Expert: Jayendra Upadhye - 8/25/2008

Question
in space if i throw somethings i move opposite, everythings is weightless, so if i hold spaceship and throw somethings again, should the spaceship moves with me then?

Answer
Hi Rajesh,
Sometimes when it is difficult for beginners to grasp (visualise, "feel" etc) what "will happen" in certain scenarios, there is a very nice way to exit the situation / dead-lock in the mind.

One must use the un-violable conservation laws to look at the situation in a "different light".

Then since the conservation laws are "always right", whatever answer you get, "will be right".

Apply this to your question.
1 - Let the mass of spaceship be Ms.
2 - the mass of man be Mm.
3 - weight of object he throws be Mo.

Now, we use the law of conservation of momentum.

By this law, momentum of system before and after the throw "shall be conserved".

We ignore the system before the throw, as THAT velocity can be made zero gaianst an arbitary system with "same velocity".

[
In any given isolated frame of refrence, the initial velocity if shared by all objects in question, namely the thrown object, the thrower (man) and the base spaceship, is SAME, it can be ignored, as after the throw, it will continue to be shared as a component velocity by all those objects.
That initial velocity then can be simply assumed to be zero as no other object of reference is mentioned, against which it can be calibrated/measured.
]

The individual velocities of the system components will be valid only for that frame of reference, and with respect to each other.

So,

equation 1:- (Ms + Mm)*V1 + Mo*V2 = 0

Since the initial momentum was zero (owing to arguments above which reduced initial system velocity to 0).

equation 2:- (Ms + Mm)*V1 = -(Mo*V2)
The -ve sign connotes that V1 and V2 velocities are opposite to each other.

The combination of spaceship + you moves away from the direction of throw of the projectile, at velocity V1 such that

equation 3:- V1 = -V2 * [Mo /(Ms + Mm)]

Knowing Mo, Mm, Ms, and the launch velocity of the object V2, we can always find V1.

Vi, and V2 depend one on the other and knowing one, the other can be found.
interesting equation is
equation 4:- V1/v2 = -[ Mo/(Ms + Mm)] = constant

No matter how fast or slow you throw, the ratio of both velocities will be a constant decided by the component masses!

Hope that suffcies.

In an aside, i will like to pint out that the same analogy works when finding "what happens" in a conservative gravitational system (potential energy against kinetic energy).
The work energy thorem helps there whereby, IN CONSERVATIVE SYSTEMS, work done against the field (going up), is stored as potential energy, and given back as kinetic energy if one is allowed to "fall".

The gravitational field is a conservative field.

Friction is dissipative, one looses energy when working against it in any direction.

Work done on gases is to some extent conservative, if "things happen fast" (adiabatic system). Otherwise, it is a dissipative process in isothermal slow processes.

Most importantly, in a incompressible fluid flow system, prerssure can be described as energy per unit volume!

I leave it to you to find out HOW, as an excercise.

Analogies help greately in understanding system behaviour say by equating voltage to pressure (work/unit charge), current to flow (rate of flow of charge), etc.

Please do rate the answer if you find it helpful.
regards
Jayen