เมนูนำทาง
กฎการเคลื่อนที่ของดาวเคราะห์
where θ ^ {\displaystyle {\hat {\boldsymbol {\theta }}}} is the tangential unit vector, and
θ ^ ˙ = − θ ˙ r ^ . {\displaystyle {\dot {\hat {\boldsymbol {\theta }}}}=-{\dot {\theta }}{\hat {\mathbf {r} }}.}So the position vector
r = r r ^ {\displaystyle \mathbf {r} =r{\hat {\mathbf {r} }}}is differentiated twice to give the velocity vector and the acceleration vector
r ˙ = r ˙ r ^ + r r ^ ˙ = r ˙ r ^ + r θ ˙ θ ^ , {\displaystyle {\dot {\mathbf {r} }}={\dot {r}}{\hat {\mathbf {r} }}+r{\dot {\hat {\mathbf {r} }}}={\dot {r}}{\hat {\mathbf {r} }}+r{\dot {\theta }}{\hat {\boldsymbol {\theta }}},} r ¨ = ( r ¨ r ^ + r ˙ r ^ ˙ ) + ( r ˙ θ ˙ θ ^ + r θ ¨ θ ^ + r θ ˙ θ ^ ˙ ) = ( r ¨ − r θ ˙ 2 ) r ^ + ( r θ ¨ + 2 r ˙ θ ˙ ) θ ^ . {\displaystyle {\ddot {\mathbf {r} }}=({\ddot {r}}{\hat {\mathbf {r} }}+{\dot {r}}{\dot {\hat {\mathbf {r} }}})+({\dot {r}}{\dot {\theta }}{\hat {\boldsymbol {\theta }}}+r{\ddot {\theta }}{\hat {\boldsymbol {\theta }}}+r{\dot {\theta }}{\dot {\hat {\boldsymbol {\theta }}}})=({\ddot {r}}-r{\dot {\theta }}^{2}){\hat {\mathbf {r} }}+(r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}){\hat {\boldsymbol {\theta }}}.}Note that for constant distance, r {\displaystyle \ r} , the planet is subject to the centripetal acceleration, r θ ˙ 2 {\displaystyle r{\dot {\theta }}^{2}} , and for constant angular speed, θ ˙ {\displaystyle {\dot {\theta }}} , the planet is subject to the coriolis acceleration, 2 r ˙ θ ˙ {\displaystyle 2{\dot {r}}{\dot {\theta }}} .
Inserting the acceleration vector into Newton's laws, and dividing by m, gives the vector equation of motion
( r ¨ − r θ ˙ 2 ) r ^ + ( r θ ¨ + 2 r ˙ θ ˙ ) θ ^ = − G M r − 2 r ^ {\displaystyle ({\ddot {r}}-r{\dot {\theta }}^{2}){\hat {\mathbf {r} }}+(r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}){\hat {\boldsymbol {\theta }}}=-GMr^{-2}{\hat {\mathbf {r} }}}Equating component, we get the two ordinary differential equations of motion, one for the radial acceleration and one for the tangential acceleration:
r ¨ − r θ ˙ 2 = − G M r − 2 , {\displaystyle {\ddot {r}}-r{\dot {\theta }}^{2}=-GMr^{-2},} r θ ¨ + 2 r ˙ θ ˙ = 0. {\displaystyle r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}=0.}r θ ˙ : {\displaystyle \ r{\dot {\theta }}:}
θ ¨ θ ˙ + 2 r ˙ r = 0 {\displaystyle {\frac {\ddot {\theta }}{\dot {\theta }}}+2{\frac {\dot {r}}{r}}=0}and integrate:
log θ ˙ + 2 log r = log ℓ , {\displaystyle \log {\dot {\theta }}+2\log r=\log \ell ,}where log ℓ {\displaystyle \log \ell } is a constant of integration, and exponentiate:
r 2 θ ˙ = ℓ . {\displaystyle r^{2}{\dot {\theta }}=\ell .}This says that the specific angular momentum r 2 θ ˙ {\displaystyle r^{2}{\dot {\theta }}} is a constant of motion, even if both the distance r {\displaystyle \ r} and the angular speed θ ˙ {\displaystyle {\dot {\theta }}} vary.
The area swept out from time t1 to time t2,
∫ t 1 t 2 1 2 ⋅ b a s e ⋅ h e i g h t ⋅ d t = ∫ t 1 t 2 1 2 ⋅ r ⋅ r θ ˙ ⋅ d t = 1 2 ⋅ ℓ ⋅ ( t 2 − t 1 ) {\displaystyle \ \int _{t_{1}}^{t_{2}}{\frac {1}{2}}\cdot base\cdot height\cdot dt=\int _{t_{1}}^{t_{2}}{\frac {1}{2}}\cdot r\cdot r{\dot {\theta }}\cdot dt={\frac {1}{2}}\cdot \ell \cdot (t_{2}-t_{1})}depends only on the duration t2−t1. This is Kepler's second law.
and get
− G M r − 2 = − ℓ 2 p − 3 u 2 {\displaystyle -GMr^{-2}=-\ell ^{2}p^{-3}u^{2}}and
θ ˙ = ℓ r − 2 = ℓ p − 2 u 2 . {\displaystyle \ {\dot {\theta }}=\ell r^{-2}=\ell p^{-2}u^{2}.} X ˙ = d X d θ ⋅ θ ˙ = d X d θ ⋅ ℓ p − 2 u 2 . {\displaystyle \ {\dot {X}}={\frac {dX}{d\theta }}\cdot {\dot {\theta }}={\frac {dX}{d\theta }}\cdot \ell p^{-2}u^{2}.}Differentiate
r = p u − 1 {\displaystyle \ r=pu^{-1}}twice:
r ˙ = d ( p u − 1 ) d θ ⋅ ℓ p − 2 u 2 = − p u − 2 d u d θ ⋅ ℓ p − 2 u 2 = − ℓ p − 1 d u d θ {\displaystyle {\dot {r}}={\frac {d(pu^{-1})}{d\theta }}\cdot \ell p^{-2}u^{2}=-pu^{-2}{\frac {du}{d\theta }}\cdot \ell p^{-2}u^{2}=-\ell p^{-1}{\frac {du}{d\theta }}} r ¨ = d r ˙ d θ ⋅ ℓ p − 2 u 2 = d d θ ( − ℓ p − 1 d u d θ ) ⋅ ℓ p − 2 u 2 = − ℓ 2 p − 3 u 2 d 2 u d θ 2 {\displaystyle {\ddot {r}}={\frac {d{\dot {r}}}{d\theta }}\cdot \ell p^{-2}u^{2}={\frac {d}{d\theta }}(-\ell p^{-1}{\frac {du}{d\theta }})\cdot \ell p^{-2}u^{2}=-\ell ^{2}p^{-3}u^{2}{\frac {d^{2}u}{d\theta ^{2}}}}Substitute into the radial equation of motion
r ¨ − r θ ˙ 2 = − G M r − 2 {\displaystyle {\ddot {r}}-r{\dot {\theta }}^{2}=-GMr^{-2}}and get
( − ℓ 2 p − 3 u 2 d 2 u d θ 2 ) − ( p u − 1 ) ( ℓ p − 2 u 2 ) 2 = − ℓ 2 p − 3 u 2 {\displaystyle (-\ell ^{2}p^{-3}u^{2}{\frac {d^{2}u}{d\theta ^{2}}})-(pu^{-1})(\ell p^{-2}u^{2})^{2}=-\ell ^{2}p^{-3}u^{2}}Divide by − ℓ 2 p − 3 u 2 {\displaystyle -\ell ^{2}p^{-3}u^{2}}
d 2 u d θ 2 + u = 1. {\displaystyle {\frac {d^{2}u}{d\theta ^{2}}}+u=1.}These solutions are
u = ϵ ⋅ cos ( θ − A ) {\displaystyle \ u=\epsilon \cdot \cos(\theta -A)}where ϵ {\displaystyle \ \epsilon } and A {\displaystyle \ A} are arbitrary constants of integration. So the result is
u = 1 + ϵ ⋅ cos ( θ − A ) {\displaystyle \ u=1+\epsilon \cdot \cos(\theta -A)}Choosing the axis of the coordinate system such that A = 0 {\displaystyle \ A=0} , and inserting u = p r − 1 {\displaystyle \ u=pr^{-1}} , gives:
p r − 1 = 1 + ϵ ⋅ cos θ . {\displaystyle \ pr^{-1}=1+\epsilon \cdot \cos \theta .}If ϵ < 1 , {\displaystyle \ \epsilon <1,} this is Kepler's first law.
where:
โดย:
เมนูนำทาง
กฎการเคลื่อนที่ของดาวเคราะห์ใกล้เคียง
แหล่งที่มา
WikiPedia: กฎการเคลื่อนที่ของดาวเคราะห์