Changing a spaceship's direction via ejecting mass

[CyborgTriceratops]

Recently I watched Ender's Game and, with out spoiling it, there is a scene in the movie where mass is ejected out of an object moving through zero-gravity in order to change the object's speed and direction. When I watched it, it struck me as too much change for the mass ejected to mass retained ration and I want to verify it.

SPOILERS: The specific scene I am refering to is the scene in the battle dome where Ender has his army form a 'wedge' of sorts around a main player and then floats their way through the battlefield, ejecting kids at certain angles to make his needed course adjustments. I know some of the work I need to do to verify if the right math was used, such as guestimating the mass of each child, determining how many children there were, the ejected children's speed and angle, and the resulting angle. That is where my knowledge stops though. I am trying to find the formula I need to use but I seemed to have hit a stand still.

NO SPOILERS: I have looked through a few different websites and pulled up my own rusty knowledge, and what I have come up with is using F=MA of the ejected mass, then use the known force into F=MA for the spaceship. We know the mass and we know the force applied to it from a certain angle, we then can determine the acceleration of the ship in that direction. What I do not know though, is how do I calculate the dropping of mass into the equation? In the example it could be around 1/10th or 1/15th of the entire system's mass being ejected, where as in traditional rockets it is a much, much smaller percentage of their mass being used to propel them.

Do I do the second F=MA equation with the lesser mass? Am I on the wrong path completely? Any help in this issue would be greatly appreciated, as it's bugged me since I watched the movie. I do not want to simply be given the answer though. Pointers and tips, maybe say, "research the oberth effect (for example)", or other such help would be appreciated more then, "Use X formula because Y."

[Sensei]

Traditional rocket engine is ejecting highly accelerated gases (H2O, when source is Hydrogen and Oxygen) all the time while they're running.

 

So the same with Hall's engine

http://en.wikipedia.org/wiki/Hall_effect_thruster

which is ionizing some gas, and then accelerating it, and ejecting from engine.

where as in traditional rockets it is a much, much smaller percentage of their mass being used to propel them.

 

 

Traditional rocket is "flying tank"..

 

Even airplane can be called so.

Airbus A380 has mass 276,000 kg.

Fuel 310,000 liters..

Total 560,000 kg

It has more fuel than anything else at beginning of travel..

[CyborgTriceratops]

I think I should clarify my statement. I apologize.

I know that fuel is the largest part of flight in terms of mass. What I meant by

where as in traditional rockets it is a much, much smaller percentage of their mass being used to propel them.

Is that at any one time the percentage is smaller, not that it is smaller overall. Five seconds on a traditional rocket will burn far less of their fuel then the movie does in it's five seconds.

 

[Strange]

Can't you just use conservation of momentum to calculate this? The velocity change of the main group will be the velocity of the ejected kid times the mass ratio of the kid and the group. (And by assuming the kids all have roughly the same mass, you can just use the number of kids to simplify this).

[swansont]

The proper equation to apply is F = dp/dt = v dm/dt + m dv/dt. The second part is the familiar F = ma, but it assumes constant mass which is not true in this example. Since the mass is not a continuously changing variable, the easiest way to evaluate this scenario is conservation of momentum, as Strange has pointed out.

[Alan McDougall]

How do you overcome the problem of inertia?

[Strange]

Inertia is what allows the simple conservation of momentum calculation to work. (You do need to do a vector sum to work out how much the "ship" will be diverted from its (inertial) course by the change in momentum at some angle.)

[swansont]

How do you overcome the problem of inertia?

 

You will have to be more specific.

[John Cuthber]

The simple answer is that, if you throw something out fast enough, you can get an arbitrarily large change in the ship's velocity.

[Alan McDougall]

 

You will have to be more specific.

 

Well you cant ignore inertia, take a fun park ride, if inertia is ignored in the design of the ride, the people could break their necks going around a too sharp corner. The spaceman could likewise die if too much mass is ejected to change the spaceship direction, resulting in maybe "a too sudden right angle change in the direction of motion" for which for the human body is not designed by evolution to withstand, killing him/her outright.

 

Inertia is the tendency of an object to resist changes in its state of motion. But what is meant by the phrase state of motion? The state of motion of an object is defined by its velocity or the speed with a direction. Thus, inertia could be redefined as follows:

 

Inertia: tendency of an object to resist changes in its velocity, the unlucky spaceman could be just such an object!

 

 

.

[John Cuthber]

If you are doing physics, inertia is just mass.

[swansont]

Well you cant ignore inertia, take a fun park ride, if inertia is ignored in the design of the ride, the people could break their necks going around a too sharp corner. The spaceman could likewise die if too much mass is ejected to change the spaceship direction, resulting in maybe "a too sudden right angle change in the direction of motion" for which for the human body is not designed by evolution to withstand, killing him/her outright.

 

Inertia is the tendency of an object to resist changes in its state of motion. But what is meant by the phrase state of motion? The state of motion of an object is defined by its velocity or the speed with a direction. Thus, inertia could be redefined as follows:

 

Inertia: tendency of an object to resist changes in its velocity, the unlucky spaceman could be just such an object!

 

 

.

That's not an issue raised in the OP.

[CyborgTriceratops]

Well, I've been trying to find a good clip of the movie to guestimate the mass, trajectory, and speed of both the ship and the ejected mass, but I can't seem to do that. I'll have to wait until I get it on DVD.

Until then, I'll make up some numbers.

Initial mass of ship - 1000kg

Speed of Ship - 3 m/s

Ejected Mass - 100kg
Angle - 45 Degrees of Aft
Speed of Ejected Mass - 1 m/s

Now, according to Swansont, I need to use "F = dp/dt = v dm/dt + m dv/dt" to solve the issue. I am still learning what all of these terms mean, and I am a bit loss. I am learning what I can from Wikipedia and reading forums here, but it is taking some bit. I will go ahead and ask for forgiveness, as my highest level of math so far is College Algebra, so this isn't something I've learned.

F= Force, which I don't know yet as I do not know the acceleration after ejecting the mass.
v=vector at the initial start
d=is determined by the rate of change of time by said object, P, M, and V.
m=mass

Would I plug it in as follows?

F = dp/d(.5 seconds) = (3 m/s) d(900)/dt + (100) d(1 m/s)/d(.5 seconds)

Or am I still off? Is this something that I need to hold off on until I get into calculus and physics?

[swansont]

Use conservation of momentum.

 

A mass of 1000 kg ejects a mass of 100 kg, leaving 900 kg. The sum of the two momenta must be equal to the original value, since the net external force is zero.

 

p = p₁ + p₂ (One must recall that these are vectors)

 

Let's solve this in the spaceship frame, and apply the fact that the frame is moving at 3 m/s afterwards.

 

In that frame, p = 0. Thus, p₁ = -p₂ Rewriting, we get m₁v₁ = -m₂v₂ or v₁= -(m₂/m₁)v₂

Thus, the larger mass moves off at 1/9 m/s in the opposite direction, which is 45º to the original motion. Finding the resultant speed is a little bit of trig - you need to add the two vectors.

[Alan McDougall]

If you are doing physics, inertia is just mass.

 

Mass plus velocity!

[CyborgTriceratops]

p stands for vector, got it.

p is equal to 0 because there is no increase or decrease of momentum in the begining of the problem, correct? If there was an increase, then it would be a non-zero number.

to get v₁=-(m₂/m1)v₂ I just apply basic math? Divide both sides by m₁? if so, shouldn't the v₂ also be divided by m₁ or no because the m's cancel each other out.

Plugging in the variables I get

1000(3 m/s)=-[100(1 m/s)]

3 m/s=-[(100/1000)(1 m/s)]

3 m/s=-[(.1)(1 m/s)]

3 m/s=-.1 m/s

and this is where I run into problems. They should both equal the same number just one should be a negative. Even if I had used 900 instead of 1000, there difference wouldn't of been -2.9 m/s. Am I doing the p=0 part to late? Should I of started out like this:

 

900(0 m/s)=-[100(1 m/s)]

0 m/s=-[(100/900)(1 m/s)]

0 m/s=-[(.1)(1 m/s)]

0 m/s=-.111 m/s

or a negative 1/9 m/s in the opposite direction of the push? Since I got your answer, I am guessing it is correct, I just want to make sure I have the math right as I still get 0=.111 which isn't true. Secondly, does the initial mass of the frame count at all? It shouldn't, since the mass being ejected is leaving the spaceship before any of it starts.

[John Cuthber]

 

Mass plus velocity!

Nope.

Mass times velocity- sometimes.

"In common usage the term "inertia" may refer to an object's "amount of resistance to change in velocity" (which is quantified by its mass), or sometimes to its momentum, depending on the context. "

from

http://en.wikipedia.org/wiki/Inertia

 

but if we mean "Inertia is the resistance of any physical object to any change in its state of motion (including a change in direction). "

(also from wiki) then it's mass.

 

The fact is that it's a poorly defined word and oughtn't really pop up in science discussions without clarification.

[Alan McDougall]

That's not an issue raised in the OP.

 

That's not an issue raised in the OP.

 

OK in the game they were ejecting kids to change direction, but the inertia of an object resists change, you need kids of almost any mass to control a spaceship in this way. I will return later with some maths.

Nope.

Mass times velocity- sometimes.

"In common usage the term "inertia" may refer to an object's "amount of resistance to change in velocity" (which is quantified by its mass), or sometimes to its momentum, depending on the context. "

from

http://en.wikipedia.org/wiki/Inertia

 

but if we mean "Inertia is the resistance of any physical object to any change in its state of motion (including a change in direction). "

(also from wiki) then it's mass.

 

The fact is that it's a poorly defined word and oughtn't really pop up in science discussions without clarification.

 

My statement was not literal or mathematically true, I just meant these two factors must be taken into consideration in calculating inertia, and off course you are correct. I will be more careful!

[swansont]

p stands for vector, got it.

 

p is equal to 0 because there is no increase or decrease of momentum in the begining of the problem, correct? If there was an increase, then it would be a non-zero number.

 

to get v₁=-(m₂/m1)v₂ I just apply basic math? Divide both sides by m₁? if so, shouldn't the v₂ also be divided by m₁ or no because the m's cancel each other out.

 

 

The m's don't cancel as they are not equal.

 

You seem to have missed where I said I analyzed the problem in the spaceship's frame of reference, where it is at rest. That's the only way the math I used works out. Otherwise you can't set the initial momentum to zero.

[CyborgTriceratops]

Yeah, I know they don't cancel out, not to sure why I said that they did...

My math would of worked out then, had the spaceship not been moving. Since it was, my calculations are in effective and I must return to using F = dp/dt = v dm/dt + m dv/dt?

[Alan McDougall]

Yeah, I know they don't cancel out, not to sure why I said that they did...

 

My math would of worked out then, had the spaceship not been moving. Since it was, my calculations are in effective and I must return to using F = dp/dt = v dm/dt + m dv/dt?

 

You would have serious difficulty using kids as mass ejection to alter the speed and direction of said spaceship. Heck they eat a lot , their mass thus, changes from moment to moment, the also excrete mass, chew mass, drink mass and come in different sizes and shapes?. How do you eject these odd shaped objects at the exact correct angle, at the exact correct velocity, do you have different ejection ports for different size kids?. Only after you have resolved those issues to exactitude, can you think of coming up with the correct mathematical equation.

[swansont]

 

You would have serious difficulty using kids as mass ejection to alter the speed and direction of said spaceship. Heck they eat a lot , their mass thus, changes from moment to moment, the also excrete mass, chew mass, drink mass and come in different sizes and shapes?. How do you eject these odd shaped objects at the exact correct angle, at the exact correct velocity, do you have different ejection ports for different size kids?. Only after you have resolved those issues to exactitude, can you think of coming up with the correct mathematical equation.

 

Quite the opposite. The equations are the easy part, and come first.

[Alan McDougall]

 

Quite the opposite. The equations are the easy part, and come first.

 

I did not say the equations were hard , in fact if you read what I said the difficult part came before I mentioned equations , "you would have serious difficulty using kids" How could you write an equation, without knowing the exact mass of the object you were going to eject from your spaceship, you must plug this data into your equation first?

Edited January 6, 2014 by Alan McDougall

[Strange]

a) You write the equation first and then determine the mass(es) to use.

 

b) "Round cows": for the purposes of the OP it is almost certainly accurate enough to approximate an average mass per kid (and the effects of a meal, excretion, etc. are going to be very small, anyway).

[swansont]

You write out the equations the way a physicist does, i.e. symbolically. You can come up with an equation that is a solution without knowing any of the numbers. In this case you will have an infinite number of solutions for any desired result, because mass, speed and angle are all variable. Or, you can restrict the values of the variables and come up with a range of possible solutions.

 

A mistake that many physics students make is substituting numbers in for symbols way too early in solving a problem. Plugging numbers in (and calculating) is often the last step in a problem.