into rotational velocity. The lost energy causes the angle of attack to increase and push the craft into the surface more forcefully. This increases heating and rotational velocity. Within seconds, the combination of high rotational velocity and frictional heating causes the craft to disintegrate, to fly apart. 1.2 km/s is just too high a velocity with which to approach the moon's surface.
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