3 ms·
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 cir
by 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.