12. Single-stage orbital insertion in polar coordinates

Reaching orbit requires more than altitude: a vehicle also needs enough horizontal speed to keep falling around Earth. How should thrust turn from upward to horizontal to deliver the greatest mass to orbit?

Item	Definition
States	Radius $R$, angle $\theta$, radial and tangential velocities $V_r,V_t$, mass $m$
Controls	Radial and tangential thrust $T_r,T_t$
Objective	Maximize final mass
Phases	One phase, 30 nodes, free final time
New concepts	Polar dynamics, two thrust components, multiple path constraints

Situation

A 100 t rocket starts at rest and must enter a 400 km circular orbit. At arrival it needs the target altitude, zero radial velocity, and circular speed about 7.67 km/s. The dry vehicle is 1 t and propellant accounts for 99% of initial mass. Maximizing final mass measures how much propellant can remain. Earth rotation is neglected. Thrust and the acceleration felt by the vehicle are limited.

Formulation

Coordinates

• $R$ is distance from Earth’s center; altitude is $h=R-R_e$.
• $\theta$ is angular distance from the launch site; downrange distance is $R_e\theta$.
• $V_r$ is outward radial (vertical) velocity; $V_t$ is tangential (local horizontal) velocity.
• $T_r,T_t$ are thrust components. Magnitude is $|T|=\sqrt{T_r^2+T_t^2}$ and elevation is $\arctan(T_r/T_t)$ in this nonnegative quadrant.

Components avoid introducing a steering angle that becomes undefined at zero thrust and adds trigonometric nonlinearities.

Dynamics

$$\begin{aligned} \dot R&=V_r, &\dot\theta&=V_t/R,\\ \dot V_r&=(T_r-D_r)/m-\mu/R^2+V_t^2/R,\\ \dot V_t&=(T_t-D_t)/m-V_rV_t/R,\\ \dot m&=-\sqrt{T_r^2+T_t^2}/(g_0I_{sp}). \end{aligned}$$

The centrifugal term $V_t^2/R$ balances gravity on a circular orbit. The term $-V_rV_t/R$ expresses conservation of angular momentum: without tangential force, $RV_t$ stays constant as radius changes.

$$V=\sqrt{V_r^2+V_t^2},\qquad D_r=\tfrac12\rho(h)V_rVC_DA,\qquad D_t=\tfrac12\rho(h)V_tVC_DA,$$ $$\rho(h)=1.225e^{-h/8500},\qquad g(R)=\mu/R^2.$$

Drag opposes velocity. The same gravitational parameter $\mu$ is used for the dynamics and target circular speed.

Boundary conditions

$$\begin{aligned} R(0)&=R_e,&\theta(0)&=0,&V_r(0)&=V_t(0)=0,&m(0)&=100\ \mathrm t,\\ R(t_f)&=R_e+400\ \mathrm{km},&V_r(t_f)&=0,&V_t(t_f)&=\sqrt{\mu/R(t_f)}. \end{aligned}$$

Target speed is about 7672.6 m/s. Final angle and time are free.

Constraints

$$\begin{aligned} R&\ge R_e,\\ 0&\le T_r,T_t\le T_{\max},\\ \sqrt{T_r^2+T_t^2}&\le T_{\max},\\ 1\ \mathrm t&\le m\le100\ \mathrm t,\\ \left|\frac{\boldsymbol T-\boldsymbol D}{m}\right|&\le8g_0. \end{aligned}$$

Thrust cannot point downward or backward. Its magnitude is limited to 1.5 times initial weight, about 1471 kN. The 8 G limit is non-gravitational acceleration, as measured by an accelerometer. All constraints apply at nodes.

Objective and parameters

$$J=-m(t_f)/m(0).$$

Parameter	Value
$R_e$	6371 km
$g_0$	9.80665 m/s²
$\mu$	$3.986\times10^{14}$ m³/s²
Initial / dry mass	100 t / 1 t
$C_D,A$	0.6, 4 m²
$I_{sp}$	300 s
Thrust-to-weight ratio	1.5
Acceleration limit	8 G

Scale radius by $R_e$, speed by $\sqrt{\mu/R_e}$, and mass by initial mass (scaling).

Initial guess

Use time 0 to 200 s. Radius rises cubically to the target, angle rises cubically to 25°, radial velocity is zero, and tangential velocity grows linearly to target speed. Mass decreases cubically from 100 t to 1 t. Both thrust components decrease cubically from $T_{\max}/2$ to zero.

Results

Final altitude is 400.0 km (target 400.0 km), and speed is 7,672.6 m/s (target 7,672.6 m/s).

Altitude versus downrange distance, with altitude exaggerated. Dots are nodes; orange arrows show thrust direction and relative magnitude at selected powered nodes.

Altitude, tangential and radial velocity, mass, thrust magnitude, thrust elevation, and non-gravitational acceleration. Elevation is hidden when the engine is off; red lines mark thrust and acceleration limits.

Ascent beside velocity, thrust magnitude, and elevation histories. Coasting is played faster.

• The first burn ends around 263 s.
• After coasting upward, a second burn from about 886 to 901 s circularizes the orbit.
• Final mass is 4,370 kg, about 4% of initial mass: 1 t of structure plus remaining propellant.
• Insertion occurs 5,105 km downrange.
• Maximum node acceleration is 8.00 G, below the inactive 8 G limit.

Explanation

Rise, then turn

Thrust starts nearly upward, around 80°, then turns toward horizontal. Early climbing escapes dense air and drag. Turning too late wastes propellant against gravity. The optimum balances these effects, resembling a gravity turn.

Burn, coast, and circularize

The first burn creates a trajectory with apogee near 400 km. Coasting raises altitude without consuming propellant. A final horizontal burn at apogee creates the circular orbit, similar to a Hohmann transfer. Tangential speed decreases during the outward coast because angular momentum is conserved.

Interpreting the plots

Constraints hold at nodes; interpolating curves may exceed limits between them. A single polynomial spans both engine switches, so fine details depend on the node count (multiple phases, checking results).

Try it

• Change the 30 nodes or the 200 s time guess and compare convergence and final mass.
• Lower MaxG from 8 to 4 and inspect throttling when the acceleration constraint becomes active.
• Change Htarget to 300 or 600 km and compare coast duration and final mass.

Run

uv run python examples/12_rocket_ascent_polar_ssto.py

View the Python source on GitHub.