Circular orbit in a central force field
$begingroup$
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
asked Jan 19 at 13:04
Peter HoferPeter Hofer
184
$endgroup$
|
show 2 more comments
$begingroup$
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
asked Jan 19 at 13:04
Peter HoferPeter Hofer
184
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
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
asked Jan 19 at 13:04
Peter HoferPeter Hofer
184
$endgroup$
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
asked Jan 19 at 13:04
Peter HoferPeter Hofer
184
asked Jan 19 at 13:04
Peter HoferPeter Hofer
184
edited Jan 19 at 15:02
Peter Hofer
asked Jan 19 at 13:04
Peter HoferPeter Hofer
184
asked Jan 19 at 13:04
Peter HoferPeter Hofer
184
asked Jan 19 at 13:04
Peter HoferPeter Hofer
184
184
$begingroup$
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Jan 19 at 16:14
ReignReign
677211
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $frac{d,U}{d,r}$
$$boxed{F(r)=frac{d,U}{d,r}=frac{2,L^2}{m}frac{1}{r^3}}$$
answered Jan 19 at 17:14
EliEli
639116
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
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$begingroup$
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.
answered Jan 19 at 16:14
ReignReign
677211
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 19 at 16:14
ReignReign
677211
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 19 at 16:14
ReignReign
677211
$endgroup$
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.
answered Jan 19 at 16:14
ReignReign
677211
edited Jan 19 at 17:55
answered Jan 19 at 16:14
ReignReign
677211
answered Jan 19 at 16:14
ReignReign
677211
answered Jan 19 at 16:14
ReignReign
677211
677211
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $frac{d,U}{d,r}$
$$boxed{F(r)=frac{d,U}{d,r}=frac{2,L^2}{m}frac{1}{r^3}}$$
answered Jan 19 at 17:14
EliEli
639116
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Unless m is dimensionless and equal to 1, something is very wrong here!
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– Bill N
Jan 19 at 19:02
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Yes I also think so, I Have to check
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– Eli
Jan 19 at 19:33
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@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 $k$.
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– Eli
Jan 19 at 20:45
add a comment |
$begingroup$
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 $frac{d,U}{d,r}$
$$boxed{F(r)=frac{d,U}{d,r}=frac{2,L^2}{m}frac{1}{r^3}}$$
answered Jan 19 at 17:14
EliEli
639116
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $frac{d,U}{d,r}$
$$boxed{F(r)=frac{d,U}{d,r}=frac{2,L^2}{m}frac{1}{r^3}}$$
answered Jan 19 at 17:14
EliEli
639116
$endgroup$
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 $frac{d,U}{d,r}$
$$boxed{F(r)=frac{d,U}{d,r}=frac{2,L^2}{m}frac{1}{r^3}}$$
answered Jan 19 at 17:14
EliEli
639116
edited Jan 19 at 20:41
answered Jan 19 at 17:14
EliEli
639116
answered Jan 19 at 17:14
EliEli
639116
answered Jan 19 at 17:14
EliEli
639116
639116
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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Are you asking about the stability of the circular orbit solution?
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– Qmechanic♦
Jan 19 at 13:19
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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.
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– Peter Hofer
Jan 19 at 13:38
1
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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.
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– WarreG
Jan 19 at 13:56
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After a small calculation the derivative seems to be zero at $k=l^2/m$, so I guess it is a valid solution.
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– WarreG
Jan 19 at 14:01
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You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
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– FGSUZ
Jan 19 at 14:03