# Planar reentry equations

The planar reentry equations are the equations of motion governing the unpowered reentry of a spacecraft, based on the assumptions of planar motion and constant mass, in an Earth-fixed reference frame.
${\begin{cases}{\frac {dV}{dt}}&=-{\frac {\rho V^{2}}{2\beta }}+g\sin \gamma \\{\frac {d\gamma }{dt}}&=-{\frac {V\cos \gamma }{r}}-{\frac {\rho V}{2\beta }}\left({\frac {L}{D}}\right)\cos \sigma +{\frac {g\cos \gamma }{V}}\\{\frac {dh}{dt}}&=-V\sin \gamma \end{cases}}$
where the quantities in these equations are:
* $V$ is the velocity
* $\gamma >0$ is the flight path angle
* $h$ is the altitude
* $\rho$ is the atmospheric density
* $\beta$ is the ballistic coefficient
* $g$ is the gravitational acceleration
* $r=r_{e}+h$ is the radius from the center of a planet with equatorial radius $r_{e}$
* $L/D$ is the lift-to-drag ratio
* $\sigma$ is the bank angle of the spacecraft.

## Simplifications
### Allen-Eggers solution

Harry Allen and Alfred Eggers, based on their studies of ICBM trajectories, were able to derive an analytical expression for the velocity as a function of altitude. They made several assumptions:
* The spacecraft's entry was purely ballistic $(L=0)$.
* The effect of gravity is small compared to drag, and can be ignored.
* The flight path angle and ballistic coefficient are constant.
* An exponential atmosphere, where $\rho (h)=\rho _{0}\exp(-h/H)$, with $\rho _{0}$ being the density at the planet's surface and $H$ being the scale height.
These assumptions are valid for hypersonic speeds, where the Mach number is greater than 5. Then the planar reentry equations for the spacecraft are:
    ${\begin{cases}{\frac {dV}{dt}}&=-{\frac {\rho _{0}}{2\beta }}V^{2}e^{-h/H}\\{\frac {dh}{dt}}&=-V\sin \gamma \end{cases}}\implies {\frac {dV}{dh}}={\frac {\rho _{0}}{2\beta \sin \gamma }}Ve^{-h/H}$
Rearranging terms and integrating from the atmospheric interface conditions at the start of reentry $(V_{\text{atm}},h_{\text{atm}})$ leads to the expression:
    ${\frac {dV}{V}}={\frac {\rho _{0}}{2\beta \sin \gamma }}e^{-h/H}dh\implies \log \left({\frac {V}{V_{\text{atm}}}}\right)=-{\frac {\rho _{0}H}{2\beta \sin \gamma }}\left(e^{-h/H}-e^{-h_{\text{atm}}/H}\right)$
The term $\exp(-h_{\text{atm}}/H)$ is small and may be neglected, leading to the velocity:
    $V(h)=V_{\text{atm}}\exp \left(-{\frac {\rho _{0}H}{2\beta \sin \gamma }}e^{-h/H}\right)$
Allen and Eggers were also able to calculate the deceleration along the trajectory, in terms of the number of g's experienced $n=g_{0}^{-1}(dV/dt)$, where $g_{0}$ is the gravitational acceleration at the planet's surface. The altitude and velocity at maximum deceleration are:
    $h_{n_{\max }}=H\log \left(-{\frac {\rho _{0}H}{\beta \sin \gamma }}\right),\quad V_{n_{\max }}=V_{\text{atm}}e^{-1/2}\implies n_{\max }=-{\frac {V_{\text{atm}}^{2}\sin \gamma }{2g_{0}eH}}$
It is also possible to compute the maximum stagnation point convective heating with the Allen-Eggers solution and a heat transfer correlation; the Sutton-Graves correlation is commonly chosen. The heat rate ${\dot {q}}''$ at the stagnation point, with units of Watts per square meter, is assumed to have the form:
    ${\dot {q}}''=k\left({\frac {\rho }{r_{n}}}\right)^{1/2}V^{3}\sim {\text{W}}/{\text{m}}^{2}$
where $r_{n}$ is the effective nose radius. The constant $k=1.74153\times 10^{-4}$ for Earth. Then the altitude and value of peak convective heating may be found:
    $h_{{\dot {q}}_{\max }''}=H\log \left(-{\frac {\beta \sin \gamma }{3H\rho _{0}}}\right)\implies {\dot {q}}_{\max }''=k{\sqrt {-{\frac {\beta \sin \gamma }{3Hr_{n}e}}}}V_{\text{atm}}^{3}$

### Equilibrium glide condition

Another commonly encountered simplification is a lifting entry with a shallow, slowly-varying, flight path angle. The velocity as a function of altitude can be derived from two assumptions:
* The flight path angle is shallow, meaning that: $\cos \gamma \approx 1,\sin \gamma \approx \gamma$.
* The flight path angle changes very slowly, such that $d\gamma /dt\approx 0$.
From these two assumptions, we may infer from the second equation of motion that:
$\left[{\frac {1}{r}}+{\frac {\rho }{2\beta }}\left({\frac {L}{D}}\right)\cos \sigma \right]V^{2}=g\implies V(h)={\sqrt {\frac {gr}{1+{\frac {\rho r}{2\beta }}\left({\frac {L}{D}}\right)\cos \sigma }}}$