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I don't understand why the answer choice is D, I thought for unbound orbits, none of these principles applied?

Could someone clear this up for me?

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    $\begingroup$ The conservation laws of the system are properties of the equations of motion, while the distinction between bound and unbound orbits is down to the initial conditions. Of course the conservation principles apply to all the orbits. Why do you think they don't? $\endgroup$ – Emilio Pisanty Dec 13 '19 at 8:54
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    $\begingroup$ so why is it that linear momentum is not conserved in this case? $\endgroup$ – student Dec 13 '19 at 8:59
  • $\begingroup$ @student Linear momentum is not conserved when a force is present (gravity is present in this case) angular momentum is not conserved when a torque is present (gravitational force is along line of action as mentioned by Farcher so no torque) and energy is conserved for conservative forces (gravity is conservative and friction is ignored) $\endgroup$ – aditya_stack Dec 13 '19 at 13:27
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    $\begingroup$ The total linear momentum of the asteroid and the star is conserved, but not the linear momentum of the asteroid alone. $\endgroup$ – G. Smith Dec 13 '19 at 17:01
  • $\begingroup$ The answer should be e). The question asks which conservation laws apply, and Energy, Linear Momentum, and Angular Momentum are always conserved. That doesn't mean all of them are constant for a defined set of objects. Conservation laws are continuity conditions which include the transfer of those quantities. $\endgroup$ – Bill N Dec 13 '19 at 18:32
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The question

Which conservation laws relate to the motion of the asteroid?

implies that it is the asteroid which is the system under consideration.

The gravitational force is an attractive central force which acts along the line joining the centre of masses of the star and the asteroid.

The asteroid has an external force acting on it, the gravitational attraction due to the star, so linear momentum of the asteroid is not conserved.

The torque, about the star, acting on the asteroid due to the star is zero because the direction of the gravitational attractive force due the star is along the line joining the star and the asteroid, so the angular momentum, about the star, of the asteroid is constant.

In classical mechanics energy is always conserved and in this example, with no frictional force etc acting, so is mechanical (gravitational potential and kinetic) energy of the asteroid and star system.

The type of orbit depends on the initial conditions and it could have been circular, elliptical, parabolic or hyperbolic.

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    $\begingroup$ "In classical mechanics energy is always conserved" it certainly is not, that claim only works for conservative forces/time independent lagrangian and not for any general classical mechanical system. $\endgroup$ – Umaxo Dec 13 '19 at 16:22
  • $\begingroup$ @ Umaxo Why do people routinely assume that conservation means constant? It doesn't. Conservation includes the transfer of quantities: $E_{new}=E_{old}+Work$ That's a conservation law: en.wikipedia.org/wiki/Conservation_law Also look at Noether's theorem which includes currents. $\endgroup$ – Bill N Dec 13 '19 at 18:38
  • $\begingroup$ @Farcher If you look at only the quantities belonging to the asteroid, then there is no potential energy. That's a system-defined quantity. Kinetic energy is not constant because the gravitational attractive force does work. But energy is still conserved because work is the transfer of energy between objects. Linear momentum is not constant, but it is conserved because momentum is neither created nor destroyed: it is transferred via impulse. $\endgroup$ – Bill N Dec 13 '19 at 18:45
  • $\begingroup$ @BillN The conservation means exactly that the current/quantity is constant over time - from your wiki "In physics, a conservation law states that a particular measurable property of an isolated physical system does not change as the system evolves over time". And Noether theorem says exactly that the energy is conserved for systems with langrangian that does not depend explicitly on time coodinate...What quantity is conserved in your equation $E_\text{new}=E_\text{old}+W$? $\endgroup$ – Umaxo Dec 13 '19 at 19:55
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The conservation laws have nothing to do with actual trajectory, but with the nature of forces acting on the object.

Take for example conservation of energy of some object in gravitational field. The only force acting on the object is (for simplicity, let us assume radial motion): $$F=-G\frac{mM}{r^2}$$ In very short time, this will produce the change in velocity: $$\frac{\Delta v}{\Delta t}=-G\frac{M}{r^2},$$ so the kinetic energy changed by: $$\Delta E_k=\frac{m(v+\Delta v)^2}{2}-\frac{mv^2}{2}\approx mv\Delta v,$$ where we neglected $\Delta v^2$, because we are assuming small time interval and thus this change is also small. Plugging in from the force law: $$\Delta E_k=-G\frac{mvM}{r^2}\Delta t,$$ Now, can we see the right hand side as negative change in potential energy? That is can we write the right hand side as simple a function of position (r)? In that case the gained/lost kinetic energy would be fully transformed into potential energy.

If you know calculus, you can calculate what kind of potential function you will get. If you don't just use the known result $E_p=-G\frac{mM}{r}$ and see wheter it works: $$\Delta E_p=-G\frac{mM}{r+\Delta r}+G\frac{mM}{r}= G\frac{mM\Delta r}{r(r+\Delta r)}\approx G\frac{mM\Delta r}{r^2},$$ But $\Delta r$ is just $v\Delta t$ and plugging it in, you will see it works $\Delta E_k=-\Delta E_p$!

Notice, the only thing we used is the form of the force acting on the object. We did not assume any initial velocity or position, so our formulas work for any scenario in which gravitational (and only gravitational) force acts on the object.

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every thing is already wrote by @Umaxo, but i prefer to write it like this:

according to NEWTON second law and your case the force is a function of r:

$$m\,\ddot{r}=F(r)\tag 1$$

multiply (1) with $\dot{r}$

$$m\,\dot{r}\,\ddot{r}=\dot{r}\,F(r)\tag 2$$

with

$$\dot{r}\,\ddot{r}=\frac{1}{2}\frac{d}{dt}\,\dot{r}^2$$

and

$$\dot{r}=\frac{d\,r}{dt}=v$$

you get:

$$\frac{m}{2}\frac{d}{dt}\,(v^2)=\frac{dr}{dt}\,F(r)$$

$\Rightarrow$

$$\int \frac{m}{2}\,d(v^2)=\int F(r)\,dr+E=-U(r)$$

or

$$\boxed{\frac{m}{2}\,v^2+U(r)=\text{E}}$$

this is the conservation of energy

the angular momentum is always conserved if the force $F$ is only function of r, for example gravitation force

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  • $\begingroup$ I think you meant if the potential U is only function of r? $\endgroup$ – Umaxo Dec 13 '19 at 16:37
  • $\begingroup$ @Umax0 the equation that I wrote are for scalar r. $F(r)=-\partial_r\,U(r)$ $\endgroup$ – Eli Dec 13 '19 at 17:42
  • $\begingroup$ I meant, you should say that "the angular momentum is always conserved if the potential energy $U$ is only function of r." Not the force. $\endgroup$ – Umaxo Dec 13 '19 at 19:45

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