Prev
Unit 9 · Lesson 9-4
Rendezvous and Intercept Targeting
55m100 XP
Next

Rendezvous and Intercept Targeting

Δv as a function of departure time and flight time, safety constraints, phasing, and correcting Lambert for real dynamics

Gemini 4, June 1965
Jim McDivitt has the spent upper stage of his own Titan rocket a few hundred metres ahead, and the flight plan says to fly formation with it. He does what every pilot's instinct says: points the nose at the target and thrusts towards it. The target drifts away — upward and ahead. He thrusts again, harder. It recedes further. After using a large share of his manoeuvring propellant the attempt is abandoned. Nothing was broken. Thrusting forward raised his orbit, and a higher orbit is slower, so he fell behind. Two years earlier an Air Force pilot named Buzz Aldrin had written an MIT doctoral thesis on exactly this — line-of-sight rendezvous guidance — and six months after McDivitt's attempt Gemini 6A and Gemini 7 met in orbit using the method. Orbital rendezvous is not flying towards something. It is solving Lambert's problem.
Watch first · presented by Shoaib Aziz
The numbers in this film are the labs' own: for a chaser at 300 km and a target at 600 km, 30° ahead, the Hohmann transfer costs 167.9 m/s and takes 46.8 min, so the target must lead by 5.8° at departure — reached after a 95.5-minute wait, then every 23.7 h — and Lambert at that wait and flight time returns the same 167.9 m/s; intercepts from 10 to 120 min reject 12 of 23 flight times for diving into the Earth, leaving 452 m/s at 70 min as the cheapest safe one; the rendezvous map's grid floor is 181.9 m/s at a 90-minute wait and a 50-minute flight; and flying the Lambert answer with J2 misses by 64.5 km after 70 min, which Newton shooting cuts to 229 m and then 3 cm with a 4.6 m/s velocity change.
Intercept versus rendezvous. An intercept only has to arrive at the target's position: one burn, Δv1=∣v⃗1−v⃗chaser∣\Delta v_1 = |\vec v_1 - \vec v_{\text{chaser}}|. A rendezvous must also match velocity: a second burn Δv2=∣v⃗target−v⃗2∣\Delta v_2 = |\vec v_{\text{target}} - \vec v_2| at arrival. Two free choices drive the cost:
  • the wait time twt_w before the first burn (it sets where both spacecraft are when the transfer starts), and
  • the flight time Δt\Delta t (it sets where the target will be at arrival).
For each pair, propagate both spacecraft, solve Lambert from the chaser's position at twt_w to the target's position at tw+Δtt_w + \Delta t, and price the burns. The result is a two-dimensional cost map.
Loading interactive…
💡Reading the map
Valleys run diagonally: leaving later and flying shorter can reach the same arrival geometry. They repeat with the synodic period of the two orbits, Tsyn=2π/∣nc−nt∣T_{syn} = 2\pi/|n_c - n_t|, because that is how long the relative geometry takes to come back. Cliffs appear where the transfer angle passes 180° (the plane flips and the cost spikes) and where short flights would need to cut through the planet. The floor of the best valley sits at the Hohmann transfer when the orbits are circular and coplanar: wait until the target leads by 180°−nttH180° - n_t t_H, then fly half the transfer ellipse. Lambert does not know about Hohmann — it simply finds it.
⚠Valid is not the same as flyable
A fast transfer between two nearby points is nearly a straight line, and a straight chord between two points in low orbit passes through the Earth. Every Lambert answer must be checked for perigee on the arc: perigee is passed between departure and arrival exactly when r⃗1⋅v⃗1<0\vec r_1\cdot\vec v_1 < 0 and r⃗2⋅v⃗2>0\vec r_2 \cdot \vec v_2 > 0. If so, require rp>R⊕+hmin⁡r_p > R_\oplus + h_{\min}. In the first lab, 12 of 23 intercept times between 300 km and 600 km orbits fail this test — all the fastest ones.
📐Worked Example: 300 km to 600 km
Chaser at 300 km, target at 600 km, coplanar circular orbits, target 30° ahead. Hohmann. aH=6828a_H = 6828 km, Δv=167.9\Delta v = 167.9 m/s, flight time 46.8 min. The target moves nttHn_t t_H during the transfer, so it must lead by 180°−nttH=5.8°180° - n_t t_H = 5.8° at departure. It starts 30° ahead and the chaser gains on it at nc−ntn_c - n_t, so the right moment comes after 95.5 min. The synodic period is 23.7 h: miss it and you wait a day. Lambert map. A 15 min × 5 min grid finds 181.9 m/s at (wait 90 min, fly 50 min) — a grid point near the floor. Solving Lambert at exactly the Hohmann timing (0.999 of the transfer time, to stay off 180°) returns 167.9 m/s. Intercept only. The cheapest safe intercept from the starting geometry is 452 m/s with a 70-minute flight; every faster option dives into the atmosphere.
Lambert is a first guess, not an answer. Lambert assumes a point-mass Earth. Fly its velocity with J2J_2 included and the arrival misses — by 64 km after 70 minutes in the third lab. Operational targeting therefore wraps Lambert in a differential corrector: propagate in the full force model, measure the miss δr⃗\delta\vec r, estimate the sensitivity matrix Φ=∂r⃗(tf)/∂v⃗0\Phi = \partial\vec r(t_f)/\partial\vec v_0 (by finite differences or the state transition matrix), and update v⃗0←v⃗0−Φ−1δr⃗\vec v_0 \leftarrow \vec v_0 - \Phi^{-1}\delta\vec r.
v⃗0(k+1)=v⃗0(k)−Φ−1(r⃗(tf;v⃗0(k))−r⃗target)\vec v_0^{(k+1)} = \vec v_0^{(k)} - \Phi^{-1}\big(\vec r(t_f;\vec v_0^{(k)}) - \vec r_{\text{target}}\big)

Newton shooting. Because Lambert's answer is already close, convergence is quadratic: 64 km → 229 m → 3 cm in two corrections, for a velocity change of 4.6 m/s.

🛰How a crew vehicle reaches the ISS
A crew or cargo vehicle is launched into a lower orbit so that it gains on the station from behind. Over the following hours it performs a sequence of phasing and height-adjust burns, each targeted with a Lambert-type solution to an aim point behind or below the station, re-solved after every burn with fresh tracking. For the last few kilometres the dynamics are linearised about the station's orbit — the Clohessy–Wiltshire equations — and the approach is flown along a corridor. Russian vehicles that once took two days now regularly reach the station about three hours after launch, by fitting the whole phasing sequence into two orbits.
💡McDivitt explained
Thrusting toward a target ahead raises your orbit; the higher orbit has a longer period, so you fall behind. To catch something ahead, drop to a lower, faster orbit and let geometry close the gap — or, equivalently, solve Lambert for the arrival point and fly the velocity it gives you, which will usually point somewhere a pilot's instinct would not.
↓But there is a problem
Two free dates, one Lambert solve per pair, a cost at every point. In Earth orbit the map repeats every synodic period of a few hours or a day. Now change the attracting body to the Sun and the two orbits to Earth's and Mars's. The synodic period becomes 26 months. The map is the same kind of object — but it is the most important chart in planetary exploration, and missions live or die by which part of it their launch date falls in.