Editing Orbit (section)

=== Newtonian analysis of orbital motion ===
{{further|Kepler orbit|orbit equation|Kepler's first law}}

The following derivation applies to such an elliptical orbit. We start only with the [[classical mechanics|Newtonian]] law of gravitation stating that the gravitational acceleration towards the central body is related to the inverse of the square of the distance between them, namely

: <math> F_2 = -\frac {G m_1 m_2}{r^2} </math>

where ''F''<sub>2</sub> is the force acting on the mass ''m''<sub>2</sub> caused by the gravitational attraction mass ''m''<sub>1</sub> has for ''m''<sub>2</sub>, ''G'' is the universal gravitational constant, and ''r'' is the distance between the two masses centers.

From Newton's second law, the summation of the forces acting on ''m''<sub>2</sub> related to that body's acceleration:

: <math>F_2 = m_2 A_2</math>

where ''A''<sub>2</sub> is the acceleration of ''m''<sub>2</sub> caused by the force of gravitational attraction ''F''<sub>2</sub> of ''m''<sub>1</sub> acting on ''m''<sub>2</sub>.

Combining Eq. 1 and 2:

: <math> -\frac {G  m_1  m_2}{r^2} = m_2  A_2 </math>

Solving for the acceleration, ''A''<sub>2</sub>:

: <math> A_2 = \frac{F_2}{m_2} = - \frac{1}{m_2} \frac{G m_1 m_2}{r^2} =  -\frac{\mu}{r^2} </math>

where <math> \mu\, </math> is the [[standard gravitational parameter]], in this case <math>G m_1</math>. It is understood that the system being described is ''m''<sub>2</sub>, hence the subscripts can be dropped.

We assume that the central body is massive enough that it can be considered to be stationary and we ignore the more subtle effects of [[general relativity]].

When a pendulum or an object attached to a spring swings in an ellipse, the inward acceleration/force is proportional to the distance <math> A = F/m  = - k r.</math> Due to the way vectors add, the component of the force in the <math> \hat{\mathbf{x}} </math> or in the <math> \hat{\mathbf{y}} </math> directions are also proportionate to the respective components of the distances, <math> r''_x = A_x = - k r_x </math>. Hence, the entire analysis can be done separately in these dimensions. This results in the harmonic parabolic equations <math> x = A \cos(t) </math> and <math> y = B \sin(t) </math> of the ellipse.

The location of the orbiting object at the current time <math> t </math> is located in the plane using [[vector calculus]] in [[polar coordinates]] both with the standard Euclidean basis and with the polar basis with the origin coinciding with the center of force. Let <math> r </math> be the distance between the object and the center and <math> \theta </math> be the angle it has rotated. Let <math> \hat{\mathbf{x}} </math> and <math> \hat{\mathbf{y}} </math> be the standard [[Euclidean space|Euclidean]] bases and let <math> \hat{\mathbf{r}} = \cos(\theta)\hat{\mathbf{x}} + \sin(\theta)\hat{\mathbf{y}} </math> and <math> \hat{\boldsymbol \theta} = - \sin(\theta)\hat{\mathbf{x}} + \cos(\theta)\hat{\mathbf{y}} </math> be the radial and transverse [[Polar coordinate system#Vector calculus|polar]] basis with the first being the unit vector pointing from the central body to the current location of the orbiting object and the second being the orthogonal unit vector pointing in the direction that the orbiting object would travel if orbiting in a counter clockwise circle. Then the vector to the orbiting object is

: <math> \hat{\mathbf{O}} = r \cos(\theta)\hat{\mathbf{x}} + r \sin(\theta)\hat{\mathbf{y}} = r \hat{\mathbf{r}} </math>

We use <math> \dot r </math> and <math> \dot \theta </math> to denote the standard derivatives of how this distance and angle change over time. We take the derivative of a vector to see how it changes over time by subtracting its location at time <math> t </math> from that at time <math> t + \delta t </math> and dividing by <math>\delta t </math>. The result is also a vector. Because our basis vector <math> \hat{\mathbf{r}} </math> moves as the object orbits, we start by differentiating it. From time <math> t </math> to <math> t + \delta t </math>, the vector <math> \hat{\mathbf{r}} </math> keeps its beginning at the origin and rotates from angle <math> \theta </math> to <math> \theta  + \dot \theta\ \delta t </math> which moves its head a distance <math> \dot \theta\ \delta t </math> in the perpendicular direction <math> \hat{\boldsymbol \theta} </math> giving a derivative of <math> \dot \theta \hat{\boldsymbol \theta} </math>.

: <math>\begin{align}
  \hat{\mathbf{r}}
    &= \cos(\theta)\hat{\mathbf{x}} + \sin(\theta)\hat{\mathbf{y}} \\
  \frac{\delta \hat{\mathbf{r}}}{\delta t} = \dot{\mathbf r}
    &= -\sin(\theta)\dot \theta \hat{\mathbf{x}} + \cos(\theta)\dot \theta \hat{\mathbf{y}} = \dot \theta \hat{\boldsymbol \theta} \\
  \hat{\boldsymbol \theta}
    &= -\sin(\theta)\hat{\mathbf{x}} + \cos(\theta)\hat{\mathbf{y}} \\
  \frac{\delta \hat{\boldsymbol \theta}}{\delta t} = \dot{\boldsymbol \theta}
    &= -\cos(\theta)\dot \theta \hat{\mathbf{x}} - \sin(\theta) \dot \theta \hat{\mathbf{y}} = -\dot \theta \hat{\mathbf r}
\end{align}</math>

We can now find the velocity and acceleration of our orbiting object.
: <math>\begin{align}
   \hat{\mathbf{O}} &= r \hat{\mathbf{r}} \\
   \dot{\mathbf{O}} &= \frac{\delta r}{\delta t} \hat{\mathbf{r}} + r \frac{\delta \hat{\mathbf{r}}}{\delta t}
                     = \dot r \hat {\mathbf r} + r \left[ \dot \theta \hat {\boldsymbol \theta} \right] \\
  \ddot{\mathbf{O}} &= \left[\ddot r \hat {\mathbf r} + \dot r \dot \theta \hat {\boldsymbol \theta}\right] +
                       \left[\dot r \dot \theta \hat {\boldsymbol \theta} +
                             r \ddot \theta \hat {\boldsymbol \theta} -
                             r \dot \theta^2 \hat {\mathbf r}
                       \right] \\
                    &= \left[\ddot r - r\dot\theta^2\right]\hat{\mathbf{r}} +
                       \left[r \ddot\theta + 2 \dot r \dot\theta\right] \hat{\boldsymbol \theta}
\end{align}</math>

The coefficients of <math> \hat{\mathbf{r}} </math> and <math> \hat{\boldsymbol \theta} </math> give the accelerations in the radial and transverse directions. As said, Newton gives this first due to gravity is <math> -\mu/r^2 </math> and the second is zero.

{{NumBlk|:|<math> \ddot r - r\dot\theta^2 = - \frac{\mu}{r^2} </math>|1}}
{{NumBlk|:|<math> r \ddot\theta + 2 \dot r \dot\theta = 0 </math>|2}}

Equation (2) can be rearranged using integration by parts.

: <math> r \ddot\theta + 2 \dot r \dot\theta = \frac{1}{r}\frac{d}{dt}\left( r^2 \dot \theta \right) = 0 </math>

We can multiply through by <math> r </math> because it is not zero unless the orbiting object crashes.
Then having the derivative be zero gives that the function is a constant.

{{NumBlk|:|<math>r^2 \dot \theta = h </math>|3}}

which is actually the theoretical proof of [[Kepler's second law]] (A line joining a planet and the Sun sweeps out equal areas during equal intervals of time). The constant of integration, ''h'', is the [[specific relative angular momentum|angular momentum per unit mass]].

In order to get an equation for the orbit from equation (1), we need to eliminate time.<ref>{{cite web|url=http://farside.ph.utexas.edu/teaching/301/lectures/node155.html |title=Planetary orbits|last=Fitzpatrick |first=Richard |date=2 February 2006 |work=Classical Mechanics – an introductory course |publisher=The University of Texas at Austin |archive-url=https://web.archive.org/web/20010303195257/http://farside.ph.utexas.edu/teaching/301/lectures/node155.html |url-status=live |archive-date=3 March 2001 }}</ref> (See also [[Binet equation]].)
In polar coordinates, this would express the distance <math> r </math> of the orbiting object from the center as a function of its angle <math> \theta </math>. However, it is easier to introduce the auxiliary variable <math> u = 1/r  </math> and to express <math> u </math> as a function of <math> \theta </math>. Derivatives of <math>r</math> with respect to time may be rewritten as derivatives of <math>u</math> with respect to angle.

: <math>u = { 1 \over r }</math>
: <math>\dot\theta = \frac{h}{r^2} = hu^2</math> (reworking (3))
: <math>\begin{align}
      \frac{\delta u}{\delta \theta} &= \frac{\delta}{\delta t}\left(\frac{1}{r}\right)\frac{\delta t}{\delta \theta } = -\frac{\dot{r}}{r^2\dot{\theta }} = -\frac{\dot{r}}{h} \\
  \frac{\delta^2 u}{\delta \theta^2} &= -\frac{1}{h}\frac{\delta \dot{r}}{\delta t}\frac{\delta t}{\delta \theta } = -\frac{\ddot{r}}{h\dot{\theta}} = -\frac{\ddot{r}}{h^2 u^2}
  \ \ \ \text{ or } \ \ \  \ddot r = - h^2 u^2 \frac{\delta^2 u}{\delta \theta^2}
\end{align}</math>

Plugging these into (1) gives

: <math>\begin{align}
                                                          \ddot r - r\dot\theta^2 &= -\frac{\mu}{r^2} \\
   -h^2 u^2 \frac{\delta^2 u}{\delta \theta^2} - \frac{1}{u} \left(h u^2\right)^2 &= -\mu u^2
\end{align}</math>
{{NumBlk|:|<math> \frac{\delta^2 u}{\delta \theta^2} + u = \frac{\mu}{h^2}</math>|4}}

So for the gravitational force – or, more generally, for ''any'' inverse square force law – the right hand side of the equation becomes a constant and the equation is seen to be the [[harmonic oscillator|harmonic equation]] (up to a shift of origin of the dependent variable). The solution is:

:<math> u(\theta) = \frac\mu {h^2} + A \cos(\theta - \theta_0) </math>

where ''A'' and ''θ''<sub>0</sub> are arbitrary constants. This resulting equation of the orbit of the object is that of an [[ellipse#Polar form relative to focus|ellipse]] in Polar form relative to one of the focal points. This is put into a more standard form by letting <math> e \equiv h^2 A/\mu </math> be the [[eccentricity (orbit)|eccentricity]], which when rearranged we see:
:<math> u(\theta) = \frac\mu {h^2} (1+ e\cos(\theta - \theta_0))</math>

Note that by letting <math> a \equiv h^2/\mu\left(1 - e^2\right) </math> be the semi-major axis and letting <math> \theta_0 \equiv 0 </math> so the long axis of the ellipse is along the positive ''x'' coordinate we yield:
:<math>r(\theta) = \frac{a \left(1 - e^2\right)}{1 + e\cos\theta}</math>

When the two-body system is under the influence of torque, the angular momentum ''h'' is not a constant. After the following calculation:
:<math>\begin{align}
                   \frac{\delta r}{\delta \theta} &= -\frac{1}{u^2} \frac{\delta u}{\delta \theta} = -\frac{h}{m} \frac{\delta u}{\delta \theta} \\
               \frac{\delta^2 r}{\delta \theta^2} &= -\frac{h^2u^2}{m^2} \frac{\delta^2 u}{\delta \theta^2} - \frac{hu^2}{m^2} \frac{\delta h}{\delta \theta} \frac{\delta u}{\delta \theta} \\
  \left(\frac{\delta \theta}{\delta t}\right)^2 r &= \frac{h^2 u^3}{m^2}
\end{align}</math>

we will get the Sturm-Liouville equation of two-body system.<ref>{{cite journal |last1=Luo |first1=Siwei |title=The Sturm-Liouville problem of two-body system |journal=Journal of Physics Communications |date=22 June 2020 |volume=4 |issue=6 |page=061001 |doi=10.1088/2399-6528/ab9c30|bibcode=2020JPhCo...4f1001L |doi-access=free }}</ref> 
{{NumBlk|:|<math> \frac{\delta}{\delta \theta} \left(h \frac{\delta u}{\delta \theta}\right) + hu = \frac{\mu}{h}</math>|5}}