M11-T6 — Entry, Descent and Landing

Learning objectives

After this lesson, you should be able to:

• List the phases of an entry, descent and landing sequence and say what each is guided by.
• Estimate the peak deceleration and its altitude from the entry speed, angle, ballistic coefficient and scale height.
• Find a parachute deployment condition from the dynamic pressure and the speed.
• Say why a parachute alone cannot land the vehicle and estimate the powered phase that completes it.

The intuition

Navigation M9-T5: Other Platforms: Legged, Aerial, Marine and Off-Road places spacecraft in the range of platforms a navigation stack serves; this lesson covers the one phase of a spacecraft mission where guidance is time-critical and the vehicle is committed. Hopper is the same final phase on a small scale: the powered descent of M9-T4: Powered-Descent Guidance by Convexification is the end of every propulsive landing.

Five phases

Entry begins at the atmospheric interface (125 km here) at 4.5 km/s and 17° below the horizontal. The vehicle is a blunt body with a heat shield, and drag removes most of the kinetic energy: the speed falls from 4500 to 330 m/s in 146 s. With no lift (a ballistic entry) the flight path is fixed by the entry angle and the ballistic coefficient $\beta = m/(C_DA)$, 56 kg/m² here; with lift it can be steered, which is how larger vehicles correct dispersions. Descent on a parachute starts when the dynamic pressure has fallen to a safe value (500 Pa here, at Mach 1.4 and 8.8 km). The parachute slows the vehicle to its terminal speed, $\sqrt{2mg/(\rho C_DA)}$, 47 m/s at 1.5 km. Landing is propulsive: the heat shield and parachute are released and engines remove the 47 m/s and carry the weight for about 9.5 s. Navigation in these phases changes: inertial only during entry (the plasma blacks out radio and the atmosphere is too thick to see through), a radar altimeter and velocimeter on the parachute, and terrain-relative navigation and radar in the powered descent.

The peak deceleration, almost without a vehicle

For a ballistic entry in an exponential atmosphere $\rho = \rho_0e^{-h/H}$, an approximate solution (Allen and Eggers) gives a peak deceleration $a_{\max} = V_E^2\sin\gamma_E/(2eH)$, 98 m/s² (10.0 g) for the example against 92.2 m/s² (9.4 g) in the simulation; it does not depend on the vehicle at all. The altitude of the peak depends on the vehicle: $h = H\ln[\rho_0H/(\beta\sin\gamma_E)]$, 29.0 km (simulation 30.0 km), 31.5, 29.0 and 27.0 km for ballistic coefficients 0.8, 1.0 and 1.2 times nominal. A 20% error in $\beta$ moves the deceleration peak by 2 to 2.5 km without changing its value. The entry angle matters for the value: 15°, 17° and 19° give 8.9, 10.0 and 11.1 g.

The parachute is not enough

Under a parachute of 135 m² drag area the mass of 700 kg (after the heat shield is dropped) has a terminal speed of 47 m/s at 1.5 km: reached after 112 s of descent, with 48 m/s at that point. A landing at 47 m/s is not survivable, so the last phase is propulsive. With a net deceleration of 5 m/s² the powered phase lasts 9.5 s, needs a delta-v of $v + gt = 83$ m/s (the second term is the gravity loss, 35 m/s), about 26 kg of propellant at a specific impulse of 220 s, and the vehicle falls 226 m while braking. The guidance that makes this phase safe and cheap is the convex powered descent of module 9, with the constraints of thrust limits and glide slope.

Entry is drag-limited and almost insensitive to the vehicle in peak load; the parachute is deployed on dynamic pressure; and a parachute leaves the vehicle at a terminal speed that only propulsion can remove. Each phase uses a different sensor, and the last one is the powered descent that the course built.

Definitions

Entry interface

The altitude (here 125 km) at which the sensible atmosphere starts and the entry state is defined.

Ballistic coefficient $\beta$

$\beta = m/(C_DA)$ in kg/m²: the mass per unit drag area; a larger value means the vehicle decelerates lower in the atmosphere.

Dynamic pressure $q$

$q = \tfrac12\rho V^2$, the quantity that sets aerodynamic loads and the parachute deployment condition.

Terminal speed

The speed at which drag equals weight: $\sqrt{2mg/(\rho C_DA)}$.

Terrain-relative navigation

Navigation by matching camera or radar data of the surface with a stored map, as in the aiding-sensors lesson of module 7.

Derivation: the peak deceleration of a ballistic entry

Allen and Eggers' approximation neglects gravity and the curvature of the path.

1. $$\dfrac{dV}{dt} = -\dfrac{\rho V^2}{2\beta},\qquad \dfrac{dh}{dt} = -V\sin\gamma_E$$

   The drag equation with a constant flight path angle; the altitude decreases at the vertical component of speed.

2. $$\dfrac{dV}{dh} = \dfrac{\rho V}{2\beta\sin\gamma_E},\quad \rho = \rho_0e^{-h/H}$$

   Divide: speed as a function of altitude.

3. $$V(h) = V_E\exp\left[-\dfrac{\rho_0H}{2\beta\sin\gamma_E}\,e^{-h/H}\right]$$

   Integrate from infinity (where $V = V_E$): the speed falls double-exponentially with falling altitude.

4. $$a = \dfrac{\rho V^2}{2\beta} = \dfrac{\rho_0e^{-h/H}\,V^2}{2\beta}$$

   The deceleration at altitude h is the drag per unit mass.

5. $$\dfrac{da}{dh} = 0 \;\Rightarrow\; \dfrac{\rho_0H}{\beta\sin\gamma_E}\,e^{-h/H} = 1$$

   Differentiate and set to zero; at the peak the exponent in the speed has the value $\tfrac12$ so that $V = V_Ee^{-1/2}$.

6. $$a_{\max} = \dfrac{V_E^2e^{-1}}{2\beta}\cdot\dfrac{\rho_0e^{-h/H}}{1} = \dfrac{V_E^2\sin\gamma_E}{2eH}$$

   Substitute the altitude of the peak: the ballistic coefficient cancels, leaving the entry speed, angle and scale height.

Parachute and landing

1. $$mg = \tfrac12\rho V_t^2C_DA \;\Rightarrow\; V_t = \sqrt{2mg/(\rho C_DA)}$$

   At the terminal speed drag equals weight. With $m = 700$ kg, $g = 3.71$, $C_DA = 135$ m² and the density at 1.5 km: 47 m/s.

2. $$t_b = v_0/a_{\mathrm{net}},\quad \Delta v = v_0 + g\,t_b$$

   Braking from v₀ at a net deceleration a_net takes t_b; during it gravity adds g t_b to the delta-v: $48 + 3.71\times9.5 = 83$ m/s.

3. $$m_p = m_f\left(e^{\Delta v/(I_{sp}g_0)} - 1\right) = 700\left(e^{83/2157} - 1\right) = 26\ \mathrm{kg}$$

   The rocket equation for the propellant at an Isp of 220 s.

The peak deceleration of a ballistic entry depends on the speed, angle and scale height, not the vehicle; the vehicle sets where it happens. The parachute only reaches a terminal speed, and the rest is propulsive.

A worked example

A Viking-like entry vehicle: 900 kg, $C_D = 1.68$ on 9.62 m² (ballistic coefficient 56 kg/m²), atmosphere $\rho = 0.02\,e^{-h/11.1\,\mathrm{km}}$ kg/m³, entry at 125 km, 4500 m/s and −17°.

1. Entry. The simulated peak deceleration is 92.2 m/s² (9.4 g) at 81 s, at 30.0 km and 2762 m/s; the peak dynamic pressure is 5119 Pa at 30.0 km. Allen-Eggers: 98.1 m/s² (10.0 g) at 29.0 km and 2729 m/s.
2. Sensitivity. A ballistic coefficient 20% below or above nominal moves the altitude of the peak by +2.5 km and −2.0 km without changing its value (98.1 m/s²). Entry angles of 15°, 17° and 19° give 8.9, 10.0 and 11.1 g at 30.4, 29.0 and 27.8 km.
3. Parachute. Deployed when the dynamic pressure falls to 500 Pa: at 8.8 km, 333 m/s (Mach 1.42), 146 s after the interface and 389 km downrange.
4. Descent. Under the 135 m² parachute (700 kg after releasing the heat shield) the vehicle reaches 1.5 km after 112 s at 48 m/s; the terminal speed at 1.5 km is 47 m/s.
5. Powered phase. A net deceleration of 5 m/s² from 48 m/s lasts 9.5 s, needs 83 m/s of delta-v, about 26 kg of propellant at an Isp of 220 s, and the vehicle falls 226 m.
6. Navigation by phase. Entry interface, blackout (inertial only), parachute (radar altimeter and velocimeter), powered descent (terrain-relative navigation, radar, then the convex guidance of module 9).

import math

# a Mars-like entry vehicle (illustrative round numbers)
MU, RM, GM = 4.2828e13, 3389.5e3, 3.71
M, CD, A = 900.0, 1.68, 9.62
BETA = M / (CD * A)                    # ballistic coefficient, kg / m^2
RHO0, HS = 0.020, 11100.0
V0, GAMMA0, H0 = 4500.0, math.radians(17.0), 125e3
print(f'entry vehicle: mass {M:.0f} kg, ballistic coefficient {BETA:.0f} kg/m^2; atmosphere rho = {RHO0} exp(-h / {HS / 1e3:.1f} km) kg/m^3')
print(f'entry interface at {H0 / 1e3:.0f} km, speed {V0:.0f} m/s, flight path angle {-math.degrees(GAMMA0):.0f} deg')

def entry(dt=0.05):
    h, v, gam = H0, V0, -GAMMA0
    r = RM + h; ang = 0.0; t = 0.0
    out = []
    while h > 4000.0 and t < 600.0:
        rho = RHO0 * math.exp(-h / HS)
        D = 0.5 * rho * v * v / BETA
        gloc = MU / (RM + h) ** 2
        vdot = -D - gloc * math.sin(gam)
        gamdot = (-gloc * math.cos(gam) + v * v * math.cos(gam) / (RM + h)) / v
        h += v * math.sin(gam) * dt; v += vdot * dt; gam += gamdot * dt
        ang += v * math.cos(gam) / (RM + h) * dt
        t += dt
        out.append((t, h, v, gam, D, 0.5 * rho * v * v, ang * RM))
    return out
tr = entry()
peak = max(tr, key=lambda p: p[4])
print(f'simulated ballistic entry: peak deceleration {peak[4]:.1f} m/s^2 = {peak[4] / 9.80665:.1f} Earth g at t = {peak[0]:.0f} s, altitude {peak[1] / 1e3:.1f} km, speed {peak[2]:.0f} m/s')
amax = V0 ** 2 * math.sin(GAMMA0) / (2 * math.e * HS)
hpk = HS * math.log(RHO0 * HS / (BETA * math.sin(GAMMA0)))
print(f'Allen-Eggers: peak deceleration V^2 sin(gamma) / (2 e H) = {amax:.1f} m/s^2 = {amax / 9.80665:.1f} Earth g, at altitude H ln(rho0 H / (beta sin gamma)) = {hpk / 1e3:.1f} km; independent of the ballistic coefficient except for the altitude')
vpk = V0 * math.exp(-0.5)
print(f'  speed at the peak: V exp(-1/2) = {vpk:.0f} m/s (simulation {peak[2]:.0f})')
peak_q = max(tr, key=lambda p: p[5])
print(f'peak dynamic pressure {peak_q[5]:.0f} Pa at {peak_q[1] / 1e3:.1f} km')
Q_CHUTE = 500.0
chute = next(p for p in tr if p[1] < 12000.0 and p[5] < Q_CHUTE)
print(f'parachute deployment when the dynamic pressure falls to {Q_CHUTE:.0f} Pa: altitude {chute[1] / 1e3:.1f} km, speed {chute[2]:.0f} m/s, time {chute[0]:.0f} s after the interface, downrange {chute[6] / 1e3:.0f} km')
vs = 235.0
print(f'  (the speed of sound near the ground is about {vs:.0f} m/s: Mach {chute[2] / vs:.2f} at deployment)')
print('what happens to the altitude at which the vehicle slows if the ballistic coefficient is 20% off:')
for scale in (0.8, 1.0, 1.2):
    BETA_s = BETA * scale
    hpk_s = HS * math.log(RHO0 * HS / (BETA_s * math.sin(GAMMA0)))
    print(f'  beta x{scale:.1f}: peak deceleration altitude {hpk_s / 1e3:.1f} km ({(hpk_s - hpk) / 1e3:+.1f} km), peak value unchanged {amax:.1f} m/s^2')
print('what happens if the entry angle is 2 degrees steeper or shallower:')
for dg in (-2.0, 0.0, 2.0):
    g_ = GAMMA0 + math.radians(dg)
    print(f'  gamma {math.degrees(g_):4.1f} deg: peak deceleration {V0 ** 2 * math.sin(g_) / (2 * math.e * HS) / 9.80665:5.1f} Earth g, peak altitude {HS * math.log(RHO0 * HS / (BETA * math.sin(g_))) / 1e3:5.1f} km')

# parachute phase
CDA_CHUTE = 135.0
def chute_phase(dt=0.05):
    h, v = chute[1], chute[2]
    t = 0.0; m = M - 200.0                             # mass after the heat shield is dropped
    log = []
    while h > 1500.0 and t < 600:
        rho = RHO0 * math.exp(-h / HS)
        d = 0.5 * rho * v * v * CDA_CHUTE / m
        v += (GM - d) * dt
        h -= v * dt
        t += dt
        log.append((t, h, v))
    return log
cp = chute_phase()
print(f'under the parachute (drag area {CDA_CHUTE:.0f} m^2, mass {M - 200:.0f} kg): reaches 1.5 km after {cp[-1][0]:.0f} s at {cp[-1][2]:.0f} m/s')
rho15 = RHO0 * math.exp(-1500.0 / HS)
vt = math.sqrt(2 * (M - 200) * GM / (rho15 * CDA_CHUTE))
print(f'  the terminal speed on the parachute at 1.5 km is sqrt(2 m g / (rho Cd A)) = {vt:.0f} m/s: the parachute alone cannot land the vehicle, so a propulsive phase is needed')
v_pd = cp[-1][2]
a_net = 5.0                                          # net deceleration chosen for the powered phase, m/s^2
tb = v_pd / a_net
dv = v_pd + GM * tb                                  # remove the speed and carry the weight while doing it
mprop = (M - 200) * (1 - math.exp(-dv / (220 * 9.80665)))
print(f'  a powered phase that decelerates at {a_net:.0f} m/s^2 net from {v_pd:.0f} m/s lasts {tb:.1f} s and needs delta-v = v + g t = {dv:.0f} m/s, about {mprop:.0f} kg of propellant at an Isp of 220 s; the vehicle falls {v_pd * tb / 2:.0f} m while braking')
print('navigation phases: entry interface -> blackout (inertial only) -> parachute (radar altimeter and velocimeter) -> powered descent (terrain-relative navigation, radar, then the convex guidance of module 9)')

entry vehicle: mass 900 kg, ballistic coefficient 56 kg/m^2; atmosphere rho = 0.02 exp(-h / 11.1 km) kg/m^3
entry interface at 125 km, speed 4500 m/s, flight path angle -17 deg
simulated ballistic entry: peak deceleration 92.2 m/s^2 = 9.4 Earth g at t = 81 s, altitude 30.0 km, speed 2762 m/s
Allen-Eggers: peak deceleration V^2 sin(gamma) / (2 e H) = 98.1 m/s^2 = 10.0 Earth g, at altitude H ln(rho0 H / (beta sin gamma)) = 29.0 km; independent of the ballistic coefficient except for the altitude
  speed at the peak: V exp(-1/2) = 2729 m/s (simulation 2762)
peak dynamic pressure 5119 Pa at 30.0 km
parachute deployment when the dynamic pressure falls to 500 Pa: altitude 8.8 km, speed 333 m/s, time 146 s after the interface, downrange 389 km
  (the speed of sound near the ground is about 235 m/s: Mach 1.42 at deployment)
what happens to the altitude at which the vehicle slows if the ballistic coefficient is 20% off:
  beta x0.8: peak deceleration altitude 31.5 km (+2.5 km), peak value unchanged 98.1 m/s^2
  beta x1.0: peak deceleration altitude 29.0 km (+0.0 km), peak value unchanged 98.1 m/s^2
  beta x1.2: peak deceleration altitude 27.0 km (-2.0 km), peak value unchanged 98.1 m/s^2
what happens if the entry angle is 2 degrees steeper or shallower:
  gamma 15.0 deg: peak deceleration   8.9 Earth g, peak altitude  30.4 km
  gamma 17.0 deg: peak deceleration  10.0 Earth g, peak altitude  29.0 km
  gamma 19.0 deg: peak deceleration  11.1 Earth g, peak altitude  27.8 km
under the parachute (drag area 135 m^2, mass 700 kg): reaches 1.5 km after 112 s at 48 m/s
  the terminal speed on the parachute at 1.5 km is sqrt(2 m g / (rho Cd A)) = 47 m/s: the parachute alone cannot land the vehicle, so a propulsive phase is needed
  a powered phase that decelerates at 5 m/s^2 net from 48 m/s lasts 9.5 s and needs delta-v = v + g t = 83 m/s, about 26 kg of propellant at an Isp of 220 s; the vehicle falls 226 m while braking
navigation phases: entry interface -> blackout (inertial only) -> parachute (radar altimeter and velocimeter) -> powered descent (terrain-relative navigation, radar, then the convex guidance of module 9)

In robotics

The numbers of EDL drive the design of the vehicle and its sensors.

• Dispersions. A 20% uncertainty in ballistic coefficient moves the peak-deceleration altitude by 2 km; an entry-angle error of 2° changes the peak by 1.1 g. The atmosphere's density varies from day to day by tens of per cent; navigation errors at entry become landing-ellipse errors tens of kilometres wide, which is why lift and guidance are added to larger vehicles.
• Parachute deployment. The deployment condition (dynamic pressure, Mach number) is a window set by the parachute's strength and its stability; the altitude at which it is reached varies with the atmosphere, and the later phases must absorb the variation. Triggering on a navigation-derived velocity is more precise than on a timer.
• Terrain-relative navigation. At the end of the parachute phase the navigation uncertainty (kilometres) is larger than hazards on the surface (metres). Matching camera images with a map reduces the position error to tens of metres, as in the terrain-matching example of module 7, and the guidance can divert to a safe site.
• Powered descent. The last phase is the fuel-optimal powered descent of module 9 (minimum fuel, a thrust floor, a glide-slope constraint), solved onboard in a fraction of a second; the 26 kg of propellant in the example is typical of the figure of merit.
• Hopper. The same chain at small scale: Hopper's ascent and descent are the propulsive phase alone, with the radar altimeter of the flare lesson and the explicit and convex guidance of modules 8 and 9.

The convex guidance is M9-T4: Powered-Descent Guidance by Convexification and the explicit law M8-T3: Explicit Guidance to a Terminal State. The capstone M12-L2: Capstone: Land Hopper puts a landing guidance law in the full loop with sensor errors and dispersions.

Common misconceptions

Key takeaways

1. EDL has five phases: entry, parachute deployment, parachute descent, powered descent, landing; each uses different sensors and guidance.
2. The peak deceleration of a ballistic entry is $V_E^2\sin\gamma_E/(2eH)$, independent of the vehicle: 98 m/s² (10 g) by the formula, 92 m/s² (9.4 g) simulated.
3. The altitude of the peak depends on the ballistic coefficient: ±20% moves it by +2.5 and −2.0 km.
4. The parachute is deployed on dynamic pressure: 500 Pa at 8.8 km and Mach 1.42, 146 s into the entry.
5. On the parachute the vehicle approaches a terminal speed of 47 m/s at 1.5 km, which only propulsion can remove.
6. The powered phase needs 83 m/s and about 26 kg of propellant, and is the convex powered descent of module 9.

References and further reading

Each source is marked free or paid. Free to read does not mean free to reuse: the course resources page lists every source with its licence.

• R. D. Braun and R. M. Manning, "Mars exploration entry, descent and landing challenges", Journal of Spacecraft and Rockets 44(2), 2007, a survey of the phases and the engineering limits. Paid (AIAA).
• H. J. Allen and A. J. Eggers, A Study of the Motion and Aerodynamic Heating of Ballistic Missiles Entering the Earth's Atmosphere at High Supersonic Speeds, NACA Report 1381, 1958, the source of the approximation. Free to read from the NASA Technical Reports Server; US government work.
• A. Steltzner, A. Miguel San Martin, T. Rivellini, A. Chen and D. Kipp, "Mars Science Laboratory entry, descent, and landing system development challenges", Journal of Spacecraft and Rockets 51(4), 2014. Paid (AIAA).
• MIT 16.851 Satellite Engineering (OpenCourseWare) , for the context of the vehicle design. Free; licence CC BY-NC-SA 4.0.

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