Unless you have orbital geometry down cold this might take a little thinking. SET UP A large independent planet with a thick atmosphere included is 100 Mega-meters in diameter. Beyond that diameter we can consider that we have a real vacuum outside that diameter. “Independent” only means we will be ignoring any affect from a sun. It has a non-wobbling spin so we can always locate “the spot” 100 Mega-meters above the North Pole. An object is propelled straight up from the North Pole so that it will just reach “the spot” before gravity will return to the North Pole starting point on the surface. When the object momentarily stops at “the spot” we can give it a velocity of X or less on any heading we choose. We will consider headings as viewed from the side measured as 0^{0} continuing straight away from the planet (north); around to 180^{0} for straight back down to the north pole surface; and around to 360^{0} for the same direction as 0^{0} straight away. The max speed of X is exactly the escape velocity for a heading of 90^{0}. 1. At what other headings can escape be achieved? 2. At what headings does X need to be reduced so the object just reaches the infinity escape without any extra speed? 3. By how much do we need to reduce X, in order to establish the maximum possible orbits that allow a free return to “the spot”. 4. And what headings can we use to achieve those free returns to “the spot”? (note: ignore "progresion") For the orbital pros that know they know, use white for the first few days. If you're guessing, no need for white. RB
Really this is not that hard a brain teaser; the orbital math is simple even for beginners. But I guess there is no reason for using “White out” Any way let’s set a deferent perspective by solving from #3 Assume you can establish a circular obit at “the spot” by giving it a speed of V_{O} at 90^{0}. What is the minimum escape speed X at that same heading in terms of V_{O}? Then at what angle could you apply that same speed and still attain escape? Where can it be reduced and still escape?
[tex]V_O \sqrt{2}[/tex] At any angle smaller than 90° or greater than 270°. Nowhere. The escape velocity is the same no matter what angle you use.
Excellent. Except remembering this is a brain teaser why would I need any different speed for a lower angle say 110^{0} or 250^{0}? Is 90^{0} really as low as I can go? Similar issue with #4.
The problem with these other angles is, with the way you set up your problem, that they all lead to trajectories that take the object into the Planet's atmosphere. While it is possible to give your object enough velocity such that it is able to dip into the atmosphere, and still come back out with greater than escape velocity after losing some velocity to drag, the initial velocity needed depends on more factors than you give. On the other hand, if "the Spot" where your object starts is some distance above the limits of the atmosphere, then you could use lower angles, as long as the periapis of the trajectory does not take it into the atmosphere. In these cases the escape velocity would be the same as for any other angle. With the return orbit scenerio, drag from dipping into the atmosphere will alter the orbit such that it will not return to "the Spot". Again, if " the Spot" was some distance above the limit of the atmosphere, you could use lower angles and still produce a return orbit. You could even use lower velocities than [itex] \sqrt{\frac{GM}{r}}[/itex] as long as the periapis doesn't drop into the atmosphere.
I’ll accept the teasers first post can be interpreted as the spot being 100 Mega-meters from the center of the planet through the North Pole. Instead of the above the surface at the North Pole – that’s my fault, so which ever is most convenient for you is fine. BUT, I do not except using the diameter of the planet as its radius. #3 is at what fixed speed in terms of X can you create the maximum number of unique orbits that successfully return to “the spot”. #4 remains. But heading can be given as approximate. And as a hint to three and four with respect to your comment about, producing return orbits using lower velocities, to make an easy one; #5 what different speeds can be used at what fixed heading to produce the maximum number of return orbits.
Sorry, force of habit. I always use radius when I do these types of problems, and so even though you said, 'diameter' in your post , I thought 'radius'. So starting over: assuming diameter of the planet is 100,000,000m. and the spot is 100,000,000m from the center of the planet. If the intial velocity is X, then 0°- 135° and 225°-360° This eliminates any trajectories that enter the atmosphere. Any amount that causes the velocity to fall into the range from slighty less than X to about 58% of X, At 58% of X, the object will be in an orbit with an apapis of 100,000,000m and a perapis of 50,000,000m. just over 45° to just under 135° and just over 225° to just under 315° At the extremes the velocity will have to equal almost X. At 90° and 270° the velocity can go as low as 58° of X.
Eliminating trajectories that enter the atmosphere is of course the key to that part of the riddle. But rather than varying speeds for #3 and #4 – isn’t there just one speed at the headings you recommend that will create successful return orbits to the spot? Kind of the inverse of the #5 question, I added as a clue.
No. At 90° for example, Any velocity v where [itex] X > v \geq .58X [/itex] will achieve a successful return orbit. At v=X the object flys off into space and at v< .58X, the object hits the atmosphere at perapis. ( When v< Vo "The spot" is the apapis of the orbit and when v>Vo, "The Spot" is the perapis of the orbit.) At say, headings of 60°, 120°, 300° and 240° you get successful return orbits at [itex]X > v \geq .707X[/itex] (.707X is equal to Vo) The closest you can come to a single valid v for a given heading is when your headings approach the limits of 45° 135° 225° and 315°. As you approach these limits, the minimum required V to maintain an orbit approaches X. So, at near 45°, 135°, 315° and 225° you get the smallest range of possible velocities for valid return orbits and at 90° and 270° you get the largest range of valid velocites.
I’m going to call this close enough, If I mess up this closing you look like you’ve got the orbital math better than I to correct me. Maybe you can help me out with “Coordinate” precession of perihelion, under the Celestial Mechanics Forum. Anyway as identified the circular orbit speed of V_{ 0} (X / sqrt 2) used at 90^{0} or its inverse establishes a round orbit. And for #5 any speed will give a return orbit establishing the spot as apapis or perapis as long as the heading is 90^{0} or its inverse. As to #3 & #4 you say for heading 60^{0} & 120^{0} and the inverse the speed V_{0} is again needed for a return orbit. At least implying V_{0} for heading between 60-90-120^{0} must all use V_{0}. In fact all headings should use the same V_{0} to set a stable return orbit including those higher angles. “All or any” limited of course to speeds and angles that don’t hit or escape the planet. Any speed at 90^{0} any heading using V_{0}.
Not any speed. It must fall between between .58X and X It doesn't need to be V_{0}. V_{0} is just the minimum velocity needed to keep perapis out of the atmosphere. For these heading the speed can be anything between V_{0} and X and still produce a return orbit. All these heading can use V_{0} for a return orbit, but they don't have to to, at 90° you could use .58X , a tad under X, or any speed between and still get a return orbit. At headings between 60-120 and their inverses V_{0} is one viable speed for a return orbit, but not the only one. At headings more than 30° from 90° or 270°, V_{0} is too low to keep the orbit out of the atmosphere. If you are looking for the one speed that produces a stable return orbit through the greatest range of headings, then that would be a speed just a tad under X.
I don't understand are you saying that the limitation I put on any and all (that you even quoted) didn't say this already. As I already said "any" speed yes for 90°. Within the limitation on "all & any" already given, that means the correct concept would be; any speed at any heading will return to the same spot with the same speed and heading. Didn’t have that – I’ll try it. Thanks