Stupid school project - Stopping the moons velocity

[Myggen]

Hello all, im a electrician student, and in school we got the task (not having anything to do with electricity) to explain about something in 5-10 min, it could be hobbies..well, also something about electricity, anything goes.

But, actually the night before i saw this video by Scott Manley:

, and was instantly thinking about the video when we got the task telling about something, i just want to try it on the real moon (the one turning red in 1 hour time, im staying awake to see it atm, at about 04.30 am danish time ).

Anyway, what does it take to stop the velocity of the REAL Moon???

My plan is, to leave out the factor of the forces of gravity in the calculations, by putting a gravitymanipulator on the surface of the moon (so it wont fall "down" when loosing speed).

But, in the video, Scott Manley just show some number without explaining where he got em from and why.

The calculation he show at 1:28, says (280x124 trillion) / (800x9.81) = .....

Well, i guess the (280x124 trillion) = (deltaV x mass of gilly), but then this / (800x9.81)

I spoke to a friend who play this game alot more than i do, and he is talking about that the 800 might be isp of the engine used, but then we are in doubt of the 9.81?

Also i wanna leave out the factor of the weight of all the fuel, as its stored in a big ballon in orbit, with a long hose connected to the engine(s) on the surface...

I wanna use this rocket engine: https://en.wikipedia.org/wiki/RD-170

And i want to stop all velocity in one single orbit (27.3 days), not Scott Manley's 19 million years.

So first im gonna figure out how much fuel is needed, then how many engines is needed to complete the stop in only 27.3 days.

So far i have these notes:

1 orbit = 2,412,517.5 km / 27,3 / 24h / 60 min / 60 sec = 1022,8 m/s

Moon mass = 7,349×10^22 kg > 7,349×10^19 t > 73.477.000.000.000.000.000 t

Orbital period = 27.321582 d (27d 7h 43.1)

delta v / masse = (1022 x 73.477.000.000.000.000.000) / (isp? x 9.81?) =

Strongest rocket engine: Russian RD-170

(https://en.wikipedia.org/wiki/RD-170)

Thrust (vac.) 7.887 MN (1,773,000 lbf) > ibf = pound force > 1 ibf = 4.448222 N

(https://da.wikipedia.org/wiki/Newton_%28enhed%29)

Thrust to weight ratio: 82

(TWR > wiki.kerbalspaceprogram.com/wiki/Thrust-to-weight_ratio)

Any ideas/help would be very much appriciated, and thanks in advance

Edited September 28, 2015 by Myggen

[NFUN]

Thrust is Isp times 9.8 (times fuel flow, but don't worry about that). That is probably what Scott was referring to.

[KerikBalm]

Step 1: find the moons orbital velocity.

https://en.wikipedia.org/wiki/Moon

1022 m/s average according to wikipedia.

Step 2: find the moon's Mass:

7.3477×10^22 kg according to wikpedia

Step 3: choose an engine and Isp... I'm going to go with the Nerva and 800 Isp

Step 4: Tsiolkovsky rocket equation

https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

1022 m/s = 9.8 * 800 * ln (M_1/M_2)

0.13036= ln (M_1/M_2)

1.1392 = M_1/M_2

M_2 = 7.3477×10^22 kg as above, so M_1 is 1.1392 x bigger = 8.37075

M_2-M_1 = 1.02306 * 10 ^22 kg

= 1.02306 * 10^19 tons

You would need 10,230,600,000,000,000,000 tons of liquid H2 to stop the Moon using a nuclear thermal rocket with an Isp of 800 seconds.

Good luck.

[federicoaa]

that is ~1/8 of the total mass of the mun.

I recommend using a laser pointer instead XD

[YNM]

Considering you're an electrician student, maybe you want something to go with static electricity... Latest what-if article: http://what-if.xkcd.com/140/ it actually ends using GR, but I can slightly go through it (as density or energy density should be an invariant...)

[K^2]

Scott Manley is using an approximation to Rocket Formula.

dV = I_SPg ln((M+m)/M) ≈ I_SPg m/M

Where M is mass of the "rocket" and m is mass of the fuel. This is ONLY true whenever M >> m. Which is equivalent to saying dV << I_SPg. Since for the rocket in question dV is only 280m/s and I_SPg is nearly 8,000m/s, the approximation is fair. And you can compute fuel as m = (dV M) / (I_SPg) = (280 x 124 trillion) / (800 x 9.8)

[ChrisSpace]

Okay, my turn. And i'm going to try to use multiple different methods for comparison

STARTING REQUIREMENTS

TWR: 2.4126×10^-4 (enough to remove all velocity in 5 days)

Dv: 1022m/s

Payload: 1 lunar mass (7.34767309 × 10^22 kilograms)

Thrust needed: 173 790 000 000 000 MN

HYPERGOLIC CHEMICAL PROPULSION

Proton RD-253 stats: 333s isp, 1260kg, 1.83MN thrust

Propellant needed: 0.37 lunar mass

No. of engines needed: 9.4967213 x 10^13

Is there even enough fuel on Earth to do this?

KEROLOX CHEMICAL PROPULSION

Saturn V F-1 stats: 304s isp, 9153kg, 7.74MN thrust

Propellant needed: 0.41 lunar masses

No. of engines needed: 2.2453488 x 10^13

See the above comment.

HYDROLOX CHEMICAL PROPULISON

Space shuttle SSME stats: 453s isp, 9531kg, 1.82MN thrust

Propellant needed: 0.26 lunar masses

No. of engines needed: 9.5489011 x 10^13

Well, at least hydrogen and oxygen are in near-limitless supply in our solar system.

MASS DRIVER

Mass Driver stats: 3058s isp, 150 000kg, 0.02MN thrust

Propellant needed: 0.035 lunar masses

No. of engines needed: 8.6895 x 10^15

We'd need lots of mass drivers, but at least we can use the moon itself as projectile 'propellant'. Or asteroids.

DUMBO NUCLEAR THERMAL SOLID-CORE ROCKET

DUMBO stats: 825s isp, 5000kg, 3.5MN thrust

Propellant needed: 0.135 lunar masses

No. of engines needed: 4.9654286 x 10^13

This assumes we use molecular Hydrogen, which our solar system as plenty of. It's the fissile material in the reactors we may have problems collecting.

LIBERTY SHIP NUCLEAR THERMAL GAS-CORE ROCKET

Liberty engine stats: 3058s isp, 378 000kg, 37.38MN thrust

Propellant needed: 0.035 lunar masses

No. of engines needed: 4.6492777 x 10^12

See the above comment.

2ND GEN ORION NUCLEAR PULSE PROPULSION

Orion propulsion module stats: 12 232s isp, 3 250 000 kg, 400 MN thrust

Propellant needed: 0.009 lunar masses

No. of engines needed: 434 475 000 000

The most high-thrust engine on Atomic Rocket's list, but we'd still need over 400 billion of them.

NUCLEAR SAT-WATER ROCKET

NSWR 20% UTB stats: 6728s isp, 33000kg, 12.9MN thrust

Propellant needed: 0.016 lunar masses

No. of engines needed: 1.3472093 x 10^13

This one may or may not irradiate everything between Mercury and Jupiter if we used it.

I excluded many other things such as ion, plasma and antimatter propulsion due to a lack of time. All info is from http://www.projectrho.com/public_html/rocket/enginelist.php

[AngelLestat]

If I have to choose a method.. I will capture big asteroids of ice, then I would place them close to the moon in free falling firing "asteroid material" to keep altitude (gravitation pull), the propulsion method will be salt water rocket (in case is possible), so we take the salt and water from the asteroid, and we just carry the plutonium or uranium from earth.

With many of these.. eventually we can slow down the moon.

[SomeGuy12]

Just boost the mass driver ISP by making the mass driver track longer. Get that ISP up to 10k or 100k and do this. Also, uh, better plan on waiting at least a century. 5 days is ridiculous.

[Shpaget]

I will capture big asteroids of ice, then I would place them close to the moon in free falling firing "asteroid material" to keep altitude (gravitation pull)

Wouldn't the asteroid material you'd be firing hit the surface of the Moon and negate the pulling?

You'd need to angle your thrust sideways, wasting a good chunk of it.

[AngelLestat]

You are right, that is a valid point

If you are far enoght, the angle is not much affected but the time it will take you increase, because the gravity would be significantly lower.

[Jovus]

Scott Manley is using an approximation to Rocket Formula.

dV = I_SPg ln((M+m)/M) ≈ I_SPg m/M

Where M is mass of the "rocket" and m is mass of the fuel. This is ONLY true whenever M >> m. Which is equivalent to saying dV << I_SPg. Since for the rocket in question dV is only 280m/s and I_SPg is nearly 8,000m/s, the approximation is fair. And you can compute fuel as m = (dV M) / (I_SPg) = (280 x 124 trillion) / (800 x 9.8)

Why is it only true when M >> m? I didn't see anything obvious in the derivation to make that the case...does it have something to do with the assumption that exhaust velocity is constant?

[K^2]

Why is it only true when M >> m? I didn't see anything obvious in the derivation to make that the case...does it have something to do with the assumption that exhaust velocity is constant?

Because ln((M+m)/M) is only approximately m/M for M >> m. Take a look at first plot here, with M = 5 as an example. Note that while m is small, the two are almost identical. But as m gets larger, they start to diverge quite significantly.

[Jovus]

Gotcha. I misread your original post and thought you were saying the rocket equation only held for M >> m.

[Kerbart]

Here's a different approach where you can demonstrate your understanding of orbital mechanics to your teacher.

Since we're way off in Lala-land, let's assume you have a magical rocket engine that doesn't need fuel. Or maybe one that somehow does total mass to energy conversion (a nice side project would be to calculate how much mass it'd need per second but I digress)

The idea would be that you'd use this magical rocket engine to prevent the moon from crashing into the earth, after stopping it. So the power of the engine is exactly equal to the gravitational pull of the earth. And then you can figure out how long it takes to stop the moon.

For more realism you can have the moon slowly rotate so that the thrust of the engine starts to offset gravity as the moon slows (though thinking about the analysis alone gives me a headache. Modelling it should be easy though). And then calculate how long it takes for the moon to slow down to, say, 5 m/s.

[Myggen]

Hey all again, thanks for all the replies, and sorry for the late reply, but had alot of stuff to do.

I look specially at this:

Step 3: choose an engine and Isp... I'm going to go with the Nerva and 800 Isp

Step 4: Tsiolkovsky rocket equation

https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

1022 m/s = 9.8 * 800 * ln (M_1/M_2)

0.13036= ln (M_1/M_2)

1.1392 = M_1/M_2

M_2 = 7.3477×10^22 kg as above, so M_1 is 1.1392 x bigger = 8.37075

M_2-M_1 = 1.02306 * 10 ^22 kg

= 1.02306 * 10^19 tons

Good luck.

But here i wanna go with the russian RD-170 with Isp on 338 https://en.wikipedia.org/wiki/RD-170

Also these i dont understand:

You write M_1 and M_2, dont you mean M_0 and M_1?

and then these:

0.13036= ln (M_1/M_2)

1.1392 = M_1/M_2

How to come up with those numbers?

Can you explain a little more? i would like to know just a little about what im talking about, not standing there sounding more stupid than the project itself haha.

Then this:

Okay, my turn. And i'm going to try to use multiple different methods for comparison

STARTING REQUIREMENTS

TWR: 2.4126×10^-4 (enough to remove all velocity in 5 days)

Dv: 1022m/s

Payload: 1 lunar mass (7.34767309 × 10^22 kilograms)

Thrust needed: 173 790 000 000 000 MN

HYPERGOLIC CHEMICAL PROPULSION

Proton RD-253 stats: 333s isp, 1260kg, 1.83MN thrust

Propellant needed: 0.37 lunar mass

No. of engines needed: 9.4967213 x 10^13

Is there even enough fuel on Earth to do this?

Yea i would need some calculations of how much thrust i need to stop the moon in one orbit, how did you come up witht the thrust needed number?

Also, shouldnt ^-4 be ^4? my calculator says syntax error using ^-4

Considering you're an electrician student, maybe you want something to go with static electricity... Latest what-if article: http://what-if.xkcd.com/140/ it actually ends using GR, but I can slightly go through it (as density or energy density should be an invariant...)

Im only a 1. year student, we havent "learned" about atoms and such stuff yet haha.

Here's a different approach where you can demonstrate your understanding of orbital mechanics to your teacher.

Since we're way off in Lala-land, let's assume you have a magical rocket engine that doesn't need fuel. Or maybe one that somehow does total mass to energy conversion (a nice side project would be to calculate how much mass it'd need per second but I digress)

The idea would be that you'd use this magical rocket engine to prevent the moon from crashing into the earth, after stopping it. So the power of the engine is exactly equal to the gravitational pull of the earth. And then you can figure out how long it takes to stop the moon.

For more realism you can have the moon slowly rotate so that the thrust of the engine starts to offset gravity as the moon slows (though thinking about the analysis alone gives me a headache. Modelling it should be easy though). And then calculate how long it takes for the moon to slow down to, say, 5 m/s.

Another good idea, and i see the point, but i think it would be pretty hardcore calculations, involving grafs.

But thanks alot so far, there is really something to think about, and i think i will consult the physic teacher tomorrow.

[ChrisSpace]

Yea i would need some calculations of how much thrust i need to stop the moon in one orbit, how did you come up with the thrust needed number?

Also, shouldnt ^-4 be ^4? my calculator says syntax error using ^-4

I came up with the thrust number using the required TWR and the mass of the moon. And yes it is ^-4.

Soon I might do some calculations on the mass of the required engines and the size of the propellant tank. And I might add some new propulsion methods.

[K^2]

I came up with the thrust number using the required TWR and the mass of the moon. And yes it is ^-4.

Without checking the math to see if that's right, just pointing out that entering ^-4 will cause a lot of parsers to throw an error. Always enter it as ^(-4) to avoid problems.

[Lord Aurelius]

This looks like a great question for XKCD's What If blog.

[PB666]

What you need here is a mass energy converter, then convert a small amount of the moons mass to energy then use the Cannae drive to stop the moon. This is a joke, only a joke, for the next sixty seconds . . . . . . .

Here's a question, if you attached a saturn v rocket to the leading surface of the moon, could you overcome the force imparted on the moon by earths tides and its rotation? How many saturn V rockets would it take I say 1000.

[Rune]

Hey all again, thanks for all the replies, and sorry for the late reply, but had alot of stuff to do.

I look specially at this:

But here i wanna go with the russian RD-170 with Isp on 338 https://en.wikipedia.org/wiki/RD-170

Also these i dont understand:

You write M_1 and M_2, dont you mean M_0 and M_1?

and then these:

0.13036= ln (M_1/M_2)

1.1392 = M_1/M_2

How to come up with those numbers?

Can you explain a little more? i would like to know just a little about what im talking about, not standing there sounding more stupid than the project itself haha.

Then this:

Yea i would need some calculations of how much thrust i need to stop the moon in one orbit, how did you come up witht the thrust needed number?

Also, shouldnt ^-4 be ^4? my calculator says syntax error using ^-4

Ok, I think a little bit of what they are saying is flying through your head. Let's explain the rocket equation a bit further:

dV=I_sp*9.81*Ln(M0/M1)

dV: the velocity change. I think you have this nailed, you need about 1022m/s.

I_sp: You also kind of have this. You want to use 333s in your example.

9.81: It's normal you don't get why that is there. The short version, I_sp is a wonky unit, but I_sp*9.81 is the effective exhaust speed of you propellant, which makes much more sense: the faster you throw stuff out the back, the more bang you get for your buck.

Ln(M0/M1): That is the term you are having problems with. That's the natural logarithm of the initial mass(Called M0, M1, or M+m by the various people that have responded, basically Moon+engines+Fuel), divided by the final mass (M1, M2, or M, the mass of the Moon and the engines, but without the fuel, HAS to be smaller). This term Ln(M1/M0), if you write it as Ln[(M+m)/M], can be rounded out to m/M if M>>m. That is the special trick Scott Manley used to approximate without doing logarithms. I think that complicates the understanding of the formula, needlessly.

Now you will notice that thrust is nowhere to be found there. That is because the rocket equation gives you mileage, and nothing else. If you want to find out how many engines you need, you have to use another equation, but luckily this time it is a REALLY easy one: F=M*a. You recognize that one, don't you? The problem is that mass here is (M+m), or M0, or in other words the mass of the moon and the fuel, and the engines. Which you don't know until you choose the mass of the engines and run the other equation. But you can't choose the mass of the engines without knowing the mass of the fuel. See the problem? You can solve both equations at once, of course, but the math gets more complicated because logarithms are involved. You can also cheat and ignore the weight of fuel, as I will do right now, and then it's much, much easier, but the answer will be wrong by about an order of magnitude:

Our desired acceleration is 0.0002416m/², or 2.416*10^-4m/², as someone calculated. Our Mass is one lunar mass because I'm cheating as I said, and that turns out to be 173.79*10¹⁸ Newtons. Dividing by the thrust of your chosen engine, 7.887*10⁶N, I get about 22*10¹² engines, which would add some 2.27¹⁰Kgs to the mass of the Moon if I haven't messed up with one or two orders of magnitude (which is really easy with these things). So maybe not that negligible after all. And wait until we add the mass of the propellant to this, then we need to re-do this again

The rest I think you can work out. A simple iterative process will quickly show how ridiculous things are getting, and show why moons are not likely to be moved around any time soon.

Rune. Which is, I think, the conclusion you should present to the class.

Edited October 1, 2015 by Rune

[ChrisSpace]

Our desired acceleration is 0.0002416m/², or 2.416*10^-4m/², as someone calculated.

Yes, my name is 'someone'.

[Myggen]

Ok, I think a little bit of what they are saying is flying through your head. Let's explain the rocket equation a bit further:

dV=I_sp*9.81*Ln(M0/M1)

dV: the velocity change. I think you have this nailed, you need about 1022m/s.

I_sp: You also kind of have this. You want to use 333s in your example.

9.81: It's normal you don't get why that is there. The short version, I_sp is a wonky unit, but I_sp*9.81 is the effective exhaust speed of you propellant, which makes much more sense: the faster you throw stuff out the back, the more bang you get for your buck.

Ln(M0/M1): That is the term you are having problems with. That's the natural logarithm of the initial mass(Called M0, M1, or M+m by the various people that have responded, basically Moon+engines+Fuel), divided by the final mass (M1, M2, or M, the mass of the Moon and the engines, but without the fuel, HAS to be smaller). This term Ln(M1/M0), if you write it as Ln[(M+m)/M], can be rounded out to m/M if M>>m. That is the special trick Scott Manley used to approximate without doing logarithms. I think that complicates the understanding of the formula, needlessly.

Now you will notice that thrust is nowhere to be found there. That is because the rocket equation gives you mileage, and nothing else. If you want to find out how many engines you need, you have to use another equation, but luckily this time it is a REALLY easy one: F=M*a. You recognize that one, don't you? The problem is that mass here is (M+m), or M0, or in other words the mass of the moon and the fuel, and the engines. Which you don't know until you choose the mass of the engines and run the other equation. But you can't choose the mass of the engines without knowing the mass of the fuel. See the problem? You can solve both equations at once, of course, but the math gets more complicated because logarithms are involved. You can also cheat and ignore the weight of fuel, as I will do right now, and then it's much, much easier, but the answer will be wrong by about an order of magnitude:

Our desired acceleration is 0.0002416m/², or 2.416*10^-4m/², as someone calculated. Our Mass is one lunar mass because I'm cheating as I said, and that turns out to be 173.79*10¹⁸ Newtons. Dividing by the thrust of your chosen engine, 7.887*10⁶N, I get about 22*10¹² engines, which would add some 2.27¹⁰Kgs to the mass of the Moon if I haven't messed up with one or two orders of magnitude (which is really easy with these things). So maybe not that negligible after all. And wait until we add the mass of the propellant to this, then we need to re-do this again

The rest I think you can work out. A simple iterative process will quickly show how ridiculous things are getting, and show why moons are not likely to be moved around any time soon.

Rune. Which is, I think, the conclusion you should present to the class.

Wow, thanks alot for explaining all that, it all makes more and more sence, still a way to go tho.

Still some things i dont understand:

9.81: is this because of this? > the acceleration due to gravity at the Earth's surface, 9.81 m/s2?`

If so, then isnt this the engines isp while flying upwards in earth gravity and atmosphere?, but in my project we are pushing at the moon in vacuum, would it then still be 9.81?

And no, you wont cheat by ignore the weight of the fuel, because this is my plan:

http://img.photobucket.com/albums/v209/total-uffe/Projekt%20Stop%20Maringnen_zps5lyer4ds.png

All the fuel is in a ballon in a stationary orbit, also with a gravity manipulator thing attached so it wont fall down...and not have to think about when spending fuel the mass gets lower etc.

But first i would really like to know the fuel needed, but need to know if i leave out the M0/M1?

So i wont be needing the M0 and M1 thing, if i leave out the weight of the fuel...and the engines, if i modify one to have all twr needed?

Then the twr to be able to stop the moon in one orbit 2,358,720 sec (60x60x24x27.3)

REALLY easy one: F=M*a. You recognize that one, don't you?

No, but i have just checked it out. Its Force=Mass*acceration right?

But then i guess i need to now the acceleration, and someone aka ChrisSpace came up with these numbers 0.0002416m/2, but why and how? and whats the /^2 mean?

I hope i make myself understandable, im very tired atm , thanks again to far.

Edited October 4, 2015 by Myggen