Hi we just studied motion under central force. we got the following question... is this possible trajectory(see attachment) under central force and force source is outside the loop? (my answer is that it is possible if force source is repulsive) whatever the answer is how can i explain it using physical arguments ?
i know that one of the angular momentum vector components is conserved so the entire motion is in the plane(perpendicular to that component) also usual energy is conserved...
Good, now what happens to the angular momentum as you move along that orbit? You don't need numbers, just imagine putting the central force somewhere outside that loop and think about how the velocity would have to change in order to maintain the angular momentum. What is an expression for the instantaneous angular momentum relative to the central body?
i already know that i was wrong... this trajectory is not possible under central force but i don't see how can i say that by the picture.
[tex] L=rmv\sin(\alpha) [/tex] so if one wants to maintain L constant then if r increases then [itex]v\sin(\alpha)[/itex] decreases and if r closer to force center the [itex]v\sin(\alpha)[/itex] is larger. i don't see any contradiction to this on the picture(or maybe the direction of L is what matters?) when particle is moving from (a) -> (b) its r vector becomes shorter and v increases and vica versa when it moves from (c) -> (d) by the way, what exactly makes such a trajectory impossible under central force? is it because two lines cross each other?
Follow up on this. If earth were going the other way around the sun, would it have the same angular momentum? Is angular momentum a vector or a scalar?
ok... so at when particle moves from (a) to (b) its velocity vector point to left while when particle moves from (c) to (d) its velocity vector points to right. so according to right-hand rule [itex]\vec{L}[/itex] changes direction... right?