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I am working on a geoengineering project that, at the moment, seeks to use many mass drivers to project lunar regolith (lunar soil) or other materials into space to create targeted shading effects on the earth for specific purposes. The material would intersect the moon allowing for easy recollection of material for quick reuse. There are other space infrastructure implications that would allow sharing of resources and a lot that can be done to increase efficiency. The problem is I can't find anywhere to help with some of the basic calculations. I know that mass drivers can be 80-90% energy efficient on the higher end and the project will be in an extremely low atmosphere environment.

I would like some help with the basic math to calculate energy needs. Even links to find the information and explore it would be welcomed.

The idea is to constantly project small units, no more than 10lb but probably smaller, approximately 3km/second from a mass driver (the small size as I understand it should allow a faster rate of projecting material which should make the whole process easier).

My question is;

How much energy would it take to take per pound of material to travel per km/second assuming 80% energy efficiency? Browny points if you can give an estimation of the refractory period between projections or point me to how to calculate that.

P.S. This is a big complex project and I always get random questions going in every direction. I'm used to it and happy to answer any questions or add them to my list of questions that need answering, but please provide useful feedback instead of blank criticism and recognize that I only have so many characters per comment to respond.

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The bare bones answer is energy = (mass x velocity²)/2. This means your projectile will have a kinetic energy, once accelerated, of e=(0.453kg x 1E3² m/sec)/2 = 226824 joules. But also, you must add the energy required to accelerate the mass. This derives from f=ma, so you have to know the launching force and time permitted to arrive at your final velocity. Note also that you are hoping to project at a velocity of 3 km/sec. The velocity is a squared term in the energy equation. Therefore don't just multiply the above answer by 3. Allow for the force of gravity (more energy). Allow for efficiency (more energy).

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  • $\begingroup$ Thank you, that gets me damn close to where I want to be and I can reasearch further far more efficiently. Thank you. $\endgroup$ – T.A. McKay Mar 4 at 2:40

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