I was thinking about about the Google x-prize and wondered if the simplest way to put a rover on the moon is to crash land it.
So the scenario is send a probe at the moon with just enough velocity to get over the Lagrange point and freefall onto the moon.
Is this possible without course correction ? And would you be going slow enough for a hardened spacecraft/rover to survive ?
If you just turned off the engines after you cross the point of gravitational equilibrium, wouldn’t you accelerate all the way to the Moon, and impact at a very high rate of speed? With no atmosphere to create drag, there is no terminal velocity, so this could be thousands of miles per hour.
No atmosphere, but gravity is only one sixth. It depends how far the L1 is from the moon, that I don’t know.
That’s exactly what would happen.
If you imagine the return trip, it’s easy to intuit that it wouldn’t have been a good idea for the Apollo missions to just cut power 150,000 or so miles out and let gravity steer them home.
An object falling to the moon from a remote distance would be accelerated to the moon’s escape velocity which is 2.38 kilometers per second. I don’t think an airbag would suffice for such an impact.
Figuring out the velocity is just a matter of calculating the difference in gravitational potential energy (due to both the Earth and the Moon) between the L1 point and the Moon’s surface; the difference between the two will be equal to the kinetic energy your hypothetical probe has when it smashes into the Moon’s surface. If I’ve done my math correctly, I get a result of about 2.28 kilometers/second, or a little over 5000 miles per hour. Better bring a helmet.
To be fair, you’d be worse off trying to do this in the other direction, since the Earth’s gravity well is “deeper” than the Moon’s and would therefore cause your probe to be going even faster when it hits atmo. Without the atmosphere you’d be impacting the surface at about 11.1 km/sec, pretty damn close to Earth’s escape velocity (cf. Lumpy.)
L1 isn’t quite “remote” enough from the Moon’s surface for that assumption to be valid, so that shaves a few percent off your impact velocity. Still not all that survivable, though.
Take a look at the Ranger missions:
http://nssdc.gsfc.nasa.gov/planetary/lunar/ranger.html
Ranger 1 through 6 failed, 7,8 and 9 returned pix and hit the moon.
The later did not survive impact.
Lunar escape velocity is about 2.37 km/sec., so your slightly lower number sounds about right.
Insert obligatory nitpick here that it’s escape speed, not escape velocity.
No it’s escape velocity not speed. It’s just that no one provided the vector. It can’t be speed since direction is important. In particular, traveling at a speed of more than escape velocity pointed directly down would hardly suffice.