# Mass and its relation to top speed in space/water

1. Sep 3, 2007

### finerty

right this might get a bit convoluted :)

the original discussion started within an online space game and whether its physics are wrong or right within its own version of physics

sorry if this all sounds a bit pointless or not real physics

first pick a spot far far far away from anything else in space so that gravity is negligible, then image a globe containing water.

this is the general physics idea of the situation, so objects come to a halt due to a drag without a continual force applied, and objects with a constant force applied have a maximum velocity.

basically the question is whether if you have objects of identical shapes but of different masses they will have the same maximum velocity tho one will take longer to reach it.

i believe this is true because mass only affects the rate of acceleration and not the force applied in any way so the point at which maximum velocity is achieved (where the force applied and the opposite force from drag are equal) will be the same

am i right and how can i prove this mathematically, the only way i can think is simply that as when Fe=Fd (Fe = force from engine, Fd = force from drag) the object is at maximum velocity. Fe is a constant as the engines always output the same force regardless of the ships mass, and Fd is based on the area exposed to the substance causing the drag and has nothing to do with mass.

am i correct?
did i put the question it correctly (not posted here for a year)?
does anyone even care? :D

Last edited: Sep 3, 2007
2. Sep 4, 2007

### Astronuc

Staff Emeritus
Objects with a constant force will accelerate to some velocity, and the maximum velocity will depend upon the duration of thrust (application for force) and the prevailing graviational force.

That depends on whether or not they travel the same trajectory, have the same thrust, and if the one with lesser thrust maintains the thrust for a longer duration.

Now in reality, thrust requires mass flow, and as thrust continues, the mass of propellant decreases, so if thrust is constant, acceleration will increase as mass of the craft+propellant decreases.

Acceleration is F/m, and that is how one would handle problem this mathematically.

Drag is related to cross-sectional area and approximately the square of velocity.
http://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html

In fluids (gases and liquids) there is drag, in space (vacuum) there is no effective drag.

3. Sep 4, 2007

### finerty

hello astronuc
I'm sorry my example wasn't particularly clear, ill try to refining it

the problem is a combination of water and space, in the sense that there is drag from the water but no gravity

at a point in space where gravity is negligible small there is a sphere of water (the container is unimportant) inside this sphere are two submarines. Both submarines have the same engine and therefor the same force is provided when they propel the water backwards. Both submarines are also identical shapes and are traveling in parallel straight lines. The only difference between the two is that submarine A has a hold containing lead and submarine B has a hold containing nothing (air).

i think thats better defined now.

i realize this, but it doesn't affect the top speed the objects (submarines now) just how long they take to reach it, to simplfy things further let us say that both submarines are running on electrical motors so there is no loss of mass from the submarines as they apply force on the water

i looked at using f=ma to prove it but i got confused about it, in the situation described F is a constant, correct? so it is A that changes when the mass is altered. Better shown in the A=F/m

so can i simply say that F is a constant, as its assumed the engines output the same force constantly or is this 'cheating' in a algebraic proof.

the problem i have is that at top speed it should be a.m=Fe=Fd (who whole drag formula) so either i just say Fe is a constant or work out how mass cancels out in a.m=Fd(drag formula)

im very sorry this is more maths than physics, its just a case of i think im right logically im just not sure how i can prove it to the person im discussing it with conclusively.

ps. if it is reasonable to stat that Fe is a constant i can see that all i would need to do is show what the velocity in the drag equation equaled if it was alone on one side of the equation. as Mass wouldn't bee on the other side of the equation this would show mass wasn't involved, correct?

thank you for time.

Last edited: Sep 4, 2007
4. Sep 5, 2007

### AbedeuS

It reaches an equilibrium, your "Fd" will increase with speed, and Fe = Fa-Fd, the force that is "Accelerating" the boat is equal to the force "Left over" after you subtract the force actually being produced by the engine from the drag being provided.

That is to say, if an engine is somehow infinitely producing a 10N force out of its back, at first the acceleration of a 1kg submarine is F/m = 10Ms^-2 but as the velocity of the sub increases, so does its drag, the acceleration therefore will decrease.

The submarine is still producing, say a 10N output, but at speed x say, the accelerative force is actually now 8N, its acceleration is thereforce 8Ms^2 and this continually decreases until the force (Fe) is equal to the drag (Fd) and the submarine moves at a constant speed.

Now to your actual question, if both submarines had exactly the same shape, and size and were in the same medium, they would in theory hit the same equilibrium if they had some sort of magical engine that produced a 10N thrust constantly, the equilibrium for the heavier submarine would take slightly longer to reach as its acceleration is lower (and the derivative of its drag is therefore lower) but they should reach a point where Fe=Fd and Fe=10N constantly.

The logical problem though is that heavier submarines need more fuel to have the same acceleration as a lighter counterpart, at a low acceleration in theory you would hit the same equilibrium, just the equilibrium might lie at a very high speed for a non-viscous medium and therefore it will take a looong time for the submarine to reach the equilibrium point of an extremely light counterpart, in this long time, the submarine in the real world would have lost mass from accelerating itself or just plain ran out of energy.