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Sandbox Physics

M075 · Relative orbit / docking corridor

Orbital Rendezvous in the Hill Frame

Drag the chaser in the target’s rotating frame and solve a two-impulse rendezvous. Compare a no-burn coast, the Hill plan and full two-body propagation; test burn trims, a docking corridor, approach-speed limits and singular flight times.

Interactive modelOrbital Rendezvous in the Hill Frame
ElapsedPending\text{Pending}
Hill separationPending\text{Pending}
Endpoint solvePending\text{Pending}
Total planned impulsePending\text{Pending}
Endpoint condition numberPending\text{Pending}
Hill position missPending\text{Pending}
Full two-body missPending\text{Pending}
Full residual speedPending\text{Pending}
Acquired model differencePending\text{Pending}
Full separation / radiusPending\text{Pending}
Acquired corridor checkPending\text{Pending}
Hill incoming speedPending\text{Pending}
Relative Hill invariant defectPending\text{Pending}
Planned flight timePending\text{Pending}

Physics tutorial

A rotating frame makes orbital rendezvous local

BackgroundA nearby chaser and a circular target both fall under gravity. In the target’s rotating frame, radial gravity differences and Coriolis terms couple the two relative coordinates. A forward impulse can initially move forward while changing the orbit so that the long-term along-track drift reverses.

Why it mattersMIT’s astrodynamics notes derive the Clohessy–Wiltshire linear equations. This Lab uses radial-outward and along-track-forward coordinates, solves the planar endpoint map, and independently propagates the same release state in full inverse-square gravity.

Start with the essentials

Focus question
Can a two-impulse plan reach the target while respecting an approach corridor?
One-sentence intuition
The endpoint solve fixes position; a separate arrival impulse matches velocity. A singular endpoint map, nonlinear model error or excessive incoming speed can still invalidate the design.

Core mathematical model

Coupled local equations

n=μrc3,x¨−2ny˙−3n2x=0,y¨+2nx˙=0n=\sqrt{\frac{\mu}{r_c^3}},\qquad \ddot x-2n\dot y-3n^2x=0,\quad\ddot y+2n\dot x=0

The target must be circular. Coriolis coupling explains why a straight-line chase intuition fails. Cross-track motion is omitted.

Endpoint design

rf=A(tf)r0+B(tf)v0+,v0+=−B(tf)−1A(tf)r0\mathbf r_f=A(t_f)\mathbf r_0+B(t_f)\mathbf v_0^+,\quad\mathbf v_0^+=-B(t_f)^{-1}A(t_f)\mathbf r_0

The departing relative velocity solves the position boundary condition. Initial velocity affects the departure cost, not the solved velocity itself.

Velocity-to-position map

nB(t)=(sin⁡nt2(1−cos⁡nt)−2(1−cos⁡nt)4sin⁡nt−3nt)nB(t)=\begin{pmatrix}\sin nt&2(1-\cos nt)\\-2(1-\cos nt)&4\sin nt-3nt\end{pmatrix}

A full target period makes this matrix rank deficient. The implementation normalises it by mean motion and rejects condition numbers above one thousand.

Position and velocity closure

Δv1=v0+−v0−,Δv2=−vf−,Δv=∥Δv1∥+∥Δv2∥\Delta\mathbf v_1=\mathbf v_0^+-\mathbf v_0^-,\quad\Delta\mathbf v_2=-\mathbf v_f^-,\qquad\Delta v=\lVert\Delta\mathbf v_1\rVert+\lVert\Delta\mathbf v_2\rVert

The second burn uses actual trimmed Hill arrival velocity. The same vector applied to the nonlinear arrival leaves a measurable residual speed.

Coast invariant

C=x˙2+y˙2−3n2x2,dCdt=0\mathcal C=\dot x^2+\dot y^2-3n^2x^2,\qquad \frac{d\mathcal C}{dt}=0

This quadratic integral belongs to the unforced Hill model. It is not the inertial spacecraft energy.

Full two-body cross-check

R=Q(nt)[(rc0)+r],V=Q(nt)[r˙+nJ((rc0)+r)],J=(0−110)\mathbf R=Q(nt)\left[\begin{pmatrix}r_c\\0\end{pmatrix}+\mathbf r\right],\quad\mathbf V=Q(nt)\left[\dot{\mathbf r}+nJ\left(\begin{pmatrix}r_c\\0\end{pmatrix}+\mathbf r\right)\right],\quad J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}

Rotation of the velocity includes the frame-rotation term. Omitting it gives the wrong inertial initial state. Units are converted before propagation.

Common difficulties

Meeting is not docking

Typical misconceptionZero position error implies zero relative speed.

Better mental modelArrival velocity is generally nonzero until the braking impulse. Check incoming speed separately.

Approximation is not exact dynamics

Typical misconceptionAn exact Hill endpoint is an exact two-body rendezvous.

Better mental modelThe purple trajectory starts from the same physical release but retains nonlinear gravity; position and residual-velocity errors grow with separation.

One orbit is not always convenient

Typical misconceptionAny selected flight time admits a unique finite burn.

Better mental modelThe endpoint map loses rank at certain times. A rejected plan displays the unforced coast instead.

Samples are not a safety proof

Typical misconceptionA green sample check guarantees continuous collision avoidance.

Better mental modelThe cone has a one-metre port allowance and checks nine hundred intervals. Unsampled violations and omitted operational physics remain outside this diagnostic.

Run the experiment

  1. 01

    Close both boundaries

    Acquire the nominal approach and compare Hill miss, full miss and residual speed.

    What to observe: Hill position closes while the nonlinear residual remains small but nonzero.
  2. 02

    Compare coasts

    Choose the forward-impulse preset and inspect the gray no-burn trajectory.

    What to observe: A forward kick produces coupled radial motion and changed along-track drift; the departure correction pays to restore the planned release.
  3. 03

    Break the plan

    Select a whole-period flight or add a small along-track trim.

    What to observe: The singular plan is rejected; a trim produces a finite position miss even after braking.
  4. 04

    Check the approach

    Tighten the speed limit, then increase release separation.

    What to observe: A zero endpoint miss does not prevent excessive approach speed or growth of the linearisation error.