Resolving the Paradox in Deriving Gravitational Potential Energy Function

[metalrose]

I'll first derive here the gravitational pot. energy by my method, and then I'll give the method that has been formally used in books. My answer differs from the actual one by a minus sign.

My derivation:

Let mass M be at the origin O. Let another mass m be at an arbitrary position r from the origin. The grav. force on m due to M is directed towards the origin. Let the mass m move a distance dr towards the origin due to this force.

The infinitisemal work done by the grav. force is, GMm/r².dr. I have expanded the dot product here, since the force and displacement dr both are in the same direction, and theta is 0 so cos(theta)=1.

Now we can integrate this infinitisemal work dW from the point a to b, and we get,

-GMm[(1/b) - (1/a)]

This is equal to the negative of the change in pot. energy. The change in pot. energy is U(b) - U(a).

So ,

-GMm[(1/b) - (1/a)] = -[U(b) - U(a)]

and thus,

GMm[(1/b) - (1/a)] = U(b) - U(a)

at a=infinity, we can chose U(a) to be 0.
We then get,

GMm/b = U(b)

--------------------------

as you see, my answer differs by a minus sign. In the book derivation, the only difference is that they have conventionally taken the infinitisemal displacement dr, to be in the outward direction.

That is, the unit vector r[tex]\widehat{}[/tex], is taken to be positive in the outward direction from the origin.

And then, according to this convention, the grav. force would be GMm/r²(-r[tex]\widehat{}[/tex])

and the answer we would get is U(r) = -GMm/r

-----------------

Though in a way, I have understood why the paradox arises, (due to the convention of taking r[tex]\widehat{}[/tex] positive in the outward dir.), I still want to know why we can't use my method?

Is it just a matter of convention?

One problem that I figured out with my method is this,

My method gives teh answer U(r)=GMm/r

This means, as the particle, travels from infinity towards the origin, under the attarctive force, r decreases, and hence acc. to my result, the pot. energy "increases" along with an increasing kinetic energy.

Please explain, what exactly is wrong with my approach and why it gives a physically non-meaningful result like the above?

Thanks

 

[Doc Al]

You found the work done by gravity, but the potential energy is the work done against gravity.

 

[D H]

Wrong answer.

Doc Al has it right; potential energy (or rather, the change in potential energy) is defined by ΔU=-W.

 

[metalrose]

@Doc al and D H

Yes, I found out the work done "by" the gravity but I never equated it to the change in P.E. I , instead equated it to the "negative of " the change in P.E.

So it becomes the same thing.

I guess the problem lies with the convention of chosing the +ve direction of the unit vector r.

 

[feynman92]

you took work done by gravity to be GMm/r2.dr which has to be positive as work done by gravity is positive.
so dr is positive.
by definition r_f=r_i+dr.
therefore r should increase as dr is positive.
but you are integrating from a->b ie in the direction of decreasing r.
hence the error.
to avoid such situations always choose direction of increasing quantities, as is done in your book.

 

[Dale]

Let the mass m move a distance dr towards the origin due to this force.

Here is the source of your sign error. If dr is a distance towards the origin then that implies that r coordinates closer to the object are larger than r coordinates further away. In other words, your r coordinate is the negative of the normal r in spherical coordinates. While there is nothing that really prevents you from using such coordinates they are a little weird and standard formulas will look slightly different in them, physically there will be no difference.

 

[metalrose]

@dalespam

physically there will be no difference

This is what even I expected. But if you look at my formula in these non standard coordinates, it is pot. energy = GMm/r. Now as the particle m moves closer and closer to mass M, 'r' goes down, and consequently, acc. to my non conventional coordinates, the "pot. energy goes up" instead of going down, as it physically does.

Therefore, the situation is not the same physically too! And this is what I don't understand.

Though I get a sense here that it's not a problem even if the pot. energy increases as the particles get closer, since what should be of concern to us is the total mech. energy
and which would be conserved acc. to my coordinates too and even if the pot. energy increases.

Am I on the right track??

Thanks!

 

[diazona]

If dr is a distance towards the origin then that implies that r coordinates closer to the object are larger than r coordinates further away.

Now as the particle m moves closer and closer to mass M, 'r' goes down

Do you see the inconsistency between these?

DaleSpam's statement (the first quote) reflects the math you did.

 

[Dale]

But if you look at my formula in these non standard coordinates, it is pot. energy = GMm/r. Now as the particle m moves closer and closer to mass M, 'r' goes down, and consequently, acc. to my non conventional coordinates, the "pot. energy goes up" instead of going down, as it physically does.

As diazona mentioned, you are mixing up your non-standard coordinates with standard coordinates. As the particle m moves closer to mass M the standard r goes down, but your non-standard 'r' goes up ('r'=-r). Consequently according to both the potential energy goes down.

 

[metalrose]

@Dalespam and diazona

Yeah you guys are totally right! Just figured out the inconsistency!

Thanks a ton!