July 18th, 2017, 04:04 PM  #1 
Newbie Joined: Jul 2017 From: Brighton Posts: 1 Thanks: 0  SmallPlanet Bullet problem A bullet, modelled by a point particle, of mass m is projected from height h above a planet, modelled by a perfect sphere in a vacuum of radius r and mass M. The initial velocity of the bullet is v, where v = v and the direction of v is tangent to the surface of the planet. See diagram. The particle therefore experiences Newtonian gravitational force, f=GMm/r^2 and so accelerates towards the centre of the planet at all times. What are the necessary and sufficient conditions, and the relationships between m, M, r, h and v such that the bullet lands on the planet exactly below where it was first projected, having completed one lap around the planet? What about if the bullet completes N laps? The attached document is the same diagram as above. 
July 22nd, 2017, 11:26 PM  #2  
Senior Member Joined: Jul 2012 From: DFW Area Posts: 607 Thanks: 83 Math Focus: Electrical Engineering Applications 
Hi vriska, and welcome to the forums. Quote:
1) M>>m, so that we can assume the center of mass is effectively at the center of the planet. I think this is reasonable since the problem is stated in terms of a planet and a bullet, and not a planet and a moon, for example. 2) r>>h, so that we can ignore changes in the acceleration due to gravity and the change in distance that the bullet travels around the planet as it loses height. However, for h=0, I would require any formula to reduce to a velocity of: $\displaystyle \large \sqrt{\frac{GM}{r}}$ Similarly, for h>0 and for very large N, I would require any formula to reduce to a velocity of: $\displaystyle \large \sqrt{\frac{GM}{r+h}}$ So here is how I worked the problem (as stated below, I am not sure that it is correct): The force due to gravity is: $\displaystyle \large F=\frac{GMm}{(r+h)^2}=mg $ So the acceleration due to gravity is: $\displaystyle \large g=\frac{GM}{(r+h)^2}$ The centripetal acceleration due to the orbit is: $\displaystyle \large a_o=\frac{v^2}{r+h}$ The difference is: $\displaystyle \Delta a = ga_0= \frac{GM}{(r+h)^2}\frac{v^2}{r+h}$ The vertical velocity in the downward direction is: $\displaystyle \large \Delta a \ t$ And the loss in height is: $\displaystyle \large \frac{\Delta a \ t^2}{2}$ So the time to lose height $h$ is: $\displaystyle \large { \begin{align*} \frac{\Delta a \ t^2}{2} &= h \\ t^2 &= \frac{2h}{\Delta a} \\ t &=\sqrt{\frac{2h}{\Delta a}} \\ t &= \sqrt{\frac{2h}{\frac{GM}{(r+h)^2}\frac{v^2}{r+h}}} \end{align*}}$ (It would have been cleaner to simplify this here but it is too late to rework it). Velocity is distance divided by time. The distance for N orbits is: $\displaystyle \large 2\pi N(r+h)$ This gives us a formula for the velocity: $\displaystyle \large { \begin{align*} v &= \frac{2\pi N(r+h)}{\sqrt{\frac{2h}{\frac{GM}{(r+h)^2}\frac{v^2}{r+h}}}} \\ v^2 &= \frac{4\pi^2 N^2 (r+h)^2}{\frac{2h}{\frac{GM}{(r+h)^2}\frac{v^2}{r+h}}} \\ v^2 &= \frac{4\pi^2 N^2 (r+h)^2}{\frac{2h}{\frac{GMv^2(r+h)}{(r+h)^2}}} \\ v^2 &= \frac{4\pi^2 N^2 (r+h)^2(GMv^2(r+h))}{2h(r+h)^2} \\ v^2 &= \frac{4\pi^2 N^2 (r+h)^2 GM}{2h(r+h)^2}v^2\frac{4\pi^2 N^2 (r+h)^3}{2h(r+h)^2} \\ v^2 &= \frac{4\pi^2 N^2 GM}{2h}v^2\frac{4\pi^2 N^2 (r+h)}{2h} \\ v^2 \left (1+\frac{4\pi^2 N^2 (r+h)}{2h} \right ) &= \frac{4\pi^2 N^2 GM}{2h} \\ v^2 \left (\frac{2h+4\pi^2 N^2 (r+h)}{2h} \right ) &= \frac{4\pi^2 N^2 GM}{2h} \\ v^2 &= \frac{4\pi^2 N^2 GM}{2h+4\pi^2 N^2 (r+h)} \end{align*}}$ I think that this might be the equation that you are looking for. Note that for h=0: $\displaystyle \large v=\sqrt{\frac{GM}{r}}$ And for h>0 and very large N: $\displaystyle \large v=\sqrt{\frac{GM}{r+h}}$ I wrote a program using the mass and radius of the Earth with adjustable height and number of orbits and it seems to work, so I think I got the algebra right, despite not simplifying the expression for $t$ first. The program outputs for a height of 100 km and 1 and 10 orbits are attached. The change in velocity from the perpetual orbit velocity is only 3.07 m/s for N=1 and 0.0307 m/s for N=10 (the difference apparently is inversely proportional to N^2). I did not quite expect these results as the change in velocity from the perpetual orbit velocity seems small to me. But the program does show that the change in height is indeed 100 km in both cases. As I said this may just mean I did the algebra right, I have no idea if my method is correct.  
July 28th, 2017, 10:11 AM  #3 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,080 Thanks: 698 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
I've been thinking a lot about this problem recently and also taking a look at JKS's solution and other solutions online. It's a very interesting problem with an extensive amount of literature on it because it gets to the heart of Keplerian orbital mechanics. I still have a fair amount of reading to do, but I'd like to mention some stuff here first. JKS's solution is pretty good and yields results that are relatively sensible and intuitive. There's an error in the trajectory of the mass, which is not precisely circular (it's a spiral), so the distance travelled for N orbits is not $\displaystyle 2\pi N(r + h)$ but I think it's a reasonable approximation given the assumptions already in play. If one relaxes the assumptions made by JKS, especially the one where the acceleration of gravity does not change with height, then you effectively have Keplerian orbits and can use the results derived from the analysis of twobody problems. The Wikipedia page on Keplerian orbits is amazing (https://en.wikipedia.org/wiki/Kepler_orbit) and I'm going to have a go at that set of maths this weekend if I get time. There are some interesting consequences of Keplerian orbits. The first is that in the absence of air resistance, tidal forces, Coriolis effect or other gravitational bodies, the orbit is exactly elliptical such that the moment a bullet is fired, its trajectory is represented exactly by an ellipse. If the initial velocity is slower than the orbital velocity required for circular motion, the initial position of the bullet represents an apocenter and the ellipse drawn may or may not intersect with the large body M. If it does, it will intersect within the first quadrant. That is, if the object travels some angle $\displaystyle \theta$ in polar coordinates before colliding, then $\displaystyle 0< \theta< \frac{\pi}{2}$. If this is not the case, then the ellipse will not intersect, the bullet will wrap around the whole body and back to its original position. Therefore, the number of possible orbits is either 1 or infinite. Consequently, the function of "number of laps made by the bullet as a function of initial speed" is $\displaystyle N = \left\{ \begin{array}{ll} 1, & u<v \\ \infty , & u \ge v \\ \end{array}\right.$ Therefore, when considering situations where it's the case where N is not 1 or infinity, the reason that's the case is because there's air resistance, tidal forces, Coriolis effect or some other acceleration that's affecting the bullet apart from gravity. I've tried experimenting with some equations of motion on paper myself and immediately ran into problems because the nonlinear second order ODEs are horrible. I also need to swot up on my vector calculus because I haven't done any for a very long time! Last edited by skipjack; July 28th, 2017 at 12:40 PM. 
July 28th, 2017, 09:31 PM  #4 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 607 Thanks: 83 Math Focus: Electrical Engineering Applications 
Benit13, Thanks for the insight, especially into how the bullet would either crash into the planet or settle into an elliptical orbit, as I was wondering about this (elliptical orbit vs. a circular orbit). I also think that my example was terrible, because at 100 km height, with the values given, g=9.51853 whereas on the ground g=9.81968 so I violated my condition. A better height would have been 10 km (g=9.78893) or 1 km (g=9.8166) but the velocities are obviously even closer to the perpetual velocities in these cases. 

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bullet, gravity, mechanics, problem, smallplanet 
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