How Does Energy Transfer Differ Between Two Airplane Crash Scenarios?

[tasp77]

Thought experiment (hopefully we don't need to crash a plane to figure this out);

Let's say we lift a commercial airliner up to 40,000 feet with a large balloon and then drop the airplane. It will hit the ground sometime later somewhere short of the speed of sound (600 mph or so, depending on it's orientation on the way down, I posit it will nose dive perpendicularly for maximum velocity).

Now let's take another plane, this one flying at 600 mph at 40,000 feet, and suddenly, the pilot shoves the stick forward, and dives the plane straight into the ground.

In the second case, the plane doesn't accelerate to 1200 mph on the way down as we might have thought at first take. The plane will reach the ground much faster in the second case and gravity has less time to further accelerate the plane. So it will hit the ground somewhat faster than in the first case, but will not be traveling twice as fast at the moment of impact.

Is there an 'energy deficit' in the second case? The plane has moved through Earth's gravity field the same distance, but in a faster time. Is there additional energy manifested in the system somewhere that I have not noted? Like the wreckage will be warmer in the second case?

Do the 'books balance' energy wise in both crashes?

 

[russ_watters]

Aerodynamic drag.

...because of it, a plane with no engines won't go anywhere close to 600mph.

 

[Antiphon]

If you consider the potential energy if the situation you will see the gravitational contribution must be the same. You would explain the higher kinetic energy and shorter time of flight in the second case as being powered by the jet fuel.

 

[AlephZero]

Forget about the planes and think about dropping a ball in a vacuum, so there is no air resistance.

In the first case, you drop the ball from height h and the velocity when it hits the ground is $v = \sqrt{2 g h}$

In the second case, you drop the ball from height h with an intial downwards velocity of $\sqrt{2 g h}$. The initial kinetic energy of the ball is $(1/2)mv^2 = mgh$.

When it hits the ground its KE has increased to $2mgh$.

So its final velocity will be given by $(1/2)mv^2 = 2mgh$ or $v = 2\sqrt{gh}$. In other words it will hit the ground about 1.414 times faster, not twice as fast.