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 10 hours ago
Peter HoferPeter Hofer
184
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Peter Hofer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
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 10 hours ago
Peter HoferPeter Hofer
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.
$endgroup$
$begingroup$
Are you asking about the stability of the circular orbit solution?
$endgroup$
– Qmechanic♦
10 hours ago
$begingroup$
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
10 hours ago
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
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
$endgroup$
– FGSUZ
10 hours ago
|
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 10 hours ago
Peter HoferPeter Hofer
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.
$endgroup$
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 10 hours ago
Peter HoferPeter Hofer
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.
asked 10 hours ago
Peter HoferPeter Hofer
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.
edited 9 hours ago
Peter Hofer
asked 10 hours ago
Peter HoferPeter Hofer
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.
asked 10 hours ago
Peter HoferPeter Hofer
184
asked 10 hours ago
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♦
10 hours ago
$begingroup$
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
10 hours ago
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
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
$endgroup$
– FGSUZ
10 hours ago
|
show 2 more comments
$begingroup$
Are you asking about the stability of the circular orbit solution?
$endgroup$
– Qmechanic♦
10 hours ago
$begingroup$
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
10 hours ago
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
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
$endgroup$
– FGSUZ
10 hours ago
$begingroup$
Are you asking about the stability of the circular orbit solution?
$endgroup$
– Qmechanic♦
10 hours ago
$begingroup$
Are you asking about the stability of the circular orbit solution?
$endgroup$
– Qmechanic♦
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
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
10 hours ago
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
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
$endgroup$
– FGSUZ
10 hours ago
$begingroup$
You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
$endgroup$
– FGSUZ
10 hours ago
|
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 7 hours ago
ReignReign
46029
$endgroup$
$begingroup$
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
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 6 hours ago
EliEli
53316
$endgroup$
$begingroup$
Unless m is dimensionless and equal to 1, something is very wrong here!
$endgroup$
– Bill N
5 hours ago
$begingroup$
Yes I also think so, I Have to check
$endgroup$
– Eli
4 hours ago
$begingroup$
@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
3 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 7 hours ago
ReignReign
46029
$endgroup$
$begingroup$
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
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 7 hours ago
ReignReign
46029
$endgroup$
$begingroup$
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
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 7 hours ago
ReignReign
46029
$endgroup$
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 7 hours ago
ReignReign
46029
edited 6 hours ago
answered 7 hours ago
ReignReign
46029
answered 7 hours ago
ReignReign
46029
answered 7 hours ago
ReignReign
46029
46029
$begingroup$
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
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 6 hours ago
EliEli
53316
$endgroup$
$begingroup$
Unless m is dimensionless and equal to 1, something is very wrong here!
$endgroup$
– Bill N
5 hours ago
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Yes I also think so, I Have to check
$endgroup$
– Eli
4 hours ago
$begingroup$
@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
3 hours ago
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 6 hours ago
EliEli
53316
$endgroup$
$begingroup$
Unless m is dimensionless and equal to 1, something is very wrong here!
$endgroup$
– Bill N
5 hours ago
$begingroup$
Yes I also think so, I Have to check
$endgroup$
– Eli
4 hours ago
$begingroup$
@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
3 hours ago
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 6 hours ago
EliEli
53316
$endgroup$
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 6 hours ago
EliEli
53316
edited 3 hours ago
answered 6 hours ago
EliEli
53316
answered 6 hours ago
EliEli
53316
answered 6 hours ago
EliEli
53316
53316
$begingroup$
Unless m is dimensionless and equal to 1, something is very wrong here!
$endgroup$
– Bill N
5 hours ago
$begingroup$
Yes I also think so, I Have to check
$endgroup$
– Eli
4 hours ago
$begingroup$
@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
3 hours ago
add a comment |
$begingroup$
Unless m is dimensionless and equal to 1, something is very wrong here!
$endgroup$
– Bill N
5 hours ago
$begingroup$
Yes I also think so, I Have to check
$endgroup$
– Eli
4 hours ago
$begingroup$
@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
3 hours ago
$begingroup$
Unless m is dimensionless and equal to 1, something is very wrong here!
$endgroup$
– Bill N
5 hours ago
$begingroup$
Unless m is dimensionless and equal to 1, something is very wrong here!
$endgroup$
– Bill N
5 hours ago
$begingroup$
Yes I also think so, I Have to check
$endgroup$
– Eli
4 hours ago
$begingroup$
Yes I also think so, I Have to check
$endgroup$
– Eli
4 hours ago
$begingroup$
@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
3 hours ago
$begingroup$
@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
3 hours ago
add a comment |
Peter Hofer is a new contributor. Be nice, and check out our Code of Conduct.
Peter Hofer is a new contributor. Be nice, and check out our Code of Conduct.
Peter Hofer is a new contributor. Be nice, and check out our Code of Conduct.
Peter Hofer is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Are you asking about the stability of the circular orbit solution?
$endgroup$
– Qmechanic♦
10 hours ago
$begingroup$
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
10 hours ago
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.
$endgroup$
– WarreG
10 hours ago
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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.
$endgroup$
– WarreG
10 hours ago
$begingroup$
You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$
$endgroup$
– FGSUZ
10 hours ago