I had a chance to talk to the flight crew of the Orbital ATK L1011 Stargazer yesterday at Cape Canaveral. You have a 110,000KG airplane flying at a velocity of Mach 0.82. At an altitude of 39,000 feet, they release a Pegasus rocket weighing 18,500 KG. The rocket is traveling with its nose in the direction of flight. The rocket is in free fall for 5 seconds before they light the rocket.
From the kinematic equation, I calculate that the rocket drops 122.5 meters from the airplane in 5 seconds.The pilots indicated that the rocket manufacturer wants them to be 152 meters from the rocket when it ignites in case of an anomaly.
The pilots explained three things are happening in addition to the kinematic equation.
First: We turn left to get off the center line of the rocket trajectory.
Second: I am quoting his not-quite scientific explanation: "We just dropped 18,500 kg of weight, so we instantly rise about 1000 feet in altitude". While I get the basic concept here, I am sure they don't "instantly" rise 1000 feet. You have equal thrust pushing an object that now weighs less so the aircraft might climb over the 5 seconds. The L1011 aircraft weighs 110,000 KG alone. The rocket is 18,500 KG. So you go from 128,500 KG to 110,000 KG in mass. The mass is now .856 as much as it was pre-drop. Assuming they don't cut the engines back, and assuming their goal is to safely get as far away from the rocket before it lights, how much height could the conceivably gain in five seconds?
Third, the airplane keeps moving at Mach 0.82 (or faster now, with less weight) and the rocket starts decelerating. I am trying to quantify how far behind the airplane the rocket will be when they light the rocket motor. I can calculate that the airplane is traveling at 0.82*1234.8 km/h or 1.4 km every 5 seconds.
My question is: How far will the rocket in free fall travel horizontally in those same five seconds?