Force in Space

0.

Would be better to have written a better question. For instance, what are the initial conditions? You never gave us the initial conditions. Also, you never gave any time for the force to be applied. Were you looking for an equation of the velocity as a function of time? The answer Lyn gave you is 100% correct for the velocity at time zero assuming the initial conditions were zero velocity with respect point of observation.

At the risk of completing the OP's homework problem for him. You are not loking at the EQUATIONS correctly.

The OBJECTS are in SPACE, and therefore, not under a GRAVITATIONAL FORCE from EARTH or any other CELESTIAL BODY. By DEFINITION, an OBJECT in SPACE is in a THEORETICAL VOID where NOTHING outside of the DEFINED OBJECTS exist. This makes the PROBLEM one of SIMPLE MATH without any unnessaccary VARIABLES.

The EQUATION the PROBLEM is looking for is the FORCE-MASS-ACCELERATION EQUATION:

F=mA

Where:

F equals the FORCE applied to the OBJECT

m equals the MASS of the OBJECT

A equals the ACCELATION of the OBJECT

Now all EQUATIONS (with few exceptions) can be written to SOLVE for any VARIABLE within. What the QUESTION is looking for is the ACCELERATION, so we SOLVE the EQUATION for A:

F=mA

F/m=mA/m

F/m=A

A=F/m

Now the FORCE applied to the OBJECTS is identical, so F can be considered a CONSTANT, and the MASS of the two OBJECTS is different, so it can be considered a VARIABLE. Since we are not given EXACT VALUES and are only looking at the RELATIVE DIFFERENCE, we can treat all UNKNOWN CONSTANTS as 1:

A=1/m

This EQUATION shows clearly that MASS and ACCELERATION are INVERSELY PROPORTIONAL, the GREATER the MASS, the slower the ACCELERATION.

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Also, have you THOUGHT about your USE of ALL-CAPS WORDS? It MAKES it SEEM like you ARE putting strange EMPHASIS on certain WORDS, which BREAKS up the INNER MONOLOG people HAVE when READING your POSTS. It AlSo MaKeS tHe SeNtEnCeS hArD tO rEaD. I would understand Capitalizing Proper Names, or even Capitalizing every Noun (the latter happens a lot in German), but the ALL-CAPS bit has traditionally been 'read' as 'shouting,' and is as jarring as someone writing in leet-speak, ^n1) \/\/e a11 h8 r3^1)1ng 1337-sp33k.

One will be slower!

I give you a GA for patience.

Thanks, Canary.

Good answer BUT you missed the bit that newbies most often confuse.

The units for force and mass are different, and in the metric system these are typically Kg for mass and Newton for force.

The ratio of the magnitude of these two units being g or 9.8 m/sec2

Your maths is correct but I am not sure about your understanding. 1 newton is the force on 1 kg of mass that causes it to accelerate at 1 m/sec. On Earth the gravitational force is such that 1 kg of mass weighs 9.8 newton...that is where gravity comes in.

The relationship between the newton and the kilogram is determined by the unit of length - 1 m in the metric system. If the French had chosen 2m instead of 1m as the basic unit of length (call it SM for super-meter) then the system would have been 1 SN (super-newton) causes 1 kg to accelerate at 1 SM/sec2, and 1 kg of mass would weigh 4.9 SN - SN in that system would be twice the size N as we know it in today's metric system.

And of course we already have that sort of shift in the imperial system with lb and lbal were 32.2 lbal equals 1 lb and acceleration due to gravity is 32.2 ft/sec2.

"... then what is the velocity of these two body..."

Relative to... what?

If the frame of reference is body M, then the velocity of body M will be zero (as pointed out in post number one).

Maybe don't be so quick with the snotty comments.

Thanks.

You are correct.

However that 0 had a dual meaning.

The second meaning was the score of the quality of the original question.

If bodies have negligible effect of gravity by distance and masses, your initial velocities are the same as your final, what ever initial you cast the 2 bodies therein.

The bodies will have perpetual motion, not unless external forces are present to influence its dynamics.

There is no force at final state, since F= m dv/dt, dv= v_f-v_f= d(const)=0.

Initially there will be a force ex. F= M dv/dt= M [v_f-0]/dt so as with mass-m

v_f is the velocity resulting from Force F, initial v=0, when F is applied the change of dv will be V_f - 0, the rest of the transition then on will be v_f of mass M and m respectively.

I suppose the OP mean a momentary Force F. Applying Force all through might not be realistic if motion is achieved by a momentary instantaneous push.

is there any other mass in the neighborhood, such as a star or giant planet, possibly a comet with an upside-down lander on it?

LOL - good job on that one!

You mean to say we DID end up flipping it over?

Oh well, it was still incredible for a first attempt, since we literally had no idea what sort od surface we were going to land on, and we probably can still get some info back, even in that position. If nothing else we can get photos from a truly unique POV, and if we can keep the gear running (or sleeping) until the comet gets closer to the sun and grows a tail, we'll be able to get some before-and-after pics to see how comets melt/sublimate/outgass/whatever.

I heard that this may turn out to be one of those bad results that turns out to be far better than what was planned. In its current position, it is far more resistant to overheating while the comet approaches the sun. They turned the biggest solar panel to a position that should let it wake up later when there is more sunlight and the temperature increases.

http://bigstory.ap.org/article/cb36c392a2454f14a8c80b05c6245147/results-comet-landers-experiments-expected

Well, cheers all around.

No experiment is a failure if it can produce good data. Even if it wasn't the data we expected to get.

They would have minimal velocity towards each other according to their gravitational mass....the object with the lesser mass would move a little faster towards the greater mass....With no reference point or gravitational field present the objects can have no defined movement except that which would be caused by the objects themselves....

http://www.physicsclassroom.com/class/circles/Lesson-3/Newton-s-Law-of-Universal-Gravitation

That may be true, but the vectors on his drawing look like the force is applied externally (upward) and mutual attraction is not considered in that diagram.

That's my guess, but once again I feel like the mushroom physicist with the lack of information here.

Yeah you answered it in #5, I was just offering a different approach based on the lack of parameters and/or conditions....

Without knowing more about the initial conditions the question is meaningless.

Poor Fredski, I feel very badly for you about that Lander, It seems you were excited about the project. I loved your question...

Wow, Fredskie might be involve in Google Lunar project competition. Big job and pretty interesting.

I wonder what could be the effect or impact on environment mining minerals outside the earth.

But, wow, big dream for the corporate world.

Anyhow who owns the moon?

The chinese had been there already.

F=ma (Newton's First Law), nothing breaks the law, that I am aware of.

If two objects of varying mass have the same force applied, then each will experience a different acceleration by the equation above. If the time of acceleration is not specified, then nothing can be said about the magnitude of the velocity, only the vector may be known (aligned with the applied positive force).

Impulse is F•dt, and translates such that ∫F•dt = p (momentum).

If however, you meant to ask if the objects (en vacuo) were subjected to the same field (gravitational or otherwise), the time the field (acceleration) is applied does not affect the relative velocity of the two objects which is, in fact, 0.