Kepler orbit

The gravitational acceleration caused by a homogenous sphere

For symmetry reasons it is clear that the gravitational force caused by a homogonous sphere is directed towards the centre of the sphere. By making a mathematical integration over all mass elements it can further be shown that this force only depends on the total mass of the sphere and on the distance to the centre but that the radius of the sphere does not matter. This implies that the gravitational force is just the same as if all mass was concentrated to the very centre of the sphere. The acceleration of a body caused by the gravitational force is therefore

${\ddot {\bar {r}}}=-\mu \cdot {\frac {\hat {r}}{r^{2}}}\ \ (1)$

where

${\bar {r}}$ is the radius vector from the centre of the homogenous sphere to the body

$\mu$ is the product of the general gravitational constant and the mass of the homogenous sphere

${\hat {r}}$ is the unit vector directed to the centre of the homogenous sphere

$r$ is the distance to the centre of the homogenous sphere

Mathematical solution of the differential equation

Like for the movement under any central force, i.e. a force aligned with ${\hat {r}}$, the angular momentum ${\bar {H}}={\bar {r}}\times {\dot {\bar {r}}}$ stays constant:

${\dot {\bar {H}}}={\dot {\overbrace {{\bar {r}}\times {\dot {\bar {r}}}} }}={\dot {\bar {r}}}\times {\dot {\bar {r}}}+{\bar {r}}\times {\ddot {\bar {r}}}={\bar {0}}+{\bar {0}}={\bar {0}}$

Introducing a coordinate system ${\hat {x}}\ ,\ {\hat {y}}$ in the plane orthogonal to ${\bar {H}}$ and polar coordinates

${\bar {r}}=r\cdot (\cos {\theta }\cdot {\hat {x}}+\sin {\theta }\cdot {\hat {y}})$ the differential equation (1) takes the form

$r^{2}\cdot {\dot {\theta }}=H\ \ (2)$

${\ddot {r}}-r\cdot {\dot {\theta }}^{2}=-{\frac {\mu }{r^{2}}}\ \ (3)$

Taking the time derivative of (2) one gets

${\ddot {\theta }}=-{\frac {2\cdot H\cdot {\dot {r}}}{r^{3}}}\ \ (4)$

Using the chain rule for differentiation one gets

${\dot {r}}={\frac {dr}{d\theta }}\cdot {\dot {\theta }}\ \ (5)$

${\ddot {r}}={\frac {d^{2}r}{d\theta ^{2}}}\cdot {\dot {\theta }}^{2}+{\frac {dr}{d\theta }}\cdot {\ddot {\theta }}\ \ (6)$

Using the expressions for ${\dot {\theta }},\ {\ddot {\theta }},\ {\dot {r}},\ {\ddot {r}}$ of equations (2), (4), (5) and (6) all time derivatives in (3) can be replaced by derivatives of $r$ as function of $\theta$. After some simplification one gets

${\ddot {r}}-r\cdot {\dot {\theta }}^{2}={\frac {H^{2}}{r^{4}}}\cdot \left({\frac {d^{2}r}{d\theta ^{2}}}-2\cdot {\frac {\left({\frac {dr}{d\theta }}\right)^{2}}{r}}-r\right)=-{\frac {\mu }{r^{2}}}\ \ (7)$

The differential equation (7) can be solved analytically by the variable substitution

$r={\frac {1}{s}}\ \ (8)$

Using the chain rule for differentiation one gets:

${\frac {dr}{d\theta }}=-{\frac {1}{s^{2}}}\cdot {\frac {ds}{d\theta }}\ \ (9)$

${\frac {d^{2}r}{d\theta ^{2}}}={\frac {2}{s^{3}}}\cdot \left({\frac {ds}{d\theta }}\right)^{2}-{\frac {1}{s^{2}}}\cdot {\frac {d^{2}s}{d\theta ^{2}}}\ \ (10)$

Using the expressions (10) and (9) for ${\frac {d^{2}r}{d\theta ^{2}}}$ and ${\frac {dr}{d\theta }}$ one gets

$H^{2}\cdot \left({\frac {d^{2}s}{d\theta ^{2}}}+s\right)=\mu \ \ (11)$

with the general solution

$s={\frac {\mu }{H^{2}}}\cdot \left(1+e\cdot \cos {(\theta }-\theta _{0})\right)\ \ (12)$

where $e$ and $\theta _{0}$ are constants of integration depending on the initial values for $s$ and ${\frac {ds}{d\theta }}$. Instead of using the integration parameter $\theta _{0}$ explicitly one introduces the convention that the unit vectors ${\hat {x}}\ ,\ {\hat {y}}$ defining the coordinate system in the orbital plane are selected such that $\theta _{0}$ takes the value zero and $e$ is positive. This then means that $\theta$ is zero at the point where $s$ is maximal and therefore $r={\frac {1}{s}}$ is minimal. Defining the parameter p as ${\frac {H^{2}}{\mu }}$ one has that

$r={\frac {1}{s}}={\frac {p}{1+e\cdot \cos {\theta }}}\ \ (13)$

This is the equation in polar coordinates for a conic section with origin in a focal point. The argument $\theta$

is called "true anomaly". For $e=0$ this is a circle with radius For $0<e<1$ this is an ellipse with

$a={\frac {p}{1-e^{2}}}\ \ (14)$

$b={\frac {p}{\sqrt {1-e^{2}}}}=a\cdot {\sqrt {1-e^{2}}}\ \ (15)$

For $e=1$ this is a parabola with focal length $p$

For $e>1$ this is a hyperbola with

$a={\frac {p}{e^{2}-1}}\ \ (16)$

$b={\frac {p}{\sqrt {e^{2}-1}}}=a\cdot {\sqrt {e^{2}-1}}\ \ (17)$

Using the chain rule for differentiation (5), the equation (2) and the definition of $p$

as ${\frac {H^{2}}{\mu }}$ one gets that the radial velocity component is

$V_{r}={\dot {r}}={\frac {H}{p}}\cdot e\cdot \sin {\theta }={\sqrt {\frac {\mu }{p}}}\cdot e\cdot \sin {\theta }\ \ (18)$

The tangential component is

$V_{t}=r\cdot {\dot {\theta }}={\frac {H}{r}}={\sqrt {\frac {\mu }{p}}}\cdot (1+e\cdot \cos {\theta })\ \ (19)$

The determination of the connection between the polar argument $\theta$ and time $t$ is slightly different for ellipses and hyperbolas. For an elliptic orbit one switches to the "eccentric anomaly" $E$ for which

$x=a\cdot (\cos {E}-e)\ \ (20)$

$y=b\cdot \sin {E}\ \ (21)$

and consequently

${\dot {x}}=-a\cdot \sin {E}\cdot {\dot {E}}\ \ (22)$

${\dot {y}}=b\cdot \cos {E}\cdot {\dot {E}}\ \ (23)$

and the angular momentum $H$ is

$H=x\cdot {\dot {y}}-y\cdot {\dot {x}}=a\cdot b\cdot (1-e\cdot \cos {E})\cdot {\dot {E}}\ \ (24)$

Integrating with respect to time $t$ one gets

$H\cdot t=a\cdot b\cdot (E-e\cdot \sin {E})\ \ (25)$

under the assumption that time $t=0$ is selected such that the integration constant is zero. As by definition of $p$ one has

$H={\sqrt {\mu \cdot p}}\ \ (26)$

this can be written

$t=a\cdot {\sqrt {\frac {a}{\mu }}}(E-e\cdot \sin {E})\ \ (27)$

For an hyperbolic orbit one uses the parameterisation

$x=a\cdot (e-\cosh {E})\ \ (28)$

$y=b\cdot \sinh {E}\ \ (29)$

for which one has

${\dot {x}}=-a\cdot \sinh {E}\cdot {\dot {E}}\ \ (30)$

${\dot {y}}=b\cdot \cosh {E}\cdot {\dot {E}}\ \ (31)$

and the angular momentum $H$ is

$H=x\cdot {\dot {y}}-y\cdot {\dot {x}}=a\cdot b\cdot (e\cdot \cosh {E}-1)\cdot {\dot {E}}\ \ (32)$

Integrating with respect to time $t$ one gets

$H\cdot t=a\cdot b\cdot (e\cdot \sinh {E}-E)\ \ (33)$

i.e.

$t=a\cdot {\sqrt {\frac {a}{\mu }}}(e\cdot \sinh {E}-E)\ \ (34)$

To find what time t that corresponds to a certain true anomaly $\theta$ one computes corresponding parameter $E$ connected to time with relation (27) for an elliptic and with relation (34) for an hyperbolic orbit.

Some additional formulas

For an elliptic orbit one gets from (20) and (21) that

$r=a\cdot (1-e\cdot \cos {E})\ \ (35)$

and therefore that

$\cos \theta ={\frac {x}{r}}={\frac {\cos {E}-e}{1-e\cdot \cos {E}}}\ \ (36)$

With some trigonometric calculations one can from this derive that

$\tan {\frac {\theta }{2}}={\sqrt {\frac {1+e}{1-e}}}\cdot \tan {\frac {E}{2}}\ \ (37)$

This relation is convenient for passing between "true anomaly" and "eccentric anomaly", the latter being connected to time through relation (27).

From relation (27) follows that the orbital period $P$ is

$P=2\pi \cdot a\cdot {\sqrt {\frac {a}{\mu }}}\ \ (38)$

As the potential energy corresponding to the force field of relation (1) is

$-{\frac {\mu }{r}}$

it follows from (13) , (14), (18) and (19) that the sum of the kinetic and the potential energy for an elliptic orbit is

$-{\frac {\mu }{2\cdot a}}\ \ (39)$

From (13) , (16), (18) and (19) follows that the sum of the kinetic and the potential energy for a hyperbolic orbit is

${\frac {\mu }{2\cdot a}}\ \ (40)$

Orbital Elements

A Kepler orbit is specified by 6 orbital element that normally are

$a$ - semi-major axis

$e$ - eccentricity

$i$ - inclination

$\Omega$ - right ascension of ascending node

$\omega$ - argument of perigee

$\theta$ - the true anomaly that corresponds to a specified time

The angles $\Omega ,i,\omega$ are the euler angles ($\alpha ,\beta ,\gamma$ with the notations of that article) characterising the orientation of the coordinate system ${\hat {x}},{\hat {y}},{\hat {z}}$ with ${\hat {z}}$ in the direction of the angular momentum ${\bar {H}}$ and with ${\hat {x}}$ directed from the centre of the sphere to the pericentre of the orbit.