Movimiento elíptico

El ejemplo más conocido de movimiento elíptico es el de los planetas del sistema solar alrededor del sol . La imagen muestra los parámetros característicos de la órbita, con los nombres de los ábsides.

En cinemática , el movimiento elíptico es el movimiento de un cuerpo , o de un punto material , a lo largo de una trayectoria elíptica . En general, un cuerpo tiende a tomar una trayectoria elíptica cuando se somete a una fuerza central .

Análisis de movimiento y derivación de trayectorias

Al definir el momento mecánico específico del vector:

\mathbf {c} =\mathbf {r} \times \mathbf {a}

{\ Displaystyle \ mathbf {c} = \ mathbf {r} \ times \ mathbf {a}}

{\ Displaystyle \ mathbf {c} = \ mathbf {r} \ times \ mathbf {a}}

En el caso del movimiento central , tenemos que $\mathbf {r}$ ${\ Displaystyle \ mathbf {r}}$ $\ mathbf r$ Y $\mathbf {a}$ ${\ Displaystyle \ mathbf {a}}$ $\ mathbf a$ son por tanto paralelos $\mathbf {c} =0$ ${\ Displaystyle \ mathbf {c} = 0}$ ${\ Displaystyle \ mathbf {c} = 0}$ . Dado que el polo con respecto al cual se calcula $\mathbf {c}$ ${\ Displaystyle \ mathbf {c}}$ $\ mathbf c$ coincide con el centro de masa , que se puede suponer estacionario, tenemos que el momento mecánico específico es igual a la primera derivada con respecto al tiempo del momento angular específico $\mathbf {h}$ ${\ Displaystyle \ mathbf {h}}$ ${\ mathbf h}$ :

{\frac {\mathrm {d} \mathbf {h} }{\mathrm {d} t}}={\frac {\mathrm {d} }{\mathrm {d} t}}(\mathbf {r} \times \mathbf {v} )=\mathbf {c} =0

{\ Displaystyle {\ frac {\ mathrm {d} \ mathbf {h}} {\ mathrm {d} t}} = {\ frac {\ mathrm {d}} {\ mathrm {d} t}} (\ mathbf {r} \ times \ mathbf {v}) = \ mathbf {c} = 0}

{\ Displaystyle {\ frac {\ mathrm {d} \ mathbf {h}} {\ mathrm {d} t}} = {\ frac {\ mathrm {d}} {\ mathrm {d} t}} (\ mathbf {r} \ times \ mathbf {v}) = \ mathbf {c} = 0}

por lo tanto tenemos eso $\mathbf {h}$ ${\ Displaystyle \ mathbf {h}}$ ${\ mathbf h}$ es constante, de acuerdo con la segunda ley de Kepler . La velocidad areolar ${\dot {\mathbf {A} }}$ ${\ Displaystyle {\ dot {\ mathbf {A}}}}$ ${\ Displaystyle {\ dot {\ mathbf {A}}}}$ es igual a:

{\dot {\mathbf {A} }}={\frac {\mathbf {r} \times \mathbf {v} }{2}}={\frac {\mathbf {r} \times ({\boldsymbol {\omega }}\times \mathbf {r} )}{2}}={\frac {(\mathbf {r} \cdot \mathbf {r} ){\boldsymbol {\omega }}-{\cancel {({\boldsymbol {\omega }}\cdot \mathbf {r} )\mathbf {r} }}}{2}}={\frac {r^{2}{\boldsymbol {\omega }}}{2}}

{\ Displaystyle {\ dot {\ mathbf {A}}} = {\ frac {\ mathbf {r} \ times \ mathbf {v}} {2}} = {\ frac {\ mathbf {r} \ times ({ \ boldsymbol {\ omega}} \ times \ mathbf {r})} {2}} = {\ frac {(\ mathbf {r} \ cdot \ mathbf {r}) {\ boldsymbol {\ omega}} - {\ cancelar {({\ boldsymbol {\ omega}} \ cdot \ mathbf {r}) \ mathbf {r}}}} {2}} = {\ frac {r ^ {2} {\ boldsymbol {\ omega}}} {2}}}

{\ Displaystyle {\ dot {\ mathbf {A}}} = {\ frac {\ mathbf {r} \ times \ mathbf {v}} {2}} = {\ frac {\ mathbf {r} \ times ({ \ boldsymbol {\ omega}} \ times \ mathbf {r})} {2}} = {\ frac {(\ mathbf {r} \ cdot \ mathbf {r}) {\ boldsymbol {\ omega}} - {\ cancelar {({\ boldsymbol {\ omega}} \ cdot \ mathbf {r}) \ mathbf {r}}}} {2}} = {\ frac {r ^ {2} {\ boldsymbol {\ omega}}} {2}}}

Dónde está ${\boldsymbol {\omega }}$ ${\ displaystyle {\ boldsymbol {\ omega}}}$ ${\ displaystyle {\ boldsymbol {\ omega}}}$ es la velocidad angular .

Sabiendo que en coordenadas polares tenemos:

{\begin{aligned}&\theta =\arctan {\frac {y}{x}},\quad r^{2}=x^{2}+y^{2};\\[4pt]&{\dot {\theta }}=\omega ={\frac {1}{1+\left({\frac {y}{x}}\right)^{2}}}\left({\frac {x{\dot {y}}-{\dot {x}}y}{x^{2}}}\right)={\frac {\cancel {x^{2}}}{x^{2}+y^{2}}}\left({\frac {x{\dot {y}}-{\dot {x}}y}{\cancel {x^{2}}}}\right)={\frac {x{\dot {y}}-{\dot {x}}y}{x^{2}+y^{2}}}\\[4pt]&r^{2}\omega ={\frac {x{\dot {y}}-{\dot {x}}y}{\cancel {(x^{2}+y^{2})}}}{\cancel {(x^{2}+y^{2})}}=x{\dot {y}}-{\dot {x}}y\\[4pt]\end{aligned}}

{\ Displaystyle {\ begin {alineado} & \ theta = \ arctan {\ frac {y} {x}}, \ quad r ^ {2} = x ^ {2} + y ^ {2}; \\ [4pt ] & {\ dot {\ theta}} = \ omega = {\ frac {1} {1+ \ left ({\ frac {y} {x}} \ right) ^ {2}}} \ left ({\ frac {x {\ dot {y}} - {\ dot {x}} y} {x ^ {2}}} \ right) = {\ frac {\ cancel {x ^ {2}}} {x ^ { 2} + y ^ {2}}} \ left ({\ frac {x {\ dot {y}} - {\ dot {x}} y} {\ cancel {x ^ {2}}}} \ right) = {\ frac {x {\ dot {y}} - {\ dot {x}} y} {x ^ {2} + y ^ {2}}} \\ [4pt] & r ^ {2} \ omega = {\ frac {x {\ dot {y}} - {\ dot {x}} y} {\ cancel {(x ^ {2} + y ^ {2})}}} {\ cancel {(x ^ {2} + y ^ {2})}} = x {\ dot {y}} - {\ dot {x}} y \\ [4pt] \ end {alineado}}}

{\ Displaystyle {\ begin {alineado} & \ theta = \ arctan {\ frac {y} {x}}, \ quad r ^ {2} = x ^ {2} + y ^ {2}; \\ [4pt ] & {\ dot {\ theta}} = \ omega = {\ frac {1} {1+ \ left ({\ frac {y} {x}} \ right) ^ {2}}} \ left ({\ frac {x {\ dot {y}} - {\ dot {x}} y} {x ^ {2}}} \ right) = {\ frac {\ cancel {x ^ {2}}} {x ^ { 2} + y ^ {2}}} \ left ({\ frac {x {\ dot {y}} - {\ dot {x}} y} {\ cancel {x ^ {2}}}} \ right) = {\ frac {x {\ dot {y}} - {\ dot {x}} y} {x ^ {2} + y ^ {2}}} \\ [4pt] & r ^ {2} \ omega = {\ frac {x {\ dot {y}} - {\ dot {x}} y} {\ cancel {(x ^ {2} + y ^ {2})}}} {\ cancel {(x ^ {2} + y ^ {2})}} = x {\ dot {y}} - {\ dot {x}} y \\ [4pt] \ end {alineado}}}

mientras que la elipse en coordenadas polares es:

\left\{{\begin{aligned}&x=a\cos \theta \\&y=b\sin \theta \\\end{aligned}}\right.\implies \left\{{\begin{aligned}&{\dot {x}}=-a\omega \sin \theta \\&{\dot {y}}=b\omega \cos \theta \\\end{aligned}}\right.

{\ Displaystyle \ left \ {{\ begin {alineado} & x = a \ cos \ theta \\ & y = b \ sin \ theta \\\ end {alineado}} \ right. \ implica \ left \ {{\ comenzar {alineado} & {\ dot {x}} = - a \ omega \ sin \ theta \\ & {\ dot {y}} = b \ omega \ cos \ theta \\\ end {alineado}} \ right. }

{\ Displaystyle \ left \ {{\ begin {alineado} & x = a \ cos \ theta \\ & y = b \ sin \ theta \\\ end {alineado}} \ right. \ implica \ left \ {{\ comenzar {alineado} & {\ dot {x}} = - a \ omega \ sin \ theta \\ & {\ dot {y}} = b \ omega \ cos \ theta \\\ end {alineado}} \ right. }

Por tanto, obtenemos que el valor de la velocidad areolar es:

{\dot {\mathbf {A} }}={\frac {r^{2}{\boldsymbol {\omega }}}{2}}={\frac {x{\dot {y}}-{\dot {x}}y}{2}}={\frac {ab{\boldsymbol {\omega }}}{2}}(\cos ^{2}\theta +\sin ^{2}\theta )={\frac {ab{\boldsymbol {\omega }}}{2}}

{\ displaystyle {\ dot {\ mathbf {A}}} = {\ frac {r ^ {2} {\ boldsymbol {\ omega}}} {2}} = {\ frac {x {\ dot {y}} - {\ dot {x}} y} {2}} = {\ frac {ab {\ boldsymbol {\ omega}}} {2}} (\ cos ^ {2} \ theta + \ sin ^ {2} \ theta) = {\ frac {ab {\ boldsymbol {\ omega}}} {2}}}

{\ displaystyle {\ dot {\ mathbf {A}}} = {\ frac {r ^ {2} {\ boldsymbol {\ omega}}} {2}} = {\ frac {x {\ dot {y}} - {\ dot {x}} y} {2}} = {\ frac {ab {\ boldsymbol {\ omega}}} {2}} (\ cos ^ {2} \ theta + \ sin ^ {2} \ theta) = {\ frac {ab {\ boldsymbol {\ omega}}} {2}}}

mientras que el valor del momento angular orbital específico $\mathbf {h}$ ${\ Displaystyle \ mathbf {h}}$ ${\ mathbf h}$ se vuelve:

\mathbf {h} =2{\dot {\mathbf {A} }}=ab{\boldsymbol {\omega }}

{\ Displaystyle \ mathbf {h} = 2 {\ dot {\ mathbf {A}}} = ab {\ boldsymbol {\ omega}}}

{\ Displaystyle \ mathbf {h} = 2 {\ dot {\ mathbf {A}}} = ab {\ boldsymbol {\ omega}}}

Ser $\mathbf {h}$ ${\ Displaystyle \ mathbf {h}}$ ${\ mathbf h}$ constante también ${\dot {\mathbf {A} }}$ ${\ Displaystyle {\ dot {\ mathbf {A}}}}$ ${\ Displaystyle {\ dot {\ mathbf {A}}}}$ Y ${\boldsymbol {\omega }}$ ${\ displaystyle {\ boldsymbol {\ omega}}}$ ${\ displaystyle {\ boldsymbol {\ omega}}}$ son constantes y esto permite obtener dos ecuaciones lineales con respecto al desplazamiento angular ${\boldsymbol {\theta }}$ ${\ displaystyle {\ boldsymbol {\ theta}}}$ ${\ displaystyle {\ boldsymbol {\ theta}}}$ y desplazamiento areolar $\mathbf {A}$ ${\ Displaystyle \ mathbf {A}}$ $\ mathbf A$ :

{\begin{aligned}&{\boldsymbol {\theta }}(t)={\boldsymbol {\theta }}_{0}+{\frac {\mathbf {h} }{ab}}t\\[4pt]&\mathbf {A} (t)=\mathbf {A} _{0}+{\frac {\mathbf {h} }{2}}t\\[4pt]\end{aligned}}

{\ displaystyle {\ begin {alineado} & {\ boldsymbol {\ theta}} (t) = {\ boldsymbol {\ theta}} _ {0} + {\ frac {\ mathbf {h}} {ab}} t \\ [4pt] & \ mathbf {A} (t) = \ mathbf {A} _ {0} + {\ frac {\ mathbf {h}} {2}} t \\ [4pt] \ end {alineado} }}

{\ displaystyle {\ begin {alineado} & {\ boldsymbol {\ theta}} (t) = {\ boldsymbol {\ theta}} _ {0} + {\ frac {\ mathbf {h}} {ab}} t \\ [4pt] & \ mathbf {A} (t) = \ mathbf {A} _ {0} + {\ frac {\ mathbf {h}} {2}} t \\ [4pt] \ end {alineado} }}

Las ecuaciones de movimiento en coordenadas cartesianas son:

\mathbf {r} =\left\{{\begin{aligned}&x=a\cos \left({\frac {h}{ab}}t+\theta _{0}\right)\\[4pt]&y=b\sin \left({\frac {h}{ab}}t+\theta _{0}\right)\\[4pt]\end{aligned}}\right.\implies \mathbf {v} =\left\{{\begin{aligned}&{\dot {x}}=-{\frac {h}{b}}\sin \left({\frac {h}{ab}}t+\theta _{0}\right)\\[4pt]&{\dot {y}}={\frac {h}{a}}\cos \left({\frac {h}{ab}}t+\theta _{0}\right)\\[4pt]\end{aligned}}\right.\implies \mathbf {a} =\left\{{\begin{aligned}&{\ddot {x}}=-{\frac {h^{2}}{ab^{2}}}\cos \left({\frac {h}{ab}}t+\theta _{0}\right)\\[4pt]&{\ddot {y}}={\frac {h^{2}}{a^{2}b}}\sin \left({\frac {h}{ab}}t+\theta _{0}\right)\\[4pt]\end{aligned}}\right.

{\ Displaystyle \ mathbf {r} = \ left \ {{\ begin {alineado} & x = a \ cos \ left ({\ frac {h} {ab}} t + \ theta _ {0} \ right) \ \ [4pt] & y = b \ sin \ left ({\ frac {h} {ab}} t + \ theta _ {0} \ right) \\ [4pt] \ end {alineado}} \ right. \ Implica \ mathbf {v} = \ left \ {{\ begin {alineado} & {\ dot {x}} = - {\ frac {h} {b}} \ sin \ left ({\ frac {h} {ab} } t + \ theta _ {0} \ right) \\ [4pt] & {\ dot {y}} = {\ frac {h} {a}} \ cos \ left ({\ frac {h} {ab} } t + \ theta _ {0} \ right) \\ [4pt] \ end {alineado}} \ right. \ Implies \ mathbf {a} = \ left \ {{\ begin {alineado} & {\ ddot {x }} = - {\ frac {h ^ {2}} {ab ^ {2}}} \ cos \ left ({\ frac {h} {ab}} t + \ theta _ {0} \ right) \\ [4pt] & {\ ddot {y}} = {\ frac {h ^ {2}} {a ^ {2} b}} \ sin \ left ({\ frac {h} {ab}} t + \ theta _ {0} \ right) \ \ [4pt] \ end {alineado}} \ right.}

{\ Displaystyle \ mathbf {r} = \ left \ {{\ begin {alineado} & x = a \ cos \ left ({\ frac {h} {ab}} t + \ theta _ {0} \ right) \ \ [4pt] & y = b \ sin \ left ({\ frac {h} {ab}} t + \ theta _ {0} \ right) \\ [4pt] \ end {alineado}} \ right. \ Implica \ mathbf {v} = \ left \ {{\ begin {alineado} & {\ dot {x}} = - {\ frac {h} {b}} \ sin \ left ({\ frac {h} {ab} } t + \ theta _ {0} \ right) \\ [4pt] & {\ dot {y}} = {\ frac {h} {a}} \ cos \ left ({\ frac {h} {ab} } t + \ theta _ {0} \ right) \\ [4pt] \ end {alineado}} \ right. \ Implies \ mathbf {a} = \ left \ {{\ begin {alineado} & {\ ddot {x }} = - {\ frac {h ^ {2}} {ab ^ {2}}} \ cos \ left ({\ frac {h} {ab}} t + \ theta _ {0} \ right) \\ [4pt] & {\ ddot {y}} = {\ frac {h ^ {2}} {a ^ {2} b}} \ sin \ left ({\ frac {h} {ab}} t + \ theta _ {0} \ right) \ \ [4pt] \ end {alineado}} \ right.}

esto significa que la aceleración coincide con la aceleración centrípeta $\mathbf {a} _{c}$ ${\ Displaystyle \ mathbf {a} _ {c}}$ ${\ Displaystyle \ mathbf {a} _ {c}}$ , que es igual a:

\mathbf {a} _{c}=\left({\frac {h}{ab}}\right)^{2}\mathbf {r}

{\ Displaystyle \ mathbf {a} _ {c} = \ left ({\ frac {h} {ab}} \ right) ^ {2} \ mathbf {r}}

{\ Displaystyle \ mathbf {a} _ {c} = \ left ({\ frac {h} {ab}} \ right) ^ {2} \ mathbf {r}}

Es posible observar que en el caso del movimiento circular , siendo $a=b=r$ ${\ Displaystyle a = b = r}$ ${\ Displaystyle a = b = r}$ , el valor de la aceleración centrípeta es igual a:

\mathbf {a} _{c}=\omega ^{2}\mathbf {r}

{\ Displaystyle \ mathbf {a} _ {c} = \ omega ^ {2} \ mathbf {r}}

{\ Displaystyle \ mathbf {a} _ {c} = \ omega ^ {2} \ mathbf {r}}

Bibliografía

P. Mazzoldi, M. Nigro, C. Voices, Physics - Volume I (segunda edición) , Nápoles, EdiSES, 2010, ISBN 88-7959-137-1 .

Movimiento elíptico

Análisis de movimiento y derivación de trayectorias

Bibliografía

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