Astronomy & Celestial Mechanics
the sky, computed
Concept

Flight-path angle — where it appears

The angle between a vehicle's velocity and the local horizontal. It is 90° at lift-off and zero at orbit insertion, and the gravity loss along an ascent is the integral of g times its sine.

Named by 2 essays across one field — each of them below, with the objects they name alongside it.

A corridor 0.78° wide. Peak deceleration against entry flight-path angle, from sixty-one integrated entries at 11 km/s and β = 250 kg/m². The steep edge is where the load reaches 12 g, at 6.05°. The shallow edge is skip-out: below 5.27° the vehicle passes through the upper atmosphere and leaves again at 8.55 km/s, having lost too little speed to be captured. The corridor between them is 0.78° wide, which at an approach speed of 11 km/s is a targeting problem measured in kilometres of periapsis, days out. The curve is steep everywhere, which is the other half of the difficulty: half a degree of aiming error is a factor of 1.46 in the load. Lift is what widens this, and no ballistic capsule has any.

A corridor a degree and a half wide

The peak deceleration of an entering vehicle contains no property of the vehicle at all. Only the speed and the angle of arrival decide how hard it is slowed — the ballistic coefficient decides where, and nothing decides whether.

spaceflight · Atmospheric drag Where the missing kilometres a second go. The two losses along a gravity-turn ascent, against the vehicle's thrust-to-weight ratio at lift-off, from an integration of the trajectory rather than from a table. Every run spends the same 9400 m/s of ideal Δv and every one is flown as the same manoeuvre: one pitch kick, solved by bisection so that the vehicle is horizontal at burnout, and thereafter zero angle of attack so that gravity alone turns it — which makes the steering loss identically zero and the comparison a fair one. What differs is how much of the Δv survives as speed. The gravity loss is ∫g sin γ dt, the part of the thrust spent holding the vehicle up rather than accelerating it, and it falls as the thrust rises because a vehicle that leaves quickly spends less time doing it: 1263 m/s at T/W = 1.15 against 381 at 2.2. The drag loss is ∫(D/m) dt and it rises, because the same haste means reaching high speed lower down where the air is: 119 m/s against 1874. The sum is least at T/W ≈ 1.5, at 1198 m/s, and the minimum is shallow — which is why real vehicles cluster between 1.2 and 1.5 and none of them is there because of this curve. At the marked 1.5 the 9400 m/s of ideal Δv leaves the vehicle at 8202 m/s and 42 km, against a circular speed of 7884 m/s there, so the 1198 m/s of loss is the whole of the answer to why orbit costs about 9.4 km/s when orbital speed is under 7.9.

Orbit costs 7.8 and a launch buys 9.4

The gap between orbital speed and the velocity change a launcher spends is not overhead. It is three integrals along the ascent, only one of which can be reduced by flying better, and the two that can be traded move in opposite directions.

spaceflight · Rocket equation

Named alongside it

The objects these essays reach for when they reach for this one.

Ballistic coefficientAerodynamic heatingAllen eggersAtmospheric entryBlunt bodyDecelerationΔvDrag lossDynamic pressureEntry corridorExponential atmosphereGravity loss

All concepts