Elliptical orbits and acceleration

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growingpains

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True of false: "For an object turning with constant speed in a non-perfectly circular path, the net force will not be exactly perpendicular to its motion"

I thought that this statement is true and that the centripetal acceleration vector always points toward the center of the orbited mass (which would result in acceleration vectors that are not perpendicular to the velocity vector in elliptical paths). EK 1001 Physics #199 implies that the above statement false.

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I am a bit rusty on this, but I think you're confusing perpendicular to its velocity with meaning it's going to be pointed directly at its center of mass at all times. Google a picture of an elliptical orbit (or just draw one), pick any spot, draw a tangent line to it's velocity vector there, and then draw a force vector normal to that that vector. It always shows how it's velocity vector will change as it goes forward.

I think the key here is constant speed. That means whatever net force it gets from gravity parallel to its velocity vector is zero for such a place.
 
True of false: "For an object turning with constant speed in a non-perfectly circular path, the net force will not be exactly perpendicular to its motion"

I thought that this statement is true and that the centripetal acceleration vector always points toward the center of the orbited mass (which would result in acceleration vectors that are not perpendicular to the velocity vector in elliptical paths). EK 1001 Physics #199 implies that the above statement false.


u6l4b7.gif


http://www.physicsclassroom.com/class/circles/u6l4b.cfm

However, the statement should be true because it is a double negative.. Answer "TRUE" means that the net force WILL NOT always be perpendicular. Answer "FALSE" means that the net force WILL always be perpendicular.

Right?
 
Doesn't the force vector always point towards the center of the orbited mass? If so, the force vector would only be perfectly perpendicular to the velocity vector at two points: when the orbiting object is closest and farthest from the orbited object.
 
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Doesn't the force vector always point towards the center of the orbited mass? If so, the force vector would only be perfectly perpendicular to the velocity vector at two points: when the orbiting object is closest and farthest from the orbited object.

Yes, I agree with you. I would think True as well because the statement is a double negative.
 
Sorry I edited my post on the end. I didn't really properly put things in the first paragraph.
 
Another way to think about this problem is to notice that speed is constant, even though the object is changing its direction.

This scenario is only possible when there are no force vectors (and any components of it) in direction of motion. Otherwise, the speed would change. Thus, the only way this is possible is by having net force only in perpendicular direction to the motion.
 
Another way to think about this problem is to notice that speed is constant, even though the object is changing its direction.

This scenario is only possible when there are no force vectors (and any components of it) in direction of motion. Otherwise, the speed would change. Thus, the only way this is possible is by having net force only in perpendicular direction to the motion.

speed of an elliptical orbit is not constant.

Orbit5.gif
 
AH! Samchik and chiddler, thank you so much. Now it makes sense.

I forgot that speed in an elliptical orbit is non-constant.

You guys are awesome.
 
Ah crap, missed the constant speed part. Nice job, guys!
 
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