Hello, Can anybody please explain me: While going through gravitation potential energy, I came across: F=G.m1.m2/r^2 From there it follows: g=-GM/r^2.r How does it follow? Specially the -G case? -- Shounak
They're defining the "gravitational field" to be: [tex]\mathbf{g}=-\frac{GM}{r^3}\mathbf{r}[/tex]. Basically this means that if you take the gravitational field at a point, [itex]\mathbf{g}(\mathbf{r})[/itex], and you multiply it by the mass [itex]m[/itex] of an object at that point, you get the force on that object: [tex]\mathbf{F}=m\mathbf{g}[/tex] The reason there's a "-" sign in the definition of the field is because the vector [itex]\mathbf{r}[/itex] points out from the "gravitating" mass. Since the gravitational field effectively tells you the acceleration that small, free particles experience, the [itex]\mathbf{g}[/itex] needs to point in the opposite direction of [itex]\mathbf{r}[/itex] (because gravity is attractive).
It explains: "We know that the further you get from an object, the higher your GPE relative to it. (As something must have done more work against gravity to get you there). Thus when you are infinitely far away, you have as high a GPE relative to it as possible. We choose (arbitrarily) to make the value of GPE of all bodies at infinity zero. Then since this is the highest value of GPE, all real values of GPE (closer than infinity) must be negative. Therefore the minus sign in the equation is NOT optional; it must always be included and all values of potential energy in a gravitational field are negative. (This is not the case when we come on to electric fields, because they can be repulsive too)." I want to understand "all real values of GPE (closer than infinity) must be negative." What does that mean? You have mentioned that the vector r points out from the gravitating mass. If I draw a diagram, a point m, the lines will be pointing out from the point......That means the vectors is not drawing inwards rather outwards, hence negative right? Is there any difference between POTENTIAL ENERGY AND GRAVITATIONAL POTENTIAL ENERGY? Thanks, -- Shounak
Hi Shounak! gravitational potential energy is potential energy (and potential energy is minus the work done by a conservative force) gravitational potential is potential energy per mass (just as electric potential is potential energy per charge) potential energy is relative we measure it relative to a test mass at infinity, whose PE we define to be zero since nothing can be at infinity (or further away!), that means that all real values of gravitational potential energy must be less than zero for electric potential energy, we use a (positive) test charge at infinity … so for any real positive charge, the electric potential energy will also be less than zero, but for any real negative charge, the electric potential energy will always be greater than zero not following you
Ok, I got it now. It is just as we want to avoid infinite, it is better even to get -ve values. If we consider all the non zero values it would be extremely difficult for us to compute. Am I right? What I am trying is just to draw a vector with points pointing outwards and hence -G.
Hi Shounak! no we're not trying to avoid infinity it's just that we want a convenient formula if we chose radius R as our "zero" level of potential energy, then the magnitude of the potential energy at a general distance r would be GM(1/R - 1/r) that's a rather cumbersome formula*, so we prefer to put R = ∞, which makes it GM(1/∞ - 1/r), = GM(0 - 1/r), = -GM/r * of course, if we're happy with an approximation … which we usually are … then we can use R, eg as the radius of the Earth here's an extract from the PF Library, to show how we get the usual mgh out of GM(1/R - 1/r) … Derivation of mgh: [tex]\Delta (PE)\ =\ \Delta(-mMG/r)[/tex] [tex]=\ \frac{-mMG}{r_{earth}\,+\,H\,+\,h}\ -\ \frac{-mMG}{r_{earth}\,+\,H}[/tex] which is approximately: [tex]\frac{-mMG(H\,-H\,-\,h)}{r_{earth}^2}\ =\ \frac{mMGh}{r_{earth}^2}\ =\ mgh[/tex] nope, still not following you
Thank you very much. It cleared me all up. Just 2 more questions: Wikipedia writes: "The singularity at r=0 in the formula for gravitational potential energy means that the only other apparently reasonable alternative choice of convention, with U=0 for r=0, would result in potential energy being positive, but infinitely large for all nonzero values of r, and would make calculations involving sums or differences of potential energies beyond what is possible with the real number system. Since physicists abhor infinities in their calculations, and r is always non-zero in practice, the choice of U=0 at infinity is by far the more preferable choice, even if the idea of negative energy in a gravity well appears to be peculiar at first." Hence I was asking about avoiding infinity. Secondly, if you can please explain H and h. Thanks.
wikipedia is simply saying that if you choose R = 0 as your "zero" of potential energy, then all other potential energies would be infinite! H is the height above Earth at which you choose the "zero" of potential energy for any particular experiment. h is the extra height above that. (so if you move a height h above H, the PE is +mgh)
It also says: "the choice of U=0 at infinity is by far the more preferable choice, even if the idea of negative energy in a gravity well appears to be peculiar at first"
where does this come from? what is it supposed to be? it's saying that you can choose U = 0 at any value of R, but that R = ∞ is (usually) the most convenient (for the reasons i gave above) it's adding that some people find negative energy peculiar … well, that's obviously correct, because you do!!
r is the (variable) distance at which you're finding the potential energy R is the (fixed) distance which you arbitrarily choose as your "zero" level for potential energy
To sum up: If r is non-zero, then there cannot be infinity, hence r is always non zero. If U i.e. GPE =0 at infinity then anything less than zero is always negative, hence it is negative. Right?
right no, r is non-zero because every massive object has a non-zero size, so you can't be zero from the centre of it
Ok, thank you very much for clearing all the doubts and answering the questions. I am on it. Thanks once again.