# Motion predicted by Newton's laws

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## Main Question or Discussion Point

Hello

The following thought is confusing me a little. Let say we have sphererical planet with a certain mass and radius fixed in space. Now we have a point particle that at time t0 has a velocity vo that is perpendicular to the vector from the center of the plant to the particle and has a magnitude so that v^2/r=Fgravitational. In this case it seems to me that the only possible behaviour of the particle is to continue in a perfectly circular orbit for eternity(in a mathematical sense).

Now heres the question. Lets say we have a uniform, rod that has one end just above the surface of the planet and whose direction is so that the other end has the maximum distance from the surface of the planet. At time t0 let each part of the rod have velocity perpendicular to the vector from the center of the planet to the rod of magnitude v0r where v0r where r is the distance from the center of the planet to the part of the rod. Let the length of the rod be just so that the total gravitational force on the rod F is equal to the sum of centripetal forces needed for each part of the rod to maintain a circular orbit at its velocity.

Im am now trying to think about the behaviour of this rod, in particular if it will maintain a circular orbit continuing to point radially outward(like in the beginning). My mathematical toolbox is not very sophisticated but here is my attempt. After time dt the center of mass will be where it should for this circular orbit to take place since it behaves as if all the mass and force was concentrated there. The rod will however still point in the same direction after time dt since there is no angular momentum in regards to the center of mass. This means that it no longer points radially outwards and the circular orbit fails to take place. Is this correct? My second thought is if it is possible to give the rod some initial spin so that the described circular orbit can take place?

I am not sure how much sense this question makes....

Kind regards James

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jbriggs444
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After time dt the center of mass will be where it should for this circular orbit to take place since it behaves as if all the mass and force was concentrated there.
This is not correct.

For a spherically symmetric mass distribution interacting by gravity with objects in its exterior, the behavior is like a point particle with the mass concentrated in the center. That's Newton's shell theorem. https://en.wikipedia.org/wiki/Shell_theorem

For any mass distribution in a uniform gravitational field, the behavior is like a point particle with the mass concentrated at the center of mass. I do not know that this principle has a name -- it is something that we are taught, that we take for granted and whose truth seems reasonably obvious.

However, for a non-spherical mass distribution in a non-uniform gravitational field, things are not that easy.

One way to see this would be to add up the gravitational potential of all the points on a uniform rod that is aligned vertically and compare to the points on a rod that is aligned horizontally. The rod will tend to rotate into a position where its potential energy is minimized.

Another approach would be to compute a "tidal torque" for a rod that is aligned at a 45 degree angle.

This is not correct.

For a spherically symmetric mass distribution interacting by gravity with objects in its exterior, the behavior is like a point particle with the mass concentrated in the center. That's Newton's shell theorem. https://en.wikipedia.org/wiki/Shell_theorem

For any mass distribution in a uniform gravitational field, the behavior is like a point particle with the mass concentrated at the center of mass. I do not know that this principle has a name -- it is something that we are taught, that we take for granted and whose truth seems reasonably obvious.

However, for a non-spherical mass distribution in a non-uniform gravitational field, things are not that easy.

One way to see this would be to add up the gravitational potential of all the points on a uniform rod that is aligned vertically and compare to the points on a rod that is aligned horizontally. The rod will tend to rotate into a position where its potential energy is minimized.

Another approach would be to compute a "tidal torque" for a rod that is aligned at a 45 degree angle.
Are you sure? In my book (Kleppner Kolenkow) it says " the motion of a systems center of mass behaves as if all the mass were concentrated there and all the external forces act at that point". edit: Oh I see the mistake now.

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So I now realise that the rod does have angular momentum in regards to the center of mass at time t0. Will the rod undergo uniform circular motion while pointing radially outward?

jbriggs444
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So I now realise that the rod does have angular momentum in regards to the center of mass at time t0. Will the rod undergo uniform circular motion while pointing radially outward?
Yes, of course. You've specified the length appropriately so that gravity is exactly right to provide the force to keep all pieces of the rod in uniform circular motion. If the pieces are in uniform circular motion then so is the center of mass.

Yes, of course. You've specified the length appropriately so that gravity is exactly right to provide the force to keep all pieces of the rod in uniform circular motion. If the pieces are in uniform circular motion then so is the center of mass.
Ok thanks. I can definetly see that if the rod is undergoing uniform circular motion it has to have that length so that gravitation and the centripedal force are balanced. I am however having trouble picturing why having the forces balanced means that uniform circular motion is the outcome(if for example the rod would not point radially outwards even for a little bit the forces would no longer be balanced). Could you please explain further why the forces beeing balanced means the road will undergo uniform circular motion.

jbriggs444
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Ok thanks. I can definetly see that if the rod is undergoing uniform circular motion it has to have that length so that gravitation and the centripedal force are balanced. I am however having trouble picturing why having the forces balanced means that uniform circular motion is the outcome(if for example the rod would not point radially outwards even for a little bit the forces would no longer be balanced). Could you please explain further why the forces beeing balanced means the road will undergo uniform circular motion.
In your post number 1, you stipulated that the rod is rotating -- you specified the velocity of every point on it. That stipulation means that it is rotating at precisely the right rate to remain forever vertical.

Are you trying to ask whether this equilibrium is stable?

In your post number 1, you stipulated that the rod is rotating -- you specified the velocity of every point on it. That stipulation means that it is rotating at precisely the right rate to remain forever vertical.

Are you trying to ask whether this equilibrium is stable?
No I am not interested in stability. So we know that at time t0 the velocity has the right direction and magnitude at each point for the desired circular motion. We also know that the total gravitational force balances the total centripedal force at time t0, both of these are definetly prerequisites. But dont we also need to know that the Force(sum of tension and gravitation) at each differential segment of the rod balances that points centripedal force? How do we know this?

jbriggs444
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But dont we also need to know that the Force(sum of tension and gravitation) at each differential segment of the rod balances that points centripedal force? How do we know this?
This thread contemplates a rigid rod. It is enough that the sum of the forces is right. If we were contemplating a rope, one would need to establish that each segment was under tension -- but it is obvious by inspection that this must be the case. The parts of the rope toward the ends are being pulled outward by the tidal variations in both gravity and in centrifugal force.

This thread contemplates a rigid rod. It is enough that the sum of the forces is right.
Could you please explain why this is true?

jbriggs444
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∑F=ma. If the rod is rigid, the acceleration of the center of mass is representative.

Could you ellaborate? How is the center of mass representative for a rigid object, dont we also need to know its orientation in space? It would suffice to show that after a time period dt the rod is still pointing radially outward, could you do that? edit: I get it now! thanks for your help

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Could you ellaborate? How is the center of mass representative for a rigid object, dont we also need to know its orientation in space? It would suffice to show that after a time period dt the rod is still pointing radially outward, could you do that? edit: I get it now! thanks for your help
the second paragraph of post #1 is ambiguous regarding the rotation of the rod. By "whose direction is so that the other end has the maximum distance from the surface of the planet", do you mean for all time or just at t0? Unless you specify it's initial rotational rate (tumbling) I don't see anything in your description of the problem that forces the rod to remain pointed radially from the center of the Earth. If the initial position and rotational rate is correct, then it's easy to see that there is never a torque on the rod. Otherwise, it will not stay in a radial orientation for any length of time, torque or no torque.

the second paragraph of post #1 is ambiguous regarding the rotation of the rod. By "whose direction is so that the other end has the maximum distance from the surface of the planet", do you mean for all time or just at t0? Unless you specify it's initial rotational rate (tumbling) I don't see anything in your description of the problem that forces the rod to remain pointed radially from the center of the Earth. If the initial position and rotational rate is correct, then it's easy to see that there is never a torque on the rod. Otherwise, it will not stay in a radial orientation for any length of time, torque or no torque.
The conditions are set only for time t0. Dont the velocitys of each segment specify the rods rotation rate?

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The conditions are set only for time t0. Dont the velocitys of each segment specify the rods rotation rate?
Sorry. I overlooked that point.

turns out I am still confused...I just dont see why we know that the rod remains pointing radially out at all times...

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turns out I am still confused...I just dont see why we know that the rod remains pointing radially out at all times...
At time t0 does it have the right rotational rate that will rotate it once in every orbit?
Suppose that is true. At t0, the rod is oriented radially. Pick any point on the rod. The gravitational force on that part of the rod is in line with the rod's center of mass. Therefore there is no rotational torque from the gravitational force on that point on the rod. That is true for all the gravitational force on every part of the rod, so there is no rotational force at all on the rod from gravity. Therefore, the rotational rate of the rod will not change. Its rotational rate will remain exactly matching the rate that will keep it oriented radially.

• jbriggs444
Ok now I definetly get it! Thanks

cjl
The rotation of the rod isn't completely determined by the initial angular momentum - due to the gravitational gradient, objects in orbit will tend to align themselves such that their axis with the minimum moment of inertia is oriented vertically. In the case of the rod, this will tend to keep it pointing radially, so if you start the object in this orientation, it is a stable position for it to maintain regardless of small perturbations. Satellite designers have used this fact in the past to develop passively-aligned satellites that do not require active systems or thrusters to maintain orientation relative to earth.

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