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one of the diagrams in principia

Actually I was going through the principia, reading Newton's derivations of the properties of ellipses. Suddenly I had this question which got stuck in my mind - Throughout Principia Newton uses the diagrams that i have drawn above. He says that if the body is moving in any orbit under a central force then if it moves from P to Q then its actual dispacement is RQ. Chandreshakhar uses this fact ( even Newton ) has to get the formula for centrepetal force. Sorry for the bad digram but in a small time interval RQ is parallel to the radius vector (fig A). Since displacement is parallel to the radius vector then it is parallel to the centrepetal force. Hence work done in that time interval should not be zero. But work done in circular motion is zero ( by the central force). So is the work done zero in a complete circle because even for a quarter or half a circle it shouldn't be zero according to this logic. Is there some flaw in my question. Sorry that its a very basic question but I am unable to find the fault within it ,if it has.

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Notice that in the finite approximation the vector PR is not perpendicular to the radius, so work is done on it. By the drawing, this work is of the opposite sign that that in QR, so they both compensate to zero. I'll leave to you to compute both if you are really interested.

Another way to see it, the gravity force is derived from a conservative field, so the total work between two pints is independent of the path. If you follow the circular path it will be zero. If you follow the drawn path is should also be zero

Update: The line PR is only perpendicular to the radius on P, if you take a segment near R you will notice that it is no longer perpendicular to the radius. So the mass moving along PR is not moving perpendicular to the radius, and thus gravity is doing work on the mass when it moves along that line. PR points outwards, so the work made by gravity along that path will be negative. The path RQ is also not perpendicular to gravity, but its orientation is such that gravity makes positive work on the mass along the path. The works are thus opposite in sign, and they are also of the same magnitude so they cancel each other. To show that they are equal you need to know differential calculus. The mistake is to assume that PR is perpendicular to the radius which is not.

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  • $\begingroup$ I am getting a feeling that your answer is correct but I am unable to understand in the full deapth what you are saying. Can you pleases explain more because yours is the only answer seeming to be correct. Please can you explain a little more pointing out the flaw in what I am doing or saying in the question $\endgroup$ – Shashaank May 27 '16 at 21:17
  • $\begingroup$ I added some comments, let me know if it helps or you need more detail. $\endgroup$ – user83548 May 28 '16 at 0:14
  • $\begingroup$ Yes but consider the case in gravity free space. We tie a ball to a string then whirl it in a circle. Now work done is defined as product of displacement in the direction of force and the force. Thats how we come to the dot product i.e w= f.d Now the the thing is that if the the tension had not been applied then the ball would have moved along PR. PR is not perpendicular to radius at P only if the motion wasn't circular or the tension was not continuous. In other cases PR is perpendicular to radius at P since the motion in circular ( tangent is perpendicular to normal). Continuing in comment $\endgroup$ – Shashaank May 28 '16 at 10:12
  • $\begingroup$ Now PR would be the path of ball had there been no tension but since there is tension , then in an infinitesimally small time the ball's net DISPLACEMENT would be QR. It deviates from PR by QR. Now QR is infinitely small and the point is that QR is parallel to radius at P. Force causes displacement in its direction. Hence f.d is not zero though it may be very small. Now integrating f.d till the limit to half circle we get a non zero value showing work is not zero though if we integrate over whole circle we will obviously get zero. But since we get a non zero value on integrating over half c $\endgroup$ – Shashaank May 28 '16 at 10:18
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    $\begingroup$ no, it is because the force does not act along RQ, but only at R, so $W=F\Delta x=0$ because $\Delta x =0$. $\endgroup$ – user83548 May 28 '16 at 16:45
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Work done by a central force is Zero. At every moment the force is perpendicular to the displacement of the test particle. If you see your diagram it's very easy to see that at the final and initial position of the particle the force is not in the same direction. What he actually does is to assume that P and Q are actually arbitrarily close. So now the assumption of

RQ is parallel to the radius vector

actually works. and the workdone is thus also vanishingly small.

Actually what we see here is a misconception that occurs while imagining calculus. Arbitrary small scales can give you pictures that are counter-intuitive. What Newton was trying is to take derivative of energy to get force. So he chose this path.

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  • $\begingroup$ Yes thats what I questioned. According to him and what you have just said I make out that the work done in a complete circle would be zero. But it shouldn't be zero in a quarter circle or half a circle because it would be only after the half of the circle has been traveled that work done would change sign and the summation of work done in small time intervals would become zero. Is what I am saying right $\endgroup$ – Shashaank May 25 '16 at 17:54
  • $\begingroup$ No I actually added a small portion to my answer. Workdone is zero at every point of the circle. At every point the displacement is parallel to the force. $\endgroup$ – Ari May 25 '16 at 17:56
  • $\begingroup$ And at P and Q the force is not in the same direction but everywhere for small time intervals the displacement is not perpendicular to the force. So we can summate all the small bit of work done $\endgroup$ – Shashaank May 25 '16 at 17:56
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    $\begingroup$ you say that work is done but negligible, that is not an argument because you can integrate it and make it larger. What actually happen is that the net work is zero if you add the portion done during PR $\endgroup$ – user83548 May 25 '16 at 18:06
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    $\begingroup$ Another way to see it, the gravity force is derived from a conservative field, so the total work between two pints is independent of the path. If you follow the circular path it will be zero. If you follow the drawn path is should also be zero $\endgroup$ – user83548 May 25 '16 at 18:12

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