Author Topic: Space Ship Q&A (more like concepts)  (Read 2350 times)

Offline arianabedi

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Space Ship Q&A (more like concepts)
« on: 08/21/2010 12:21 PM »
Dear readers,
Well the title is very vague and i apologize, i was trying to be like the rest of the topics here. Never the less lets get to the point!

My first question is regarding force applied via propulsion system of "spaceships" in non orbital manner  (where gravity isnt a factor).
Since there is no friction or gravity, does mass of a space ship determine the amount of energy needed for it to move?
So lets say we have this huge ship that is as big as a modern oil tanker that we have in oceans and we also have a ship as big as a normal truck, since they are considered weight-less (am i wrong?) , wouldn't they need the same amount of energy to for example reach the speed of 200 km per hour?

I am asking because (sorry for the non-realistic reference) in concepts and designs you see huge space ships having massive propulsion systems on their back (the HUGE exhaust) and when they are animated they are shown to move slowly compared to smaller "fighter" ships while the smaller ships actually have smaller engines (visually at least, since fuel would usually take up some space)

So all of this question revolves around matter being weightless in space, if this is not the case then my whole question is invalid.
And also does E=mc2 come in this since mass and energy are related here? im not sure maybe inertia comes into play?


My second question is regarding weaponry on say a to be built spaceship. Weapon is actually an example since the main law i am involving here is the "action-reaction" law of newton. "for every action there is an equal but opposite reaction"

Well my question is that for example if a spaceship shoots a bullet, does the whole, say a death-star (a massive semi-planet like ship) move backwards with the same force that was applied to the bullet?
Now if we exaggerate this to a massive nuclear bomb being shot out form a space station in orbit towards a planet, would the space station be blown backwards as well?

Please note that both objects shot out are pushed forward by the station/spaceship so the law would be triggered.

Please excuse me for my lack of knowledge, im just a curious teenage  :)

Offline Jim

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« Last Edit: 08/21/2010 01:18 PM by Jim »

Offline arianabedi

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Re: Space Ship Q&A (more like concepts)
« Reply #2 on: 08/21/2010 02:57 PM »
KE=1/2mv^2

Thank you for the reply Jim,
but one question i had is that we assume that a space walker pushes the space station with his hand, according to the action-reaction rule, wouldnt both masses move to the opposite of each other with the same speed?

If so then the same force moved a 80kg mass and a big station (2-3 tons?)

I've just done high school physics so im not completely familiar with everything that goes into account.

Offline Antares

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Re: Space Ship Q&A (more like concepts)
« Reply #3 on: 08/21/2010 04:05 PM »
Conservation of momentum is the easiest law to use in almost every case

m1v1 + m2v2 = 0

So in the space walker (m1) pushing on the space station (m2):
m1 << m2
Therefore v1 >> v2 and the directions are opposite.

In the Star Destroyer vs Shuttle energy comparison, for the same speed
EVader's ship >> ELuke's ship since
mVader's ship >> mLuke's ship even though
vVader's ship = vLuke's ship

E=mc2 means you can liberate E if you annihilate m.

In the equations of motion, it means that m actually increases the faster that m is moving.  If I recall correctly,
mfast = m0 / sqrt (1-v2/c2)

So that added mass is only noticeable when v approaches c or for long-term orbital calculations, like on probes escaping the solar system.
« Last Edit: 08/21/2010 04:07 PM by Antares »
If I like something on NSF, it's probably because I know it to be accurate.  Every once in a while, it's just something I agree with.  Facts generally receive the former.

Offline arianabedi

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Re: Space Ship Q&A (more like concepts)
« Reply #4 on: 08/21/2010 04:22 PM »
Conservation of momentum is the easiest law to use in almost every case

m1v1 + m2v2 = 0

So in the space walker (m1) pushing on the space station (m2):
m1 << m2
Therefore v1 >> v2 and the directions are opposite.

In the Star Destroyer vs Shuttle energy comparison, for the same speed
EVader's ship >> ELuke's ship since
mVader's ship >> mLuke's ship even though
vVader's ship = vLuke's ship

E=mc2 means you can liberate E if you annihilate m.

In the equations of motion, it means that m actually increases the faster that m is moving.  If I recall correctly,
mfast = m0 / sqrt (1-v2/c2)

So that added mass is only noticeable when v approaches c or for long-term orbital calculations, like on probes escaping the solar system.

Dear antares,
thank you for your reply, now I'm starting to imagine a more clear image of how the mechanics work in space. i will press that "thank you" button as soon as i find it :)  if there is any that is.

So then i see that mass does come into the equation since the same physic rules on earth work elsewhere.

Thank you.

Tags: Space ship