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Qmechanic ♦
Qmechanic ♦
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Circular Stability of circular orbit in aattractive inverse cube central force field

Considering a motion of a body under aan attractive inverse cube 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 stationarystable 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?

Circular orbit in a central force field

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?

Stability of circular orbit in attractive inverse cube central force field

Considering a motion of a body under an attractive inverse cube 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 stable 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?

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Peter Hofer
Peter Hofer
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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?

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

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

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?

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?

Source Link
Peter Hofer
Peter Hofer
  • 184
  • 1
  • 6

Circular orbit in a central force field

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

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

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?