# Accelating a Planet

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If convert 238U into 239Pu, we can get 1/0.0072 = 139 times more.

Also we have thorium, 8..13 g/t of crust, according to wiki, i.e. 0.001% of crust mass.
About U wiki says ~3*10-4 %.
So, there is ~3 times more Th than U, and it can be turned into 233U which is nice, too.

So, with converted Th + U we can get even 400 times moar and dV = 140 * 20 = 2800 m/s.

(Still not enough to fly away).

***

But we also have an ocean. It's full of D and Li.

The ocean contains 1.33*109 km3 of water = 1.3*1021 kg.
Li = 0.17 mg/l = 1.7*10-7 kg/kg.
7.5% of Li is 6Li. We can irradiate it and turn into tritium, 3 kg of T /6 kg of 6Li.

So, we can get 1.3*1021 * 1.7*10-7 * 0.075 * 3 / 6 = 8.3*1012 kg of tritium.

***

The ocean contains 1.3*1021 * 2 / 18 = 1.45*1020 kg of hydrogen.

About 0.011..0.016 % of hydrogen atoms (on the Earth) are deuterium atoms.
I.e. ~0.0135 * (2 / 1) ~= 0.03 % of hydrogen mass.
So, we can get ~3*10-4 * 1.45*1020  = 4*1016 kg of deuterium,

It's enough for 8.3*1012 kg of tritium.

This will give (3+2)/3 * 8.3*1012 = 1.4*1013 kg of D+T mixture.

Energy of D+T fusion is 3.4*1014 J/kg.
So, we can additionallly get 3.4*1014 * 1.4*1013 = 4.8*1027 J.

That's just 4.8*1027 / (4 * 6*1028) = 2% of what the fission gives.
Very, very poor is the Earth in sense of lithium.

***

But what if we have a pure deuterium fusion reactor?

We have 4*1016 kg of deuterium.
Of course H+D gives ~2.5..3 times less energy than D+T per mass, but there is a lot of D. Much more than Li.

So, spending all terrestrial deuterium we can get ~4.8*1027 * (100/0.03) / (2.5..3) = 6*1030 J.

That's 100 times greater than U+Th+Li give together, and delta-V is ~28 km/s.

At last we can fly.

***

P.S.
What conclusion should we make now?
Deuterium-based fusion reactors are what we must have asap.
There is a lot of water everywhere except most weird places. Water has deuterium. Deuterium gives energy to an extraplanetary colony.

P.P.S.
What another conclusion should we make?

6*1030 J is what it radiates every 4.5 hours.

The closest reasonable near-Sun orbit is ~0.1 AU = 15 mln km.
If place there 10000x10000 km  = 108 km2 = 1014 m2 of light collectors, this energy will be collected every (4*pi*(1.5*1011)2) / 1014 / (86400 * 365.2422) = 90 years.

So, we should make a swarm of orbital light collectors near the Sun asap.
Then we need less mining.

Edited by kerbiloid

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1 hour ago, kerbiloid said:

But we also have an ocean. It's full of D and Li.

The ocean contains 1.33*109 km3 of water = 1.3*1021 kg.
Li = 0.17 mg/l = 1.7*10-7 kg/kg.
7.5% of Li is 6Li. We can irradiate it and turn into tritium, 3 kg of T /6 kg of 6Li.

So, we can get 1.3*1021 * 1.7*10-7 * 0.075 * 3 / 6 = 8.3*1012 kg of tritium.

You replicated almost exactly the omission made by the Castle Bravo scientists!

Approx 60%(**) of Li is 7Li. On absorption of a high-energy neutron, it splits into an alpha particle, a tritium nucleus and another neutron.

This led to the Castle Bravo burst yielding approx 15Mt instead of the predicted 5Mt.

**edit

Slight error, approx 93% of natural Li is 7Li, the 60% figure came  from the 6Li-enriched lithium used for the Castle Bravo test.

Edited by p1t1o

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9 hours ago, kerbiloid said:

[...]

What another conclusion should we make?

6*1030 J is what it radiates every 4.5 hours.﻿

The closest reasonable near-Sun orbit is ~0.1 AU = 15 mln km.
If place there 10000x10000 km  = 108 km2 = 1014 m2 of light collectors, this energy will be collected every (4*pi*(1.5*1011)2) / 1014 / (86400 * 365.2422) = 90 years.﻿﻿

So, we should make a swarm of orbital light collectors near the Sun asap.
Then we need less mining.

This bring up in interesting point about energy from the sun:

The entire global nuclear arsenal contains ~6*1018 joules, but the amount of solar energy that hits the earth every second is ~4*1018 joules.

The entire estimated remaining fossil fuels on earth contain ~4*1022 joules.

All solar energy released by the sun every second is ~4*1026 joules.

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5 hours ago, Mad Rocket Scientist said:

The﻿ entire global nuclear arsenal contains ~6*1018 joules, but the amount of solar energy that hits the earth every second is ~4*1018 joules.

It's for a reason that it moves all that moves on Earth.

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You could move a planet in only a century using dynamic compression tethers and solar sails.

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I leave with discussions of earth moving and I come back to the break room of Los Alamos National Lab. This forum is fascinating

Interesting people very interesting

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Some further thoughts - some discussion of the energy required, but not much on the mass requirement. By conservation of momentum:

The Earth weighs  ~6e24kg. If you flung the entire atmosphere (~8.5e20kg) off at the speed of light that's a DV of 42km/s.

The entire oceans (~1.4e21kg) gets you 70km/s.

The entire crust (~2.5e22kg) gets you 1250km/s.

If you can only manage 0.1c then reduce dv by 10. If you can only manage 0.01c reduce by 100. A very good ion thruster might have an exhaust velocity of 300,000m/s, so reduce by 1000.

In the case of a very good ion thruster, throwing the entire mass of the crust gets you just 1.25km/s. That's hardly anything and what's left wouldn't be recognisable as earth.

*Effects of relativity ignored (as unlikely exhaust will actually approach c)

**Rocket equation not used as in all cases the mass of the earth isn't significantly changed by expending the started fuel quantities.

Edited by RCgothic

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Not exaclty the question, but Kurde…Kiuzedg…Kurzgemahhozit…these guys touched on the “would it change the Earth’s orbit” question in this video. Didn’t go into any of the details the preceding posters touched on, though.

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