The Boeing Delta II has been a very successful workhorse launch vehicle since 19

Question

The Boeing Delta II has been a very successful workhorse launch vehicle since 1989 (derived from an earlier Thor-Delta vehicle, itself active since the 1960's). The "7320" configuration has been used for various missions, including, for example, the Orbiting Carbon Observatory 2 (OCO-2) satellite launched on July 2nd, 2014. The orbit was a relatively high inclination LEO with around 14.5 orbits per day. The vehicle configuration for this mission resembled: Approximate total lift-off mass = 152,000 kg 3 strap-on GEM-40 solid propellant boosters, HTPB propellant. Total mass = 13, 080 kg each; Propellant mass = 11, 770 kg each; Isp = 245.4 (sea level) or 274 (vacuum); burn time = 63.3 s First-stage, single RS-27A engine, RP-l/LOX. Total mass = 101, 800 kg; propellant mass = 96, 120 kg; Isp 254.2 (sea level) or 301.7 (vacuum); burn time = 260.5 s First-second stage adapter Mass = 2680 kg Second stage, single Aerojet AJ10 engine, A-50/N_2O_4). Total mass = 6, 950 kg; propellant mass = 6,000 kg; lsp=319.2 (vacuum); burn time=431.6 s Payload fairing Mass = 880 kg OCO-2 satellite Mass = 453 kg Estimate the theoretical delta V for this configuration (2-1/2 stages). Make reasonable assumptions as to the effective Isp during the strap-on booster and first stage burns. Find (by internet search?) the required delta V for a high inclination orbit, launching from Vandenberg, CA.

Explanation / Answer

1) The specific impulse of a rocket is given by,

Isp = Total impulse / Weight of the rocket

Cosidering the two stage rocket with varied Isp conditions during flight,

The change in momemtum due to Starp on booster ignition is found using the Isp at sea level condition,

Isp (sea level) = 245.4 sec^-1

Isp = Total impulse / Weight of propellent

Total impulse (starp on) = total propellent mass * Veq

Thrust = dot m * Veq (assuming that the pressure diff is negligible)

Now, Isp = m * Veq / dot m * g

245.4 = 11770 * Veq / (11770 / 63.3) * 9.81

Veq = 38.03 m / s (stap on velocity)

Similarly for the First stage ignition the change in momentum or Veq is,

Isp =Total impulse / Weight of propellent

The Isp for the first stage is assumed to in the vaccum as the density of the atmosphere is negligebile at 38.03 * 63.3 = 2407.34 m)

Isp = 96120 * Veq / (96120/260.5) * 9.81

301.7 = 96120 * Veq / (96120/260.5) * 9.81

Veq = 11.38 m / s (first stage)

Second stage:

sp =Total impulse / Weight of propellent

The Isp for the second stage is assumed to in the vaccum as the density of the atmosphere is negligebile at 11.38*260.5 + 38.03 * 63.3 = 5372.70 m above sea level)

Isp = 6000 * Veq / (6000/431.6) * 9.81

319.2 = 6000 * Veq / (6000/431.6) * 9.81

Veq = 7.255 m / s (second stage)

The delta V for the rocket after stage 2 is given by,

Del V = Ve ln (mi / mf) where Mi - initial mass and Mf - final mass

Del V = (7.255 + 11.38 + 38.03) ln (152000 / (6950 + 880 +453 ) )

del V = 56.66 ln 18.35 = 164.86 m/s

2) The Vandenberg lanch site in California is suitable for high inclination orbits ie. 90 - 98 degree inclined orbits. The polar orbit (90 - 98degree) transfer from Vanderberg under the above flight regime ie. Vf = 164.86 m/s Vi = 0

Using the orbital plane change impluse eqn,

Del V = sqrt (Vi^2 + Vf^2 - 2Vi * Vf * cos theta)

Assuming theta to be 97 deg and Vi to be 0, Vf as 164.86 m/s

Del V = sqrt 3312.35 = 57.55 m/s

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