Derive Eq$.437$; state all your assumptions.Elliptical Orbits

The circular orbit described above is a special case of the more general elliptical orbit

shown in Fig. $4-8$; here, the Earth $($or any other heavenly body around which another

body is moving$)$ is located at one of the focal points of this ellipse. The relevant

equations of motion come from Kepler's laws and elliptical orbits may be described

as follows, when expressed in polar coordinates:

$u=[(\frac{2}{R}-\frac{1}{a}){]}^{\frac{1}{2}}$

where $u$ is the velocity of the body in the elliptical orbit, $R$ is the instantaneous radius

from the center of the Earth $($a vector quantity, which changes direction as well as

magnitude$),a$ is the major axis of the ellipse, and $$ is the Earth's gravitational con$-$

stant, $3\mathrm{.}986{10}^{14}\frac{m^3}{se{c}^2}.$ These symbols are defined in Fig. $4-8.$ From Eq$.4-29$ it

can be seen that the velocity has its maximum value $u_p$ when the moving body comes

closest to its focal point at its orbit's perigee and the minimum value $u_a$ at its apogee.

By substituting for $R$ in Eq$.4-29,$ and by defining the ellipse's shape factor $e$ as the

eccentricity of the ellipse, $e=\frac{\sqrt[\phantom{2}]{a^2-b^2}}{a},$ the apogee and perigee velocities can be

expressed as

$u_a=\sqrt[\phantom{2}]{\frac{(1-e)}{a(1+e)}}$

$u_p=\sqrt[\phantom{2}]{\frac{(1+e)}{a(1-e)}}$

Another property of an elliptical orbit is that the product of velocity and instanta$-$

neous radius remains constant for any location $x$ or $y$ on the ellipse, namely, $u_x{R}_x=$

$u_y{R}_y=uR.$ The exact path that a satellite takes depends on the velocity $($magnitude

and vector orientation$)$ with which it is started or was injected into orbit.

FIGURE $4-8.$ Elliptical orbit; the attracting body is at one of the focal points of the ellipse.