Scifi Moon Solar Power Base...

[Spacescifi]

In a scifi setting electrical power can be converted directly into thrust by way of repulsor rays (emitting exotic visible light with extremely repulsive force). Thrust is increased by 25% by using mirrored nozzles without power input.

The only issue is power. Since no propellant is actually used, only electrical power, a lot of power will be needed.

Good news: We have scifi mass to energy batteries... which unlike normal batteries can literally store their own mass's worth of electrical power. Meaning a kilogram fully charged would weigh 2 kilograms. Don't overcharge or it will explode... same goes if you overheat it.

Generating the power: Biggest power source I know of is the sun.

Trouble is... I honestly think even if you covered the surface of the moon with super efficient solar panels it would take probably decades before the mass to energy batteries would store enough power for high thrust applications.

So you would need massive space industry to support a smaller space fleet.

 

 

[farmerben]

Solar panels on the moon are in darkness 28 days in a row.  Satellites orbiting the sun work continously.  Even orbiting the Earth or Moon near continuous power can be reached.

[linuxgurugamer]

Just now, farmerben said:

Solar panels on the moon are in darkness 28 days in a row.  Satellites orbiting the sun work continously.  Even orbiting the Earth or Moon near continuous power can be reached.

Actually, about 14 days in a row.  The moon orbits the earth in about that amount of time, the panels would be in darkness about half that

[KSK]

OK, I’m maybe missing something here but this scenario appears to be assuming both viable energy-to-mass conversion (to charge the battery) and also mass-to-energy conversion (to discharge the battery) 

That being the case, why not just dump a kilogram of regolith into a mass-energy converter (probably not in one go) and let the resultant energy release charge the battery?

Even if the conversion isn’t 100% efficient, it would be easy enough  to dig up another few kilograms of  regolith.

Sure, you’re eventually going to run out of Moon to dig up, but 7.35 x10^22 kg should last a while.

 

[KSK]

But if you insist on using solar panels:

Surface area of Moon is about 38 million square kilometres or 3.8x10^13 square metres. Dividing by two because only half the moon is pointing at the Sun at any given time (to a rough approximation), gives us 1.9x10^13 square metres of usable ground.

The moon receives around 1.3-1.5W per square metre of sunlight, depending where you’re standing. I’m going to lowball that to 1 for simplicity.

So that’s 1.9x10^13 W available for harvest. Assume a 5% efficiency for your solar panels (because we’re necessarily going for quantity over quality here) and that gives us 0.95x10^12 W to play with.

Now for Einstein’s kicker. Dividing by 9x10^16 gives us a mass equivalent of 0.1x10^-4 kg or 10 milligrams per second.

So, back of an envelope calculation, turning the entire Moon into a giant disco ball of solar panels lets you charge up a single super battery in about a day. Assuming I haven’t fouled up my powers of ten along the way.

Which is a nice illustration of the ridiculous amount of energy your super batteries are storing.

Edit. On the upside 1 day is a lot better than a few decades.

Edited February 16, 2023 by KSK

[kerbiloid]

If heat the Moon hot enough, we don't need the Sun anymore.

[Abel Military Services]

Worse than fusion. 

[linuxgurugamer]

43 minutes ago, KSK said:

Surface area of Moon is about 38 million square kilometres or 3.8x10^13 square metres. Dividing by two because only half the moon is pointing at the Sun at any given time (to a rough approximation), gives us 1.9x10^13 square metres of usable ground.

The moon receives around 1.3-1.5W per square metre of sunlight, depending where you’re standing. I’m going to lowball that to 1 for simplicity.

So that’s 1.9x10^13 W available for harvest. Assume a 5% efficiency for your solar panels (because we’re necessarily going for quantity over quality here) and that gives us 0.95x10^12 W to play with.

Now for Einstein’s kicker. Dividing by 9x10^16 gives us a mass equivalent of 0.1x10^-4 kg or 10 milligrams per second.

So, back of an envelope calculation, turning the entire Moon into a giant disco ball of solar panels lets you charge up a single super battery in about a day. Assuming I haven’t fouled up my powers of ten along the way.

I love math!!!

[AckSed]

This is the kind of thing that would be ideal for a Dyson swarm. I can imagine an automated fleet of solar super-battery charging stations near the orbit of Mercury, with pickup drones gathering a continuous stream of charged packs. Propellantless thrusters solve a lot of problems.

49 minutes ago, Abel Military Services said:

Worse than fusion. 

I don't think such a ridiculous power density would be worse than fusion, it would be different than fusion. For an idea of how ridiculous, this What If? about an electron moon and proton earth has the moon's mass in electrons collapsing into a black hole the size of the known universe.

Storywise, this would lead to a different focus - harvesting energy instead of burning fuel. No-one says you can't have a whole mess of fusion, fission, solar power satellites all trying to charge up these batteries to approach the near-antimatter levels of power density.
 

[farmerben]

If we had a mass-energy thruster that could move the Moon.  Then we could use it to tow the Earth out to a more distant orbit and survive the red giant phase of our sun.  If the side of the Moon locked to us was covered in lights it would replace any variations in the sun.

[Shpaget]

4 hours ago, KSK said:

Assuming I haven’t fouled up my powers of ten along the way.

Well, since you ask... I noticed two errors.

You should use Moon's cross section not surface area, and insolation is about 1300 kW/m^2, not W, so about 4x10^15 W of incoming energy.

[kerbiloid]

23 minutes ago, Shpaget said:

and insolation is about 1300 kW/m^2, not W,

1370 W/m²

[KSK]

27 minutes ago, kerbiloid said:

1370 W/m²

This. Or at least that's consistent with the source I found. Turns out I misread 1,370 Wm^-2 for 1.370 Wm^-2 on my phone screen.

Punctuation. It's the difference between knowing your mulch and knowing you're mulch.

Thanks both.

OK then, adjust by a factor of 1000 one way and then divide by two (half the surface area of a sphere vs cross section, approximating cross section to area of a great circle). 

So 5gs^-1 or 200 seconds to charge up a superbattery. 

[Shpaget]

46 minutes ago, kerbiloid said:

1370 W/m²

Uh, I was debating with myself whether I should write 1,3 kW or 1300 W, and managed to mess it up. Bed time for me, I guess.

In my defense, 4x10^5 W is correct.

Edited February 16, 2023 by Shpaget

[K^2]

If you have access to batteries powerful enough to handle a photon drive at any sort of thrust levels you could even measure, your tech level allows you to go with at least fusion power.

Ideally, yeah, we want matter-to-energy conversion, which black holes can do with pretty good efficiency. A rotating black hole can be used with a ram scoop to get both energy and propellant from just ambient interstellar material.

This is less of the commentary on how plausible all of the above is, and more of how implausible the batteries being discussed are. Even nuclear isomer batteries are going to be on the wimpy side for a photon drive.

[Spacescifi]

17 hours ago, KSK said:

But if you insist on using solar panels:

Surface area of Moon is about 38 million square kilometres or 3.8x10^13 square metres. Dividing by two because only half the moon is pointing at the Sun at any given time (to a rough approximation), gives us 1.9x10^13 square metres of usable ground.

The moon receives around 1.3-1.5W per square metre of sunlight, depending where you’re standing. I’m going to lowball that to 1 for simplicity.

So that’s 1.9x10^13 W available for harvest. Assume a 5% efficiency for your solar panels (because we’re necessarily going for quantity over quality here) and that gives us 0.95x10^12 W to play with.

Now for Einstein’s kicker. Dividing by 9x10^16 gives us a mass equivalent of 0.1x10^-4 kg or 10 milligrams per second.

So, back of an envelope calculation, turning the entire Moon into a giant disco ball of solar panels lets you charge up a single super battery in about a day. Assuming I haven’t fouled up my powers of ten along the way.

Which is a nice illustration of the ridiculous amount of energy your super batteries are storing.

Edit. On the upside 1 day is a lot better than a few decades.

 

Oh but I do! Too easy otherwise. Getting energy is work and should never be too easy.

 

I guess this finally completes my search for a star wars type drive that can casually SSTO anywhere and back a few times at least without replacing it's batteries (because obviously recharging is never going to happen as you need mega structures or satelite swarms charging across a broad area... none of which a lone spaceship can do.

 

What I do like is it totally eliminates the headache of ISRU... which is complex because it requires planning all your trips and knowing exactly where you are going. It does not really allow for much in the way of unpredictability outside of where you are supposed to go or fuel up. Since you literally design engines around the fuel you expect to find where you are going... and if you screw that up you are doomed to die in space if no one comes to save you.

Also IRL you get a distress call from another ship you will probably say "Yeah we heard ya, but we do not have enough fuel to get you and get back." Or "We cannot take all of you guys because we do not have enough supplies so decide who can come over to us before we get there."

 

 

I do have one more question you may enjoy calculating.

 

A starship is 200 tons, and has enough energy stored in it's mass energy batteries to do 1g for 4 hours, but can do 5g at reduced travel time since it will use up it's energy faster.

 

The Question: How heavy would the mass energy batteries altogether weigh?

Remember unlike normal batteries they can store a maximum of electricity that equals their mass... no more or boom!

I suspect that overall this should be quite space efficient and well in line with the Star Trek/Star Wars ethos of having relatively small power packs compared to what they power.

In contrast to real rockets which are like giant fuel tanks with stuff strapped on as an afterthought.

 

I guess the main difference is that due to the sheer energy density of the super batteries, you really do not want to crash the ship as it would give off a blast rivaling the high yield of an antimatter bomb.

 

And you can forget escape pods saving you if your ship is about to blow up... since the sheer radiant energy would fry even escape pods flying away from the blast.

 

 

 

 

Edited February 17, 2023 by Spacescifi

[kerbiloid]

In Five Weeks in a Balloon by Jules Verne they crossed Africa spending one small non-rechargable electric battery to replenish the balloon with hydrogen by electrolyzing the water from rivers, and using it also for other needs like illumination.

In Twenty Thousand Leagues Under the Seas by Jules Verne a whole electric submarine performs a whole circumnavigation trip around the Earth, visiting the South Pole, and diving to the 16 km depth, recharging the battery just once (in their volcanic island with coal mine).

So, why not.

[KSK]

12 hours ago, Spacescifi said:

I do have one more question you may enjoy calculating.

A starship is 200 tons, and has enough energy stored in it's mass energy batteries to do 1g for 4 hours, but can do 5g at reduced travel time since it will use up it's energy faster.

The Question: How heavy would the mass energy batteries altogether weigh?

OK, first off, I can tell you right now that the answer is going to be '1kg plus a small amount more' because of that c² part of E=mc².  I can also tell you that this is going to be a zeroth level hack approximation but here goes.

Assumptions.

1. Spacecraft starts away from any gravitational field, so that I can ignore gravitational potential energy.
2. Spacecraft engines are 100% efficient at converting energy from battery into kinetic energy of spacecraft.
3. We're using sensible units so 1 ton = 1000 kg.
4. g=10 because I can't be bothered fiddling around with 9.8 given that this is a hack job anyway.

1g acceleration for four hours raises spacecraft velocity by 144,000 ms-2.

Gain in spacecraft kinetic energy is therefore:  2.07 x 10¹⁵J

Mass equivalent = 0.023 kg or 23g.

So your battery will weigh 1.023 kg assuming that the discharged battery weighs 1 kg.  Feel free to multiply by whatever fudge factor seems appropriate - it's still going to be a pretty small amount of mass in the battery.

 

 

[Spacescifi]

3 hours ago, KSK said:

OK, first off, I can tell you right now that the answer is going to be '1kg plus a small amount more' because of that c² part of E=mc².  I can also tell you that this is going to be a zeroth level hack approximation but here goes.

Assumptions.

1. Spacecraft starts away from any gravitational field, so that I can ignore gravitational potential energy.
2. Spacecraft engines are 100% efficient at converting energy from battery into kinetic energy of spacecraft.
3. We're using sensible units so 1 ton = 1000 kg.
4. g=10 because I can't be bothered fiddling around with 9.8 given that this is a hack job anyway.

1g acceleration for four hours raises spacecraft velocity by 144,000 ms-2.

Gain in spacecraft kinetic energy is therefore:  2.07 x 10¹⁵J

Mass equivalent = 0.023 kg or 23g.

So your battery will weigh 1.023 kg assuming that the discharged battery weighs 1 kg.  Feel free to multiply by whatever fudge factor seems appropriate - it's still going to be a pretty small amount of mass in the battery.

 

 

 

About how powerful of a bomb would the battery be if nearly fully charged and the ship crashed on Earth?

 

Why do I ask?

 

My intuition told me that even though massive fully charged batteries could be made, designers avoid that since vessels mainly use sublight engines for hopping on and off world. In space they either cruise off inertia or use some variation of warp or hyperdrive to get around faster.

 

So the idea is that four hours of charge at 1g for a 200 ton vessel is considered an acceptable risk.

 

A 1 ton battery? No way. Maybe on a moon military industry base but that's it.

[KSK]

Very. 

Assuming the battery explodes it'll release the energy equivalent of annihilating 0.5 kg of antimatter.

So:    9x10¹⁶ J of energy.

1 kiloton = 4.184 x 10¹² J

So roughly equivalent to a 21.5 megaton nuclear explosion. For comparison Tsar Bomba, the most powerful nuclear weapon ever built, had a yield of 50 megatons.

One ton battery = 21.5 gigaton explosion if it goes pop. Probably a bad idea to let that happen.

Edited February 18, 2023 by KSK