Weightlessness in Space: Understanding the Concept and Its Effects on Astronauts

[sarthak sharma]

Homework Statement

If two bodies, each of mass M and radius R, initially r (r>>>R) distant away from each other start approaching each other with negligible speed then what is their speed which they collide??

Homework Equations

The Attempt at a Solution

initial energy,IE = (-G m^2) / r

final energy,FE = { (-G m^2) / 2R } + (1/2) m v^2

solving...

v = ( 2Gm [ (1/2R) - (1/r) ] ) ^ (1/2)

 

[nasu]

Do you have a question?

 

[sarthak sharma]

Do you have a question?

the answer that i got doesn't matches with that of my book...
i don't know what wrong step i have done...
pleasez help me if u can...

 

[nasu]

What is the answer in the book?
Yours looks OK to me. Only that should be M and not m.

 

[collinsmark]

Hello sarthak sharma,

Welcome to Physics Forums! :)

Homework Statement

If two bodies, each of mass M and radius R, initially r (r>>>R) distant away from each other start approaching each other with negligible speed then what is their speed which they collide??

Homework Equations

The Attempt at a Solution

initial energy,IE = (-G m^2) / r

final energy,FE = { (-G m^2) / 2R } + (1/2) m v^2

solving...

v = ( 2Gm [ (1/2R) - (1/r) ] ) ^ (1/2)

Yep*.

*(By that I mean, "yes, it looks correct to me.")

[Edit: On further thought, strike that. See posts below.]

 

[sarthak sharma]

What is the answer in the book?
Yours looks OK to me. Only that should be M and not m.

yup its M only i typed it wrong by mistake...
in my book it shows
v = ( Gm [ (1/2R) - (1/r) ] ) ^ (1/2)
i.e. 2 is not there and it gives a reason for this and according to it we should count the KE of both bodies while calculating the final energy that is it should be 2*(1/2) m v^2 in place of (1/2) m v^2

btw i was also a bit confident that i was right and the stuff in my book is wrong so thanks for ur support
but pls do for once look out for the reason by the book as above

 

[nasu]

Yes, I think they are right. I did not pay enough attention to your derivation.

 

[sarthak sharma]

Yes, I think they are right. I did not pay enough attention to your derivation.

but pls can u tell me that why should we count KE of both...?

 

[sarthak sharma]

Yes, I think they are right. I did not pay enough attention to your derivation.

if we needed to count the KE for both then why should we don't count the PE for both...?

pleasez clarify my doubt asap i m very much confused by now...

 

[collinsmark]

yup its M only i typed it wrong by mistake...
in my book it shows
v = ( Gm [ (1/2R) - (1/r) ] ) ^ (1/2)
i.e. 2 is not there and it gives a reason for this and according to it we should count the KE of both bodies while calculating the final energy that is it should be 2*(1/2) m v^2 in place of (1/2) m v^2

btw i was also a bit confident that i was right and the stuff in my book is wrong so thanks for ur support
but pls do for once look out for the reason by the book as above

Right. If the frame of reference is the initial velocity (recall, initially, the bodies had negligible speed) then yes, you need to consider the kinetic energy of both bodies. On the other hand if you were calculating the relative velocity (the speed of one body with respect to the other), then your answer is correct. So it depends on the frame of reference. But what your book says makes sense.

[Edit: What I said above might be a little misleading. It's true that the frame of reference matters when considering velocities, but what I said about the answer being correct for relative velocity (the speed of one body with respect to the other) is not correct in this case. Sorry for the confusion.]

 

[sarthak sharma]

Right. If the frame of reference is the initial velocity (recall, initially, the bodies had negligible speed) then yes, you need to consider the kinetic energy of both bodies. On the other hand if you were calculating the relative velocity (the speed of one body with respect to the other), then your answer is correct. So it depends on the frame of reference. But what your book says makes sense.

pleasez can u explain a bit more cause i could not get you...

 

[collinsmark]

if we needed to count the KE for both then why should we don't count the PE for both...?

The potential energy already implies both bodies exist. The $U = -\frac{GMm}{r}$ formula is the potential energy of the system considering both bodies, $m$ and $M$.

The kinetic energy formula, $T = \frac{1}{2} mv^2$ only considers mass $m$.

So if the frame of reference is neither of the two bodies then one must consider the kinetic energies of both bodies.

pleasez clarify my doubt asap i m very much confused by now...

Hang in there! :)

 

[collinsmark]

Looking at it another way, suppose the frame of reference moved with one of the bodies, or suppose that one of the bodies was held stationary somehow so that only the second body was allowed to move: in that case, your original answer is correct.

[Edit: Strike my comment about the frame of reference moving with one of the bodies. The part about one of the bodies held in place is okay though.]

But if the frame of reference was the initial frame, when the bodies were not moving, then you must consider the kinetic energies of both bodies since both of them are considered in motion. In that case your book's answer is correct.

 

[sarthak sharma]

So if the frame of reference is neither of the two bodies then one must consider the kinetic energies of both bodies.

u know i m getting confused due to "frame of reference"

 

[sarthak sharma]

bro i don't have much knowledge but with just a limited knowledge i own...i think that frame of reference should not be applied in here as the bodies initially approach each other with negligible speed...

you know i don't have much and proper knowledge about initial frame of reference and how it is applied in different types of motion...

 

[collinsmark]

u know i m getting confused due to "frame of reference"

Yes, it can get confusing sometimes.

Just remember that velocity is always relative. It doesn't make sense to define a velocity without specifying what that velocity is relative to. Whatever has zero velocity is the frame of reference.

By extension, that means that kinetic energy is also relative. Things which have kinetic energy in one frame of reference might not have the same kinetic energy in other frames.

By the way, an inertial frame of reference is one that is not rotating, and is not accelerating. Picking one of the bodies as a frame of reference means the frame of reference is not an inertial frame, since both bodies are accelerating.

Your book's choice of picking the frame of references as the initial velocity frame is a good choice since it is an inertial frame. So while both answers are correct in a sense, in retrospect I prefer your book's answer.

 

[sarthak sharma]

hmm thanks collinsmark that one surely helped me...:)

 

[collinsmark]

Rereading my posts, I feel I need to make a clarification. Some of what I said might be a little misleading.

Your original answer is only correct if one of the bodies was held in place by some external force. But that's not part of the problem.

So go with your book's answer. That's a better way to approach the problem for more than one reason.

 

[sarthak sharma]

So go with your book's answer. That's a better way to approach the problem for more than one reason.

thanks for that...
can u plss help me out with this https://www.physicsforums.com/conversations/small-query-about-weightlessness-in-space.5223/