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Isaac Physics Elliptical orbit

The question:

https://isaacphysics.org/questions/speed_in_elliptical_orbit?board=31035b97-ac55-4866-96a5-eb0457387a29

(So i have attempted to construct an equation relating the speed at any given point in the orbit using the facts that at the perihelion and aphelion, the angular momentum is conserved as well as the mechanical energy being equal at both points, equating them at both points using the accompanying distances to the star and the velocities. Then using the conservation of angular momentum got the velocity at the aphelion as well as the aphelion and perihelion distances to the star and plugged them into the KE GPE eqtn but what I end up with I can`t solve for V as its too complicated for me, however I looked up the Vis-Viva equation which is what I should end up with (so kinda halfway there). So i tried subbing in the relative values in the equation and ended up with a speed of 2.35x10^6kms^-1 which Isaac physics says is wrong. Am I missing something else? I can`t think what else to do.
Any help appreciated thanks
Yo buddy! You're overthinking this a little, well, I think, either ways. yeah. We can go about all these fancy equations/etc, but, in the end there is another way: just consider conservation of energy.

I've never done rotational dynamics for stars/etc. Well, I have, and I know all this other stuff you're talking about. But, the first method that hit my head was just using conservation of energy, it's a lot simpler.

It's in eccentric elliptic orbit: it is still in orbit. The circular centripetal force at this point must be enough such that it provides acceleration to continue the orbit. Hence, use centripetal force to create an energy equation for initial kinetic energy, i.e. the non relativistic version (v<<c)

So, with initial energy, our gravitational potential equation, and a change in radius, you should be able to take it from there :smile:

If that isn't a satisfying answer, I'll grudgingly have to go whip out my old physics book and figure this out the more complicated way :_;

Also, I'll just quickly give a hint: your order of magnitude is off, i.e. by a factor of three I think.
(edited 6 years ago)
Original post by Callicious
Yo buddy! You're overthinking this a little, well, I think, either ways. yeah. We can go about all these fancy equations/etc, but, in the end there is another way: just consider conservation of energy.

I've never done rotational dynamics for stars/etc. Well, I have, and I know all this other stuff you're talking about. But, the first method that hit my head was just using conservation of energy, it's a lot simpler.

It's in eccentric elliptic orbit: it is still in orbit. The circular centripetal force at this point must be enough such that it provides acceleration to continue the orbit. Hence, use centripetal force to create an energy equation for initial kinetic energy, i.e. the non relativistic version (v<<c)

So, with initial energy, our gravitational potential equation, and a change in radius, you should be able to take it from there :smile:

If that isn't a satisfying answer, I'll grudgingly have to go whip out my old physics book and figure this out the more complicated way :_;

Also, I'll just quickly give a hint: your order of magnitude is off, i.e. by a factor of three I think.

Thanks for the input, this is where I have got up to in general, I tried to eliminate all terms in past by using the semi major axis relation with the distances but I know this will only give velocity at aphelion and perihelion. Not sure how to get v at any point
Attachment not found
(edited 6 years ago)
Original post by skyyyatteinstein
The question:

https://isaacphysics.org/questions/speed_in_elliptical_orbit?board=31035b97-ac55-4866-96a5-eb0457387a29

(So i have attempted to construct an equation relating the speed at any given point in the orbit using the facts that at the perihelion and aphelion, the angular momentum is conserved as well as the mechanical energy being equal at both points, equating them at both points using the accompanying distances to the star and the velocities. Then using the conservation of angular momentum got the velocity at the aphelion as well as the aphelion and perihelion distances to the star and plugged them into the KE GPE eqtn but what I end up with I can`t solve for V as its too complicated for me, however I looked up the Vis-Viva equation which is what I should end up with (so kinda halfway there). So i tried subbing in the relative values in the equation and ended up with a speed of 2.35x10^6kms^-1 which Isaac physics says is wrong. Am I missing something else? I can`t think what else to do.
Any help appreciated thanks



The first step is that you need both the conservation of angular momentum and energy to derive total mechanical energy


E=GMm2a E = - \dfrac{GMm}{2a}


Make sure you can derive the above expression or else you are just doing what is called “plug and play” kind of physics.

From this expression, you should be able to find the speed based on conservation of energy.
Original post by Callicious
Yo buddy! You're overthinking this a little, well, I think, either ways. yeah. We can go about all these fancy equations/etc, but, in the end there is another way: just consider conservation of energy.

......


I disagree that you just need to consider conservation of energy.
Original post by Eimmanuel
The first step is that you need both the conservation of angular momentum and energy to derive total mechanical energy


E=GMm2a E = - \dfrac{GMm}{2a}


Make sure you can derive the above expression or else you are just doing what is called “plug and play” kind of physics.

From this expression, you should be able to find the speed based on conservation of energy.


My algebra is a bit rusty for the last line but solving for E which i assume will be Ra+Rp =a on the denominator, I then used that and equated to the Mech energy at the required point and got a value of 2.35x10^6 but it is still wrong? Am I missing something else DEE4C56A-82E5-4F1B-96DE-D7FD77645911.jpg.jpeg
Original post by skyyyatteinstein
Thanks for the input, this is where I have got up to in general, I tried to eliminate all terms in past by using the semi major axis relation with the distances but I know this will only give velocity at aphelion and perihelion. Not sure how to get v at any point
Attachment not found


Seeing as we're not going for any particular directionality here, i.e. not trying to find a vector for the velocity based on a specific coordinate system, just for the sake of which a polar one would be pretty useful for this, it's a lot simpler just to consider energy.

We know that at aphelion the centripetal force is the way to go: you can't use this at other points, but at aphelion it works out dandy. The proof is a bore and a pain and I really don't like it, but, yeah. Either ways, you can just figure this via common sense anyway. Calculate an expression for kinetic energy at aphelion. This can be done using circular motion, by approximating it to be circular at aphelion and not anywhere else. Using mv^2 /R = GMm/R^2 , solve for 0.5mv^2. That's our initial kinetic energy. Move forward to when we're at our new orbit. Calculate change in gravitational potential energy, i.e. GMm(1/R2 - 1/R1): normally it is the other way around, but we want a positive magnitude, since we are going for the change in KE, which will be an increase. So. The final KE is GMm(1/R2 - 1/R1) + 0.5GMm/R1. Eliminate m and other constants. solve for v.

The answer is correct, and provided there aren't any other forces/etc at play and you have the aphelion distance and the mass parameter of the sun, you will be able to solve this for any distance, inclusive of perihelion.

Original post by Eimmanuel
I disagree that you just need to consider conservation of energy.


The answer using just conservation of energy satisfies the question, which is why I stated that it's the only necessity. If you were to go about considering the need for a vector to describe the motion, you'd need to consider angular momentum to form another set of equations that could be used in forming a vector for the velocity. It'd be a heck of a lot harder though, as you'd need to consider functions of the energies involved and what have you, at least that's how I'd do it. There's probably an easier way to parametrise the question if you had to do it in vector form, but I'd have no clue.
Original post by skyyyatteinstein
My algebra is a bit rusty for the last line but solving for E which i assume will be Ra+Rp =a on the denominator, I then used that and equated to the Mech energy at the required point and got a value of 2.35x10^6 but it is still wrong? Am I missing something else DEE4C56A-82E5-4F1B-96DE-D7FD77645911.jpg.jpeg



I am not sure what you are trying to achieve.

If you are trying to derive the expression that I mention, then you need to subtract the total mechanical energy at C and D to find an expression for L2/2m.

If you are trying to solve the problem, you just need conservation of energy with the given expression.


GMm2a=GMmr+12mv2 - \dfrac{GMm}{2a} = - \dfrac{GMm}{r} + \dfrac{1}{2}mv^2


And solve for v.
Original post by Callicious
The answer using just conservation of energy satisfies the question, which is why I stated that it's the only necessity. If you were to go about considering the need for a vector to describe the motion, you'd need to consider angular momentum to form another set of equations that could be used in forming a vector for the velocity. It'd be a heck of a lot harder though, as you'd need to consider functions of the energies involved and what have you, at least that's how I'd do it. There's probably an easier way to parametrise the question if you had to do it in vector form, but I'd have no clue.


Thanks will try now
Original post by Callicious
Seeing as we're not going for any particular directionality here, i.e. not trying to find a vector for the velocity based on a specific coordinate system, just for the sake of which a polar one would be pretty useful for this, it's a lot simpler just to consider energy.

We know that at aphelion the centripetal force is the way to go: you can't use this at other points, but at aphelion it works out dandy. The proof is a bore and a pain and I really don't like it, but, yeah. Either ways, you can just figure this via common sense anyway. Calculate an expression for kinetic energy at aphelion. This can be done using circular motion, by approximating it to be circular at aphelion and not anywhere else. Using mv^2 /R = GMm/R^2 , solve for 0.5mv^2. That's our initial kinetic energy. Move forward to when we're at our new orbit. Calculate change in gravitational potential energy, i.e. GMm(1/R2 - 1/R1): normally it is the other way around, but we want a positive magnitude, since we are going for the change in KE, which will be an increase. So. The final KE is GMm(1/R2 - 1/R1) + 0.5GMm/R1. Eliminate m and other constants. solve for v.

The answer is correct, and provided there aren't any other forces/etc at play and you have the aphelion distance and the mass parameter of the sun, you will be able to solve this for any distance, inclusive of perihelion.



The answer using just conservation of energy satisfies the question, which is why I stated that it's the only necessity. If you were to go about considering the need for a vector to describe the motion, you'd need to consider angular momentum to form another set of equations that could be used in forming a vector for the velocity. It'd be a heck of a lot harder though, as you'd need to consider functions of the energies involved and what have you, at least that's how I'd do it. There's probably an easier way to parametrise the question if you had to do it in vector form, but I'd have no clue.


It seems that you have not even look at the question to know what is/are given.
Original post by Eimmanuel
I am not sure what you are trying to achieve.

If you are trying to derive the expression that I mention, then you need to subtract the total mechanical energy at C and D to find an expression for L2/2m.

If you are trying to solve the problem, you just need conservation of energy with the given expression.


GMm2a=GMmr+12mv2 - \dfrac{GMm}{2a} = - \dfrac{GMm}{r} + \dfrac{1}{2}mv^2


And solve for v.


Yeah I was trying to get The E expression from it but I couldn't do the algebra to tidy it up, I tried using the equation above to solve for v at the point assuming r is the distance to the star however I still again got 2.35x10^6kms-1 which Isaac says is wrong
Original post by skyyyatteinstein
Yeah I was trying to get The E expression from it but I couldn't do the algebra to tidy it up, I tried using the equation above to solve for v at the point assuming r is the distance to the star however I still again got 2.35x10^6kms-1 which Isaac says is wrong


The Gravitational constant is usually given in SI, so you need to convert the distance to m instead of just using km.
Original post by skyyyatteinstein
Yeah I was trying to get The E expression from it but I couldn't do the algebra to tidy it up, I tried using the equation above to solve for v at the point assuming r is the distance to the star however I still again got 2.35x10^6kms-1 which Isaac says is wrong


Your answer is out by a factor of root(1000) or somewhere in that region. Check the numbers they gave, other than that, good work :biggrin:

Just a note, I don't really know if it is a prerequisite of Isaac Physics to use the notes they give you, I mean, it doesn't seem like it's necessary, but just use a method that works for you. If you're given a way to do it, try and think up another way to do it that applies your own knowledge, that'll help you understand the situation better.

Original post by Eimmanuel
It seems that you have not even look at the question to know what is/are given.


I'm just here to help the OP get an answer to the question. The website doesn't ask that a specific method be used, and it just provides an example way to go about the question. The method I've used doesn't presuppose any knowledge, and you can figure it out based on thinking about the forces at play and the critical points of the system, perihelion and aphelion, and then go about deriving equations for any point in-between.

In the end he ended up using the method they gave anyway, and the OP was fine with that, so I don't see a problem. We're both here to provide some help to the OP on how to answer the question, and in the end we provided two unique methods that could be used. That's what counts. Both methods work physically and each provide a good avenue to the answer. I'd like to leave it at that.
Original post by Eimmanuel
The Gravitational constant is usually given in SI, so you need to convert the distance to m instead of just using km.


Wow how did I miss that, changed my distances into metres and got the correct answer, thanks for the help
Original post by Callicious
I'm just here to help the OP get an answer to the question. The website doesn't ask that a specific method be used, and it just provides an example way to go about the question. The method I've used doesn't presuppose any knowledge, and you can figure it out based on thinking about the forces at play and the critical points of the system, perihelion and aphelion, and then go about deriving equations for any point in-between.

In the end he ended up using the method they gave anyway, and the OP was fine with that, so I don't see a problem. We're both here to provide some help to the OP on how to answer the question, and in the end we provided two unique methods that could be used. That's what counts. Both methods work physically and each provide a good avenue to the answer. I'd like to leave it at that.


---
(edited 6 years ago)
Original post by Callicious
.... I'm just here to help the OP get an answer to the question. The website doesn't ask that a specific method be used, and it just provides an example way to go about the question. The method I've used doesn't presuppose any knowledge, and you can figure it out based on thinking about the forces at play and the critical points of the system, perihelion and aphelion, and then go about deriving equations for any point in-between. ....


I understand what you said. What I have replied above is to mean that the method that you propose will not work for this problem. This is because we are not given either R𝗉 is the distance of closest approach ("perihelion") or R𝖺 the distance of farthest approach ("aphelion") but only the semi-major axis of elliptical orbit.

If we are given the distance of closest approach ("perihelion"), I would agree that we can solve the problem via conservation of energy only without the need of conservation of angular momentum.

Original post by Callicious
.... In the end he ended up using the method they gave anyway, and the OP was fine with that, so I don't see a problem. We're both here to provide some help to the OP on how to answer the question, and in the end we provided two unique methods that could be used. That's what counts. Both methods work physically and each provide a good avenue to the answer. I'd like to leave it at that.


I remembered working on something similar decade ago and our lecturer mentioned that there is not enough information to solve the problem via only conservation of energy. I am not sure if this is the one.

I would be very happy that if you can work out the answer via your method without conservation of angular momentum and post it online.

In Isaac Physics, if I am not wrong, when there are more than one method of solving the problem, they would usually state them in the hints.
Original post by Eimmanuel
I understand what you said. What I have replied above is to mean that the method that you propose will not work for this problem. This is because we are not given either R𝗉 is the distance of closest approach ("perihelion":wink: or R𝖺 the distance of farthest approach ("aphelion":wink: but only the semi-major axis of elliptical orbit.

If we are given the distance of closest approach ("perihelion":wink:, I would agree that we can solve the problem via conservation of energy only without the need of conservation of angular momentum.



I remembered working on something similar decade ago and our lecturer mentioned that there is not enough information to solve the problem via only conservation of energy. I am not sure if this is the one.

I would be very happy that if you can work out the answer via your method without conservation of angular momentum and post it online.

In Isaac Physics, if I am not wrong, when there are more than one method of solving the problem, they would usually state them in the hints.


Using my method does give the correct answer according to Isaac Phys (numerically), though it is a slight approximation based on the mass of the star exceeding the mass of the planet by a large margin. I suppose I did actually misread the question though, I took the semi-major as being the aphelion distance, which would only work to a certain approximation and would depend highly on the issue at hand.

I'm actually in awe right now, I didn't realise that I've been misreading this whole time as having been given the semi-major and not the aphelion distance. How the heck did my conservation of energy calculation give the correct answer o,o
Original post by Callicious
Using my method does give the correct answer according to Isaac Phys (numerically), though it is a slight approximation based on the mass of the star exceeding the mass of the planet by a large margin. I suppose I did actually misread the question though, I took the semi-major as being the aphelion distance, which would only work to a certain approximation and would depend highly on the issue at hand.

I'm actually in awe right now, I didn't realise that I've been misreading this whole time as having been given the semi-major and not the aphelion distance. How the heck did my conservation of energy calculation give the correct answer o,o



I am actually not surprised that a wrong method can give the correct answer. It is NOTHING due to the approximation or whatever so given by the question.

When an object of mass m orbit around a massive object M in a circular motion of radius r, the total energy of the system, is

Ecircular=GMm2r E_{circular} = - \dfrac{GMm}{2r} ------ (1)


As for an elliptical orbit, the total energy of the system becomes

Eelliptical=GMm2a E_{elliptical} = - \dfrac{GMm}{2a} ------ (2)


where a is the semi-major axis.

The two equations look very alike but to derive the equation (2), more work is needed. We need the both the conservation of angular momentum and energy to derive the equation (2).


If you look at the expression that you had written in post 7,

Final KE = GMm(1/R2 - 1/R1) + 0.5GMm/R1 ------ (3)


and compare the expression in posting 8:

GMm2a=GMmr+12mv2 - \dfrac{GMm}{2a} = - \dfrac{GMm}{r} + \dfrac{1}{2}mv^2 ------ (4)


You should not be surprised that your method would give the “correct numerical answer” as equation (3) and (4) are deceiving alike and equation (3) would become (4) when R1 is mistaken to take as semi-major axis.
Original post by Eimmanuel
I am actually not surprised that a wrong method can give the correct answer. It is NOTHING due to the approximation or whatever so given by the question.

When an object of mass m orbit around a massive object M in a circular motion of radius r, the total energy of the system, is


Ecircular=GMm2r E_{circular} = - \dfrac{GMm}{2r} ------ (1)



As for an elliptical orbit, the total energy of the system becomes


Eelliptical=GMm2a E_{elliptical} = - \dfrac{GMm}{2a} ------ (2)



where a is the semi-major axis.

The two equations look very alike but to derive the equation (2), more work is needed. We need the both the conservation of angular momentum and energy to derive the equation (2).


If you look at the expression that you had written in post 7,


Final KE = GMm(1/R2 - 1/R1) + 0.5GMm/R1 ------ (3)



and compare the expression in posting 8:


GMm2a=GMmr+12mv2 - \dfrac{GMm}{2a} = - \dfrac{GMm}{r} + \dfrac{1}{2}mv^2 ------ (4)



You should not be surprised that your method would give the “correct numerical answer” as equation (3) and (4) are deceiving alike and equation (3) would become (4) when R1 is mistaken to take as semi-major axis.


I suppose you'd have to go about using the angular momentum in the end if you'd want the correct way to get the answer :-:

This hurts my soul, I just misguided some poor fellow on the forum with a false method :_:
Original post by Callicious
I suppose you'd have to go about using the angular momentum in the end if you'd want the correct way to get the answer :-:

This hurts my soul, I just misguided some poor fellow on the forum with a false method :_:


Nah it's good I stuck to using cons. of angular momentum anyway and can see reading eimmanuels post how they differ

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