1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Hi Bill, I edited your post to make your actual question clearer. Although your post is fairly long, I would recommend a (small) diagram, to show the forces involved. The homework tag is because it is a h/w type question. Best of luck with it. $\endgroup$
    – user108787
    Commented Dec 11, 2016 at 14:48
  • $\begingroup$ Please show your attempt to calculate the distance. $\endgroup$ Commented Dec 12, 2016 at 19:08

1 Answer 1

1
$\begingroup$

For any aero object there are the 4 forces: lift, drag, thrust, and weight. So let's look at each of the objects (the Stargazer and the Pegasus) independently you see what the resulting forces are.

Stargazer: - Lift remains the same - Weight decreases suddenly which means that an aircraft that was in steady level flight now has a large excess of Lift (that remained constant) which results in a large acceleration in the Lift direction - Drag remains roughly the same - Thrust remains the same

So for the Stargazer the largest change is the weight which consequently yields an acceleration.

Pegasus: - Assume no lift - Weight is known - No thrust - Drag coefficient can be estimated from empirical methods.

Therefore the Pegasus will initially be travelling horizontally at the same velocity as the Stargazer but the drag will decelerate the Pegasus such that there develops a relative velocity horizontally between the Stargazer and Pegasus.

$\endgroup$
3
  • $\begingroup$ Despite the homework tag, it has been 28 years since I was in a physics classroom. I am trying to fact-check the pilot's assertion that the Stargazer is 1500 feet away from the rocket five seconds after release. 32 feet per/second^2 gets me 400 of the 1500 feet. For the coefficient of drag equation: Cd = D / (A * .5 * r * V^2), all I appear to have is Velocity. $\endgroup$
    – Bill Jelen
    Commented Dec 11, 2016 at 22:16
  • $\begingroup$ @BillJelen : The rocket drops 400ft. The crew claimed the airplane rises "about 1000 ft" within 5s. Total close to the claimed 1500 ft. So the issue is really to verify the 1000 ft rise, isn't it? If the rise is 1000 ft, it is hardly worth bothering about the horizontal separation. $\endgroup$ Commented Dec 12, 2016 at 19:18
  • $\begingroup$ For drag we need to estimate the drag coefficient Cd and then use the area using diameter of the rocket. This is where it gets a little complicated but search "drag coefficient of cylindrical bodies in axial flow" to estimate - based on the length to diameter ratio - the drag coefficient. Then you can use density at that altitude (searchable through standard atmospheric tables) with the cross-sectional area of the Pegasus which will give you the resulting drag force. $\endgroup$
    – nrabbit
    Commented Dec 12, 2016 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.