A launch pad located on the moon

Fiddling around with idea to use beamed energy to get a lander/ascender to/from the lunar surface.  It would be about the size of the lunar module from the Apollo Era.  To guide me in the calculations, I used the link below:

Here's a link to calculate the energy

Potential energy (PE) is just weight multiplied by distance upward; this is the same weight multiplied by distance upward, which is the same as mass multiplied by acceleration of gravity, multiplied by distance:

PE = w * h = m * g * h

g for earth is about 6 times that of the moon
g for moon is 1.622 meters/sec2

 Kinetic energy is just half the mass, multiplied by the square of the velocity:

KE = ½ * m * v2

calculations for 100 km above moon:

PE = (1 kg) x (100,000 m) x ( 1.622 meters/sec2 ) equal 162200 joules
KE for escape velocity moon 2.38 km/s equal 5664400 joules
KE for 2/3 escape velocity 1271375 joules, added together to get total energy required:  equal 1,433,575 joules

A Kilowatt-Hour is 1000 Watts for one hour, which is 1000 Joules per second for 3600 seconds, or 3.6 million Joules.

Therefore to roughly approximate, to launch 1 kg in lunar orbit at 100 km is equal to   1,433,575 joules /   3,600,000  Joules or about 0.4 kilowatt hour per kg.

That equals 1810 kilowatt hours energy for lunar orbit approximately for a mass about the size of the lunar module ascent stage which was 10000 pounds or 4500 kg.

Now guessing that to deliver 20 megawatt/hr of power for .1 hr or 6 minutes would be enough energy to get to lunar orbit--- to deliver that much energy would require 200 fuel cell units that could deliver 100 kilowatts each.

The idea here is to use Parkin's concept of microwave beam energy to land and take off from the moon.  It would have a performance ISP equal to a nuclear thermal rocket, about 850 sec., which would save mass in landing and taking off.  You would have to have a fuel supply on hand for taking off, of course.  Having the fuel depot located on the moon obviates the need to bring the landing fuel with you, simplifying matters greatly.

To execute this idea, you would need a crash landing trajectory that would be interrupted by the beamed energy to slow the spacecraft down to a soft landing.  That is to say, there would be no orbiting first.  That would require even more energy than the calculations above--- but not too much more.

You would place a microwave unit on the moon with an energy source to run it.  The energy would be beamed to the incoming lander as it gets into range.  You would have to slow down from about 5000k mph to zero in less than a few minutes ( without doing the calculations).  That's true because the beam may not have a range more than say 150 miles.  At 5000k mph, you will cover that distance quickly.

Taking off, just reverse the process.  After filling up the fuel tank of course.

The next question is: how big must the microwave unit(s) be to deliver say 30 Megawatts of power?  The current guess is that you could deliver these microwave dishes in a number of landings to a yet to be determined location on the lunar surface.  Each unit itself would be about the size of the lunar lander in the Apollo Era.  The microwave dish would be on top instead of the ascent stage.  Each dish would be an integrated unit complete with power source.

How many dishes to deliver the required power?

The rest of the topic is for later discussion.

Update:

You may need multiple launches of a the new big ass rocket being developed for a pretty penny.



Source: SPACE.com: All about our solar system, outer space and exploration

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