Circular orbit in a central force field












3












$begingroup$


Considering a motion of a body under a attractive central force,



$textbf{F}(textbf{r}) = -frac{k}{r^3} hspace{1mm}hat{textbf{r}}$ with $k>0$.



Is it possible for a body to move in an at least stationary circular orbit? Since the derivation of the effective potential



$U_{eff}(r) = frac{l^2}{2mr^2}+U(r)$



(where $l$ is the angular momentum)



has to be $0$ for a circular orbit, the only solution would be that $k = frac{l^2}{m}$. But that would lead to an effective potential $U_{eff}(r) = 0$ for any $r$ (except $r = 0$). Is this a valid solution?






forces orbital-motion







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Peter Hofer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$












  • $begingroup$
    Are you asking about the stability of the circular orbit solution?
    $endgroup$
    – Qmechanic♦
    Jan 19 at 13:19












  • $begingroup$
    I'm asking if there is any solution for a circular orbit (stable or unstable) since I'm not to sure about the validity of my solution.
    $endgroup$
    – Peter Hofer
    Jan 19 at 13:38






  • 1




    $begingroup$
    The derivative of the potential has to be zero I think. This way a test particle will stay inside the 'potential well' and so it can be a stable orbit.
    $endgroup$
    – WarreG
    Jan 19 at 13:56










  • $begingroup$
    After a small calculation the derivative seems to be zero at $k=l^2/m$, so I guess it is a valid solution.
    $endgroup$
    – WarreG
    Jan 19 at 14:01










  • $begingroup$
    You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
    $endgroup$
    – FGSUZ
    Jan 19 at 14:03
















3












$begingroup$


Considering a motion of a body under a attractive central force,



$textbf{F}(textbf{r}) = -frac{k}{r^3} hspace{1mm}hat{textbf{r}}$ with $k>0$.



Is it possible for a body to move in an at least stationary circular orbit? Since the derivation of the effective potential



$U_{eff}(r) = frac{l^2}{2mr^2}+U(r)$



(where $l$ is the angular momentum)



has to be $0$ for a circular orbit, the only solution would be that $k = frac{l^2}{m}$. But that would lead to an effective potential $U_{eff}(r) = 0$ for any $r$ (except $r = 0$). Is this a valid solution?






forces orbital-motion







share|cite|improve this question









New contributor




Peter Hofer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Are you asking about the stability of the circular orbit solution?
    $endgroup$
    – Qmechanic♦
    Jan 19 at 13:19












  • $begingroup$
    I'm asking if there is any solution for a circular orbit (stable or unstable) since I'm not to sure about the validity of my solution.
    $endgroup$
    – Peter Hofer
    Jan 19 at 13:38






  • 1




    $begingroup$
    The derivative of the potential has to be zero I think. This way a test particle will stay inside the 'potential well' and so it can be a stable orbit.
    $endgroup$
    – WarreG
    Jan 19 at 13:56










  • $begingroup$
    After a small calculation the derivative seems to be zero at $k=l^2/m$, so I guess it is a valid solution.
    $endgroup$
    – WarreG
    Jan 19 at 14:01










  • $begingroup$
    You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
    $endgroup$
    – FGSUZ
    Jan 19 at 14:03














3












3








3


1



$begingroup$


Considering a motion of a body under a attractive central force,



$textbf{F}(textbf{r}) = -frac{k}{r^3} hspace{1mm}hat{textbf{r}}$ with $k>0$.



Is it possible for a body to move in an at least stationary circular orbit? Since the derivation of the effective potential



$U_{eff}(r) = frac{l^2}{2mr^2}+U(r)$



(where $l$ is the angular momentum)



has to be $0$ for a circular orbit, the only solution would be that $k = frac{l^2}{m}$. But that would lead to an effective potential $U_{eff}(r) = 0$ for any $r$ (except $r = 0$). Is this a valid solution?






forces orbital-motion







share|cite|improve this question









New contributor




Peter Hofer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Considering a motion of a body under a attractive central force,



$textbf{F}(textbf{r}) = -frac{k}{r^3} hspace{1mm}hat{textbf{r}}$ with $k>0$.



Is it possible for a body to move in an at least stationary circular orbit? Since the derivation of the effective potential



$U_{eff}(r) = frac{l^2}{2mr^2}+U(r)$



(where $l$ is the angular momentum)



has to be $0$ for a circular orbit, the only solution would be that $k = frac{l^2}{m}$. But that would lead to an effective potential $U_{eff}(r) = 0$ for any $r$ (except $r = 0$). Is this a valid solution?






forces orbital-motion




forces orbital-motion






share|cite|improve this question









New contributor




Peter Hofer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Peter Hofer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 15:02







Peter Hofer













New contributor




Peter Hofer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Jan 19 at 13:04









Peter HoferPeter Hofer

184




184




New contributor




Peter Hofer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Peter Hofer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Peter Hofer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Are you asking about the stability of the circular orbit solution?
    $endgroup$
    – Qmechanic♦
    Jan 19 at 13:19












  • $begingroup$
    I'm asking if there is any solution for a circular orbit (stable or unstable) since I'm not to sure about the validity of my solution.
    $endgroup$
    – Peter Hofer
    Jan 19 at 13:38






  • 1




    $begingroup$
    The derivative of the potential has to be zero I think. This way a test particle will stay inside the 'potential well' and so it can be a stable orbit.
    $endgroup$
    – WarreG
    Jan 19 at 13:56










  • $begingroup$
    After a small calculation the derivative seems to be zero at $k=l^2/m$, so I guess it is a valid solution.
    $endgroup$
    – WarreG
    Jan 19 at 14:01










  • $begingroup$
    You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
    $endgroup$
    – FGSUZ
    Jan 19 at 14:03


















  • $begingroup$
    Are you asking about the stability of the circular orbit solution?
    $endgroup$
    – Qmechanic♦
    Jan 19 at 13:19












  • $begingroup$
    I'm asking if there is any solution for a circular orbit (stable or unstable) since I'm not to sure about the validity of my solution.
    $endgroup$
    – Peter Hofer
    Jan 19 at 13:38






  • 1




    $begingroup$
    The derivative of the potential has to be zero I think. This way a test particle will stay inside the 'potential well' and so it can be a stable orbit.
    $endgroup$
    – WarreG
    Jan 19 at 13:56










  • $begingroup$
    After a small calculation the derivative seems to be zero at $k=l^2/m$, so I guess it is a valid solution.
    $endgroup$
    – WarreG
    Jan 19 at 14:01










  • $begingroup$
    You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
    $endgroup$
    – FGSUZ
    Jan 19 at 14:03
















$begingroup$
Are you asking about the stability of the circular orbit solution?
$endgroup$
– Qmechanic♦
Jan 19 at 13:19






$begingroup$
Are you asking about the stability of the circular orbit solution?
$endgroup$
– Qmechanic♦
Jan 19 at 13:19














$begingroup$
I'm asking if there is any solution for a circular orbit (stable or unstable) since I'm not to sure about the validity of my solution.
$endgroup$
– Peter Hofer
Jan 19 at 13:38




$begingroup$
I'm asking if there is any solution for a circular orbit (stable or unstable) since I'm not to sure about the validity of my solution.
$endgroup$
– Peter Hofer
Jan 19 at 13:38




1




1




$begingroup$
The derivative of the potential has to be zero I think. This way a test particle will stay inside the 'potential well' and so it can be a stable orbit.
$endgroup$
– WarreG
Jan 19 at 13:56




$begingroup$
The derivative of the potential has to be zero I think. This way a test particle will stay inside the 'potential well' and so it can be a stable orbit.
$endgroup$
– WarreG
Jan 19 at 13:56












$begingroup$
After a small calculation the derivative seems to be zero at $k=l^2/m$, so I guess it is a valid solution.
$endgroup$
– WarreG
Jan 19 at 14:01




$begingroup$
After a small calculation the derivative seems to be zero at $k=l^2/m$, so I guess it is a valid solution.
$endgroup$
– WarreG
Jan 19 at 14:01












$begingroup$
You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
$endgroup$
– FGSUZ
Jan 19 at 14:03




$begingroup$
You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
$endgroup$
– FGSUZ
Jan 19 at 14:03










2 Answers
2






active

oldest

votes


















2












$begingroup$

For stable orbit, we need to have $d^2V_{eff}/dr^2>0$ at $r=r_0$ and we need to find $r_0$ from the solution of $dV_{eff}/dr=0$



We can find the potential by $$V=-int Fdr$$



Hence $$V=-frac {k} {2r^2}$$ so $$V_{eff}=frac {l^2} {2mr^2}-frac {k} {2r^2}$$



and at $r=r_0$, $dV_{eff}/dr=0$ hence



$$dV_{eff}/dr=frac {-l^2} {mr_0^{3}}+frac {k} {r_0^3}=0$$



so we have,



$$l^2=mk$$



Now we need to find $d^2V_{eff}/dr^2$ at $r=r_0$



$$d^2V_{eff}/dr^2=frac {3l^2} {mr_0^{4}}-frac {3k} {r_0^4}$$
Using the above relationship we find that,



$$d^2V_{eff}/dr^2=frac {3k} {r_0^{4}}-frac {3k} {r_0^4}=0$$ which is exactly zero. So there cannot be any stable circular orbit.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what descriptively happens if there are small deviations from the circular orbit with radius $r_0$? Would the body just describe other unstable circular orbits since $V_{eff} = 0$ (due to our chosen value for $k$)?
    $endgroup$
    – Peter Hofer
    Jan 19 at 16:34










  • $begingroup$
    There cannot be any stable circular motion as I shown. The criteria for circular orbit is not satisfied. To have a stable orbit for perturbation we definitly need $V^2_{eff}/dr^2>0$ but since we have $V^2_{eff}/dr^2=0$ any perturbation from the orbit cannot lead any stable circular motion.
    $endgroup$
    – Reign
    Jan 19 at 17:43












  • $begingroup$
    Well yes as you said the body would do unstable circular motion according. If you draw the $V-r$ graph you ll also see that there cannot be any stable circular motion.
    $endgroup$
    – Reign
    Jan 19 at 17:50










  • $begingroup$
    I find the same question that you asked in my textbook, Marion-Thornton Classical Dynamic Question 8.22. If you can find the solution manuel you ll see a more comprehensive answer, but the basic idea is the same as I explained here.
    $endgroup$
    – Reign
    Jan 19 at 17:53





















0












$begingroup$

for a circle motion die circle radius $r(t)$ must be constant.



we can calculate the EOM's with Euler-Lagrange method.



kinetic energy:



$$ T=frac{1}{2},mleft(dot{r}^2+r^2,dot{phi}^2right)$$



potential energy



$$U=frac{L^2}{2,m,r^2}+ U(r)$$



where $L$ is the angular momentum and
$U(r)$ the unknown potential for a circle motion



The equation of motions are:



$$ddot{r}=r,dot{phi}^2+frac{L^2}{m^2,r^3}-frac{1}{m}frac{d,U}{d,r}tag 1$$



$$m,rleft(ddot{phi},r+2,dot{r},dot{phi}right)=0quad Rightarrow$$



$$m,frac{d}{dt}left(r^2,dot{phi}right)=0tag 2$$



we obtain from equation (2) that $dot{phi}=h/r^2quad$ with $h=L/m$
we put this result in equation (1) and get:



$$ddot{r}=frac{2,L^2}{m^2,r^3},-frac{1}{m},frac{d,U}{d,r}tag 3$$



for a circle motion $r(t)=constquad Rightarrowquad ddot{r}=0$ ;
we solve equation (3) for $frac{d,U}{d,r}$



$$boxed{F(r)=frac{d,U}{d,r}=frac{2,L^2}{m}frac{1}{r^3}}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Unless m is dimensionless and equal to 1, something is very wrong here!
    $endgroup$
    – Bill N
    Jan 19 at 19:02










  • $begingroup$
    Yes I also think so, I Have to check
    $endgroup$
    – Eli
    Jan 19 at 19:33










  • $begingroup$
    @Bill N I found my error i forgot the mass in the kinetic energy, I think this solution is o.k ? but if i compare force sign in the quotation i got deference sing, and anther value for $k$.
    $endgroup$
    – Eli
    Jan 19 at 20:45













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

For stable orbit, we need to have $d^2V_{eff}/dr^2>0$ at $r=r_0$ and we need to find $r_0$ from the solution of $dV_{eff}/dr=0$



We can find the potential by $$V=-int Fdr$$



Hence $$V=-frac {k} {2r^2}$$ so $$V_{eff}=frac {l^2} {2mr^2}-frac {k} {2r^2}$$



and at $r=r_0$, $dV_{eff}/dr=0$ hence



$$dV_{eff}/dr=frac {-l^2} {mr_0^{3}}+frac {k} {r_0^3}=0$$



so we have,



$$l^2=mk$$



Now we need to find $d^2V_{eff}/dr^2$ at $r=r_0$



$$d^2V_{eff}/dr^2=frac {3l^2} {mr_0^{4}}-frac {3k} {r_0^4}$$
Using the above relationship we find that,



$$d^2V_{eff}/dr^2=frac {3k} {r_0^{4}}-frac {3k} {r_0^4}=0$$ which is exactly zero. So there cannot be any stable circular orbit.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what descriptively happens if there are small deviations from the circular orbit with radius $r_0$? Would the body just describe other unstable circular orbits since $V_{eff} = 0$ (due to our chosen value for $k$)?
    $endgroup$
    – Peter Hofer
    Jan 19 at 16:34










  • $begingroup$
    There cannot be any stable circular motion as I shown. The criteria for circular orbit is not satisfied. To have a stable orbit for perturbation we definitly need $V^2_{eff}/dr^2>0$ but since we have $V^2_{eff}/dr^2=0$ any perturbation from the orbit cannot lead any stable circular motion.
    $endgroup$
    – Reign
    Jan 19 at 17:43












  • $begingroup$
    Well yes as you said the body would do unstable circular motion according. If you draw the $V-r$ graph you ll also see that there cannot be any stable circular motion.
    $endgroup$
    – Reign
    Jan 19 at 17:50










  • $begingroup$
    I find the same question that you asked in my textbook, Marion-Thornton Classical Dynamic Question 8.22. If you can find the solution manuel you ll see a more comprehensive answer, but the basic idea is the same as I explained here.
    $endgroup$
    – Reign
    Jan 19 at 17:53


















2












$begingroup$

For stable orbit, we need to have $d^2V_{eff}/dr^2>0$ at $r=r_0$ and we need to find $r_0$ from the solution of $dV_{eff}/dr=0$



We can find the potential by $$V=-int Fdr$$



Hence $$V=-frac {k} {2r^2}$$ so $$V_{eff}=frac {l^2} {2mr^2}-frac {k} {2r^2}$$



and at $r=r_0$, $dV_{eff}/dr=0$ hence



$$dV_{eff}/dr=frac {-l^2} {mr_0^{3}}+frac {k} {r_0^3}=0$$



so we have,



$$l^2=mk$$



Now we need to find $d^2V_{eff}/dr^2$ at $r=r_0$



$$d^2V_{eff}/dr^2=frac {3l^2} {mr_0^{4}}-frac {3k} {r_0^4}$$
Using the above relationship we find that,



$$d^2V_{eff}/dr^2=frac {3k} {r_0^{4}}-frac {3k} {r_0^4}=0$$ which is exactly zero. So there cannot be any stable circular orbit.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what descriptively happens if there are small deviations from the circular orbit with radius $r_0$? Would the body just describe other unstable circular orbits since $V_{eff} = 0$ (due to our chosen value for $k$)?
    $endgroup$
    – Peter Hofer
    Jan 19 at 16:34










  • $begingroup$
    There cannot be any stable circular motion as I shown. The criteria for circular orbit is not satisfied. To have a stable orbit for perturbation we definitly need $V^2_{eff}/dr^2>0$ but since we have $V^2_{eff}/dr^2=0$ any perturbation from the orbit cannot lead any stable circular motion.
    $endgroup$
    – Reign
    Jan 19 at 17:43












  • $begingroup$
    Well yes as you said the body would do unstable circular motion according. If you draw the $V-r$ graph you ll also see that there cannot be any stable circular motion.
    $endgroup$
    – Reign
    Jan 19 at 17:50










  • $begingroup$
    I find the same question that you asked in my textbook, Marion-Thornton Classical Dynamic Question 8.22. If you can find the solution manuel you ll see a more comprehensive answer, but the basic idea is the same as I explained here.
    $endgroup$
    – Reign
    Jan 19 at 17:53
















2












2








2





$begingroup$

For stable orbit, we need to have $d^2V_{eff}/dr^2>0$ at $r=r_0$ and we need to find $r_0$ from the solution of $dV_{eff}/dr=0$



We can find the potential by $$V=-int Fdr$$



Hence $$V=-frac {k} {2r^2}$$ so $$V_{eff}=frac {l^2} {2mr^2}-frac {k} {2r^2}$$



and at $r=r_0$, $dV_{eff}/dr=0$ hence



$$dV_{eff}/dr=frac {-l^2} {mr_0^{3}}+frac {k} {r_0^3}=0$$



so we have,



$$l^2=mk$$



Now we need to find $d^2V_{eff}/dr^2$ at $r=r_0$



$$d^2V_{eff}/dr^2=frac {3l^2} {mr_0^{4}}-frac {3k} {r_0^4}$$
Using the above relationship we find that,



$$d^2V_{eff}/dr^2=frac {3k} {r_0^{4}}-frac {3k} {r_0^4}=0$$ which is exactly zero. So there cannot be any stable circular orbit.






share|cite|improve this answer











$endgroup$



For stable orbit, we need to have $d^2V_{eff}/dr^2>0$ at $r=r_0$ and we need to find $r_0$ from the solution of $dV_{eff}/dr=0$



We can find the potential by $$V=-int Fdr$$



Hence $$V=-frac {k} {2r^2}$$ so $$V_{eff}=frac {l^2} {2mr^2}-frac {k} {2r^2}$$



and at $r=r_0$, $dV_{eff}/dr=0$ hence



$$dV_{eff}/dr=frac {-l^2} {mr_0^{3}}+frac {k} {r_0^3}=0$$



so we have,



$$l^2=mk$$



Now we need to find $d^2V_{eff}/dr^2$ at $r=r_0$



$$d^2V_{eff}/dr^2=frac {3l^2} {mr_0^{4}}-frac {3k} {r_0^4}$$
Using the above relationship we find that,



$$d^2V_{eff}/dr^2=frac {3k} {r_0^{4}}-frac {3k} {r_0^4}=0$$ which is exactly zero. So there cannot be any stable circular orbit.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 19 at 17:55

























answered Jan 19 at 16:14









ReignReign

530210




530210












  • $begingroup$
    So what descriptively happens if there are small deviations from the circular orbit with radius $r_0$? Would the body just describe other unstable circular orbits since $V_{eff} = 0$ (due to our chosen value for $k$)?
    $endgroup$
    – Peter Hofer
    Jan 19 at 16:34










  • $begingroup$
    There cannot be any stable circular motion as I shown. The criteria for circular orbit is not satisfied. To have a stable orbit for perturbation we definitly need $V^2_{eff}/dr^2>0$ but since we have $V^2_{eff}/dr^2=0$ any perturbation from the orbit cannot lead any stable circular motion.
    $endgroup$
    – Reign
    Jan 19 at 17:43












  • $begingroup$
    Well yes as you said the body would do unstable circular motion according. If you draw the $V-r$ graph you ll also see that there cannot be any stable circular motion.
    $endgroup$
    – Reign
    Jan 19 at 17:50










  • $begingroup$
    I find the same question that you asked in my textbook, Marion-Thornton Classical Dynamic Question 8.22. If you can find the solution manuel you ll see a more comprehensive answer, but the basic idea is the same as I explained here.
    $endgroup$
    – Reign
    Jan 19 at 17:53




















  • $begingroup$
    So what descriptively happens if there are small deviations from the circular orbit with radius $r_0$? Would the body just describe other unstable circular orbits since $V_{eff} = 0$ (due to our chosen value for $k$)?
    $endgroup$
    – Peter Hofer
    Jan 19 at 16:34










  • $begingroup$
    There cannot be any stable circular motion as I shown. The criteria for circular orbit is not satisfied. To have a stable orbit for perturbation we definitly need $V^2_{eff}/dr^2>0$ but since we have $V^2_{eff}/dr^2=0$ any perturbation from the orbit cannot lead any stable circular motion.
    $endgroup$
    – Reign
    Jan 19 at 17:43












  • $begingroup$
    Well yes as you said the body would do unstable circular motion according. If you draw the $V-r$ graph you ll also see that there cannot be any stable circular motion.
    $endgroup$
    – Reign
    Jan 19 at 17:50










  • $begingroup$
    I find the same question that you asked in my textbook, Marion-Thornton Classical Dynamic Question 8.22. If you can find the solution manuel you ll see a more comprehensive answer, but the basic idea is the same as I explained here.
    $endgroup$
    – Reign
    Jan 19 at 17:53


















$begingroup$
So what descriptively happens if there are small deviations from the circular orbit with radius $r_0$? Would the body just describe other unstable circular orbits since $V_{eff} = 0$ (due to our chosen value for $k$)?
$endgroup$
– Peter Hofer
Jan 19 at 16:34




$begingroup$
So what descriptively happens if there are small deviations from the circular orbit with radius $r_0$? Would the body just describe other unstable circular orbits since $V_{eff} = 0$ (due to our chosen value for $k$)?
$endgroup$
– Peter Hofer
Jan 19 at 16:34












$begingroup$
There cannot be any stable circular motion as I shown. The criteria for circular orbit is not satisfied. To have a stable orbit for perturbation we definitly need $V^2_{eff}/dr^2>0$ but since we have $V^2_{eff}/dr^2=0$ any perturbation from the orbit cannot lead any stable circular motion.
$endgroup$
– Reign
Jan 19 at 17:43






$begingroup$
There cannot be any stable circular motion as I shown. The criteria for circular orbit is not satisfied. To have a stable orbit for perturbation we definitly need $V^2_{eff}/dr^2>0$ but since we have $V^2_{eff}/dr^2=0$ any perturbation from the orbit cannot lead any stable circular motion.
$endgroup$
– Reign
Jan 19 at 17:43














$begingroup$
Well yes as you said the body would do unstable circular motion according. If you draw the $V-r$ graph you ll also see that there cannot be any stable circular motion.
$endgroup$
– Reign
Jan 19 at 17:50




$begingroup$
Well yes as you said the body would do unstable circular motion according. If you draw the $V-r$ graph you ll also see that there cannot be any stable circular motion.
$endgroup$
– Reign
Jan 19 at 17:50












$begingroup$
I find the same question that you asked in my textbook, Marion-Thornton Classical Dynamic Question 8.22. If you can find the solution manuel you ll see a more comprehensive answer, but the basic idea is the same as I explained here.
$endgroup$
– Reign
Jan 19 at 17:53






$begingroup$
I find the same question that you asked in my textbook, Marion-Thornton Classical Dynamic Question 8.22. If you can find the solution manuel you ll see a more comprehensive answer, but the basic idea is the same as I explained here.
$endgroup$
– Reign
Jan 19 at 17:53













0












$begingroup$

for a circle motion die circle radius $r(t)$ must be constant.



we can calculate the EOM's with Euler-Lagrange method.



kinetic energy:



$$ T=frac{1}{2},mleft(dot{r}^2+r^2,dot{phi}^2right)$$



potential energy



$$U=frac{L^2}{2,m,r^2}+ U(r)$$



where $L$ is the angular momentum and
$U(r)$ the unknown potential for a circle motion



The equation of motions are:



$$ddot{r}=r,dot{phi}^2+frac{L^2}{m^2,r^3}-frac{1}{m}frac{d,U}{d,r}tag 1$$



$$m,rleft(ddot{phi},r+2,dot{r},dot{phi}right)=0quad Rightarrow$$



$$m,frac{d}{dt}left(r^2,dot{phi}right)=0tag 2$$



we obtain from equation (2) that $dot{phi}=h/r^2quad$ with $h=L/m$
we put this result in equation (1) and get:



$$ddot{r}=frac{2,L^2}{m^2,r^3},-frac{1}{m},frac{d,U}{d,r}tag 3$$



for a circle motion $r(t)=constquad Rightarrowquad ddot{r}=0$ ;
we solve equation (3) for $frac{d,U}{d,r}$



$$boxed{F(r)=frac{d,U}{d,r}=frac{2,L^2}{m}frac{1}{r^3}}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Unless m is dimensionless and equal to 1, something is very wrong here!
    $endgroup$
    – Bill N
    Jan 19 at 19:02










  • $begingroup$
    Yes I also think so, I Have to check
    $endgroup$
    – Eli
    Jan 19 at 19:33










  • $begingroup$
    @Bill N I found my error i forgot the mass in the kinetic energy, I think this solution is o.k ? but if i compare force sign in the quotation i got deference sing, and anther value for $k$.
    $endgroup$
    – Eli
    Jan 19 at 20:45


















0












$begingroup$

for a circle motion die circle radius $r(t)$ must be constant.



we can calculate the EOM's with Euler-Lagrange method.



kinetic energy:



$$ T=frac{1}{2},mleft(dot{r}^2+r^2,dot{phi}^2right)$$



potential energy



$$U=frac{L^2}{2,m,r^2}+ U(r)$$



where $L$ is the angular momentum and
$U(r)$ the unknown potential for a circle motion



The equation of motions are:



$$ddot{r}=r,dot{phi}^2+frac{L^2}{m^2,r^3}-frac{1}{m}frac{d,U}{d,r}tag 1$$



$$m,rleft(ddot{phi},r+2,dot{r},dot{phi}right)=0quad Rightarrow$$



$$m,frac{d}{dt}left(r^2,dot{phi}right)=0tag 2$$



we obtain from equation (2) that $dot{phi}=h/r^2quad$ with $h=L/m$
we put this result in equation (1) and get:



$$ddot{r}=frac{2,L^2}{m^2,r^3},-frac{1}{m},frac{d,U}{d,r}tag 3$$



for a circle motion $r(t)=constquad Rightarrowquad ddot{r}=0$ ;
we solve equation (3) for $frac{d,U}{d,r}$



$$boxed{F(r)=frac{d,U}{d,r}=frac{2,L^2}{m}frac{1}{r^3}}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Unless m is dimensionless and equal to 1, something is very wrong here!
    $endgroup$
    – Bill N
    Jan 19 at 19:02










  • $begingroup$
    Yes I also think so, I Have to check
    $endgroup$
    – Eli
    Jan 19 at 19:33










  • $begingroup$
    @Bill N I found my error i forgot the mass in the kinetic energy, I think this solution is o.k ? but if i compare force sign in the quotation i got deference sing, and anther value for $k$.
    $endgroup$
    – Eli
    Jan 19 at 20:45
















0












0








0





$begingroup$

for a circle motion die circle radius $r(t)$ must be constant.



we can calculate the EOM's with Euler-Lagrange method.



kinetic energy:



$$ T=frac{1}{2},mleft(dot{r}^2+r^2,dot{phi}^2right)$$



potential energy



$$U=frac{L^2}{2,m,r^2}+ U(r)$$



where $L$ is the angular momentum and
$U(r)$ the unknown potential for a circle motion



The equation of motions are:



$$ddot{r}=r,dot{phi}^2+frac{L^2}{m^2,r^3}-frac{1}{m}frac{d,U}{d,r}tag 1$$



$$m,rleft(ddot{phi},r+2,dot{r},dot{phi}right)=0quad Rightarrow$$



$$m,frac{d}{dt}left(r^2,dot{phi}right)=0tag 2$$



we obtain from equation (2) that $dot{phi}=h/r^2quad$ with $h=L/m$
we put this result in equation (1) and get:



$$ddot{r}=frac{2,L^2}{m^2,r^3},-frac{1}{m},frac{d,U}{d,r}tag 3$$



for a circle motion $r(t)=constquad Rightarrowquad ddot{r}=0$ ;
we solve equation (3) for $frac{d,U}{d,r}$



$$boxed{F(r)=frac{d,U}{d,r}=frac{2,L^2}{m}frac{1}{r^3}}$$






share|cite|improve this answer











$endgroup$



for a circle motion die circle radius $r(t)$ must be constant.



we can calculate the EOM's with Euler-Lagrange method.



kinetic energy:



$$ T=frac{1}{2},mleft(dot{r}^2+r^2,dot{phi}^2right)$$



potential energy



$$U=frac{L^2}{2,m,r^2}+ U(r)$$



where $L$ is the angular momentum and
$U(r)$ the unknown potential for a circle motion



The equation of motions are:



$$ddot{r}=r,dot{phi}^2+frac{L^2}{m^2,r^3}-frac{1}{m}frac{d,U}{d,r}tag 1$$



$$m,rleft(ddot{phi},r+2,dot{r},dot{phi}right)=0quad Rightarrow$$



$$m,frac{d}{dt}left(r^2,dot{phi}right)=0tag 2$$



we obtain from equation (2) that $dot{phi}=h/r^2quad$ with $h=L/m$
we put this result in equation (1) and get:



$$ddot{r}=frac{2,L^2}{m^2,r^3},-frac{1}{m},frac{d,U}{d,r}tag 3$$



for a circle motion $r(t)=constquad Rightarrowquad ddot{r}=0$ ;
we solve equation (3) for $frac{d,U}{d,r}$



$$boxed{F(r)=frac{d,U}{d,r}=frac{2,L^2}{m}frac{1}{r^3}}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 19 at 20:41

























answered Jan 19 at 17:14









EliEli

527116




527116












  • $begingroup$
    Unless m is dimensionless and equal to 1, something is very wrong here!
    $endgroup$
    – Bill N
    Jan 19 at 19:02










  • $begingroup$
    Yes I also think so, I Have to check
    $endgroup$
    – Eli
    Jan 19 at 19:33










  • $begingroup$
    @Bill N I found my error i forgot the mass in the kinetic energy, I think this solution is o.k ? but if i compare force sign in the quotation i got deference sing, and anther value for $k$.
    $endgroup$
    – Eli
    Jan 19 at 20:45




















  • $begingroup$
    Unless m is dimensionless and equal to 1, something is very wrong here!
    $endgroup$
    – Bill N
    Jan 19 at 19:02










  • $begingroup$
    Yes I also think so, I Have to check
    $endgroup$
    – Eli
    Jan 19 at 19:33










  • $begingroup$
    @Bill N I found my error i forgot the mass in the kinetic energy, I think this solution is o.k ? but if i compare force sign in the quotation i got deference sing, and anther value for $k$.
    $endgroup$
    – Eli
    Jan 19 at 20:45


















$begingroup$
Unless m is dimensionless and equal to 1, something is very wrong here!
$endgroup$
– Bill N
Jan 19 at 19:02




$begingroup$
Unless m is dimensionless and equal to 1, something is very wrong here!
$endgroup$
– Bill N
Jan 19 at 19:02












$begingroup$
Yes I also think so, I Have to check
$endgroup$
– Eli
Jan 19 at 19:33




$begingroup$
Yes I also think so, I Have to check
$endgroup$
– Eli
Jan 19 at 19:33












$begingroup$
@Bill N I found my error i forgot the mass in the kinetic energy, I think this solution is o.k ? but if i compare force sign in the quotation i got deference sing, and anther value for $k$.
$endgroup$
– Eli
Jan 19 at 20:45






$begingroup$
@Bill N I found my error i forgot the mass in the kinetic energy, I think this solution is o.k ? but if i compare force sign in the quotation i got deference sing, and anther value for $k$.
$endgroup$
– Eli
Jan 19 at 20:45












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