Dear all, Suppose we have a uniform spherical planet, of density p and radius R, and we drill a straight hole from one point to its antipode (neglecting any changes in the gravitational force that this will entail). Then, we drop a point mass from one of the ends, and it will travel through the hole. It is simple enough to show that the particle will oscillate with SHM, and the half-period of this motion will be sqrt(3pi/4Gp). It turns out that, even if the two ends of the hole are not antipodal, the particle will oscillate with the same half-period, and proving this is not much harder than in the previous case. I talk about the half-period because that is the time taken by the particle to cross the planet. In the case in which the points are not antipodal, this might not be the shortest time our particle could take to travel between the two points. For example, if the path travels towards the centre and then away, in a "curvy" shape, then the particle will be travelling faster for the journey, and this might compensate for the fact that the particle has a longer distance to travel. So the problem is to decide what the path is which will lead to the shortest time of travel. We'll assume air resistance and friction don't exist, and that no time is wasted turning (if that were possible). Regarding my own efforts on this problem, I've managed to prove some ostensibly wrong results, so am interested to see what you think. -- James
Hello, I am not certain I understand correctly but isn't this curve the brachistochrone ? http://en.wikipedia.org/wiki/Brachistochrone_curve Which is the minimal time curve, is it what you are looking for ? Or another one with constraint ? : http://en.wikipedia.org/wiki/Tautochrone_curve Best regards, simon plouffe
I see no reason why it has to be the same as the brachistochrone. That assumes we're in a uniform gravitational field, whereas the field works differently here; the field at a distance r from the centre of the planet can be given by GM'/r^2, where M' is the mass of planet which is at a distance less than r from the centre. On Wed, May 16, 2012 at 2:56 PM, Simon Plouffe <simon.plouffe@gmail.com>wrote:
Hello,
I am not certain I understand correctly but isn't this curve the brachistochrone ?
http://en.wikipedia.org/wiki/Brachistochrone_curve
Which is the minimal time curve, is it what you are looking for ?
Or another one with constraint ? : http://en.wikipedia.org/wiki/Tautochrone_curve
Best regards, simon plouffe _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- James
Wikipedia states baldly that the curves that solve these two completely different optimization problems have the same cycloid shape. No proof is given for the tunnel-through-the-earth problem. On Wed, May 16, 2012 at 10:14 AM, James Aaronson <jamesaaaronson@gmail.com>wrote:
I see no reason why it has to be the same as the brachistochrone. That assumes we're in a uniform gravitational field, whereas the field works differently here; the field at a distance r from the centre of the planet can be given by GM'/r^2, where M' is the mass of planet which is at a distance less than r from the centre.
On Wed, May 16, 2012 at 2:56 PM, Simon Plouffe <simon.plouffe@gmail.com
wrote:
Hello,
I am not certain I understand correctly but isn't this curve the brachistochrone ?
http://en.wikipedia.org/wiki/Brachistochrone_curve
Which is the minimal time curve, is it what you are looking for ?
Or another one with constraint ? : http://en.wikipedia.org/wiki/Tautochrone_curve
Best regards, simon plouffe _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- James _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Here is an easier puzzle to get started. Imagine a frictionless, gravity-driven track between two points at the same height, separated by a horizontal distance X. The track is in the shape of a squared-off U, with a vertical drop, a horizontal traverse of length X, and a vertical rise. Suppose our car can negotiate the right-angle turns without losing energy. We start our car at one endpoint, at zero velocity, and we are interested in how long it takes the car to arrive at the other endpoint. If the verticals are very short, the car will only accelerate for a brief time, and thus it will move very slowly along the horizontal segment, so a sufficiently shallow track will take an arbitrarily long time to traverse. If, on the other hand, the verticals are very long, the car can take an arbitrary amount of time falling down and up the vertical segments, and so a sufficiently deep track will also take an arbitrarily long time to traverse. Somewhere between these two extremes is the optimal rectangular-U-shaped track. Calculus easily gives the optimal proportions for this rectangle. I was slightly surprised by the answer. I think this would be a fine exercise for elementary differential calculus. Is there a way to get the answer quickly without calculus? On Wed, May 16, 2012 at 9:56 AM, Simon Plouffe <simon.plouffe@gmail.com>wrote:
Hello,
I am not certain I understand correctly but isn't this curve the brachistochrone ?
http://en.wikipedia.org/wiki/Brachistochrone_curve
Which is the minimal time curve, is it what you are looking for ?
Or another one with constraint ? : http://en.wikipedia.org/wiki/Tautochrone_curve
Best regards, simon plouffe _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Excellent! What about the corresponding problem where the track is V-shaped? It's a similarly easy exercise with a similarly nice answer -- again, I'd love to know if one can derive it directly without resorting to calculus. On Wed, May 16, 2012 at 11:22 AM, Veit Elser <ve10@cornell.edu> wrote:
On May 16, 2012, at 10:21 AM, Allan Wechsler wrote:
I was slightly surprised by the answer. I think this would be a fine exercise for elementary differential calculus.
Your problem was on the midterm of my mechanics course this semester.
-Veit
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Both the U and the V problem can be solved using inequality techniques, in slightly different ways. I'll do them below: (spoiler below) (spoiler below) Suppose the downward drop length is a (and the horizontal length is x, and the acceleration by gravity is g). We can work out v at the bottom (sqrt(2ga)) and the time taken to drop (sqrt(2a/g)) using suvat. This gives a time taken of x/sqrt(2g) * (1/sqrt(a)) + 2sqrt(2/g) * sqrt(a) overall. Now, if a = K^4, x/sqrt(2g) = M^2 and 2sqrt(2/g) = N^2, then this quantity is (M^2K^-2 + N^2K^2 = (M/K - NK)^2 + 2MN, which is at least sqrt(x/g) * 2sqrt(2) with equality when a = (M/N)^2 = x/4. Suppose that the angle at the base of the V is 2a, and the horizontal length is x (and the acceleration due to gravity is g). The component of g in the direction of motion is g cos a, and the distance to travel is x/2 cosec a. Now, the time on each branch of the V is given by the suvat equation s = 1/2 at^2 so t^2 = x/g * (1/sin(a)cos(a)) = x/g * (2/sin(2a)). So t(total) = sqrt(8x/g) /sin(2a) >= sqrt(8x/g) with equality iff the angle of the V is a right angle. On Wed, May 16, 2012 at 4:42 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Excellent! What about the corresponding problem where the track is V-shaped? It's a similarly easy exercise with a similarly nice answer -- again, I'd love to know if one can derive it directly without resorting to calculus.
On Wed, May 16, 2012 at 11:22 AM, Veit Elser <ve10@cornell.edu> wrote:
On May 16, 2012, at 10:21 AM, Allan Wechsler wrote:
I was slightly surprised by the answer. I think this would be a fine exercise for elementary differential calculus.
Your problem was on the midterm of my mechanics course this semester.
-Veit
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-- James
there are several interactive mathematica demonstrations on this general subject. coincidentally, one of them is new as of today: tunnel through the earth http://demonstrations.wolfram.com/SphereWithTunnelBrachistochrone/ this one is fairly simple, but has some references http://demonstrations.wolfram.com/MotionInATunnelThroughTheEarth/ tunnel through the earth and other planets, with rotation http://demonstrations.wolfram.com/PlanetaryTunnelsWithCoriolisEffect/ bob baillie --- James Aaronson wrote:
Dear all,
Suppose we have a uniform spherical planet, of density p and radius R, and we drill a straight hole from one point to its antipode (neglecting any changes in the gravitational force that this will entail). Then, we drop a point mass from one of the ends, and it will travel through the hole. It is simple enough to show that the particle will oscillate with SHM, and the half-period of this motion will be sqrt(3pi/4Gp). It turns out that, even if the two ends of the hole are not antipodal, the particle will oscillate with the same half-period, and proving this is not much harder than in the previous case. I talk about the half-period because that is the time taken by the particle to cross the planet.
In the case in which the points are not antipodal, this might not be the shortest time our particle could take to travel between the two points. For example, if the path travels towards the centre and then away, in a "curvy" shape, then the particle will be travelling faster for the journey, and this might compensate for the fact that the particle has a longer distance to travel.
So the problem is to decide what the path is which will lead to the shortest time of travel. We'll assume air resistance and friction don't exist, and that no time is wasted turning (if that were possible).
Regarding my own efforts on this problem, I've managed to prove some ostensibly wrong results, so am interested to see what you think.
participants (5)
-
Allan Wechsler -
James Aaronson -
Robert Baillie -
Simon Plouffe -
Veit Elser