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Why is angular momentum conserved when a planet revolves about sun in an elliptical orbit? Why is linear momentum not conserved in this case?

Please use the minimum amount of equations and try to explain using pure physics logic if possible.

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    $\begingroup$ Equations are "physics logic". Don't discriminate against formulae! $\endgroup$ – ACuriousMind Apr 7 '15 at 13:16
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    $\begingroup$ Note that, although the sun is often depicted as stationary, it is not! Perhaps your understanding will get better if you consider two equal masses rotating around each other in an elliptical orbit. $\endgroup$ – Sanchises Apr 7 '15 at 13:34
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Why is angular momentum conserved when a planet revolves about sun in an elliptical orbit? Why is linear momentum not conserved in this case?

$$\rm \text{no external }\color{red}{torque}\to\color{red}{angular}\text{ momentum conserved}\\ \text{no external }\color{red}{force}\to\color{red}{linear} \text{ momentum conserved}\\$$

There is no external torque about the sun since the force of the sun and position vector are always at an angle $180^\circ$ since $\bar \tau =\bar r\times\bar F$, so angular momentum is conserved.

But since the path in not circular but elliptical, the position vector is not perpendicular to direction of motion hence some work is done, which changes the momentum indirectly by changing magnitude of velocity directly.

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Both are conserved if you consider the whole system: If the earth looses linear momentum, the sun will gain it and vice versa. Subsystems may violate conservation laws (e.g by transfering energy/momentum). This is called local violation. But globally conservation laws will always hold.

The question why they hold globally in the first place, can be answered by Noether's theorem if the laws of physics (that is the equations of motion) are form-invariant under a continious transformation.

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  • $\begingroup$ It might be worth pointing that the central potential of a sun is rotationally, but not translationally invariant, hence angular momentum is conserved along an orbit in it, while linear momentum is not. (You are of course right about the global conversation, too) $\endgroup$ – ACuriousMind Apr 7 '15 at 13:14
  • $\begingroup$ @ACuriousMind: the term $|\vec{r}_{\text{sun}} - \vec{r}_{\text{earth}} |$ of the gravitational potential is invariant und translation. but most idealization don't translate the sun's postion under a coordinate transformation, because it is assumed to be fixed. this idealization breaks the invariance. $\endgroup$ – image Apr 7 '15 at 13:19
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Since torque=dL/dt and and there is no external torque about the sun as force and position vector are at an angle of 180 degree accordingly torque vector=position vector ×force vector which means dL/dt=0 implies L=constant I.e angular momentum of heavenly body remains constant over it's entire orbit

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  • $\begingroup$ Welcome on Physics SE and thank you for your contribution :) You might want to see here for help with typesetting formula :) $\endgroup$ – Sanya Oct 31 '16 at 9:22
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The Gravitational Force between the two masses mutually interacting depends on the separation distance ($r$) between them.

i.e, $\vec{F}=f(r)\hat{r}$

So the External Torque acting on the system is zero, $$\vec{\tau}=\vec{r}\times\vec{F}$$

$$\vec{\tau}=\vec{r}\times f(r)\hat{r} =0$$

So the Total angular momentum of the system is constant.

$\frac{d\vec{L}}{dt}=0$. So $\vec{L}=constant$

As the direction of the radial velocity of the planet changes, the linear momentum of the planet is not constant in the planetary system (refer figure below). Since there is no external force on the Sun+Planet system, so the total linear momentum of Sun+planet system is constant. enter image description here

The arrows points in the direction of linear velocity at that instant.

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protected by rob Oct 31 '16 at 11:57

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