The radius of the earth + 30,000 ft is about 0.14% more than just the radius of the earth. Basically ignore it, unless you are programming guidance control systems.
This reminds me of a math puzzle I read many years ago. It went something like this:
"Assume the Earth is a perfect sphere, and its circumference is 25,000 miles. You have a 25,000 mile long rope looped all the way around the Equator, so it is tight against the surface. Now you want to raise the rope one foot in the air, all the way around. How much more rope do you need?"
I thought about this and said, "You need a lot more rope! I don't know how much, but it will be a lot. It has to reach all the way around the Earth, only higher!"
Of course the real answer is "You only need about three more feet. To be specific, you need another 3.14159... feet."
That's all you need to raise a rope a foot in the air, all the way around the Earth.
I had a science test in high school with a question that involved something to do with calculating the air pressure inside an airplane at some high altitude. I went up to to the teacher and said "is this a trick question? Airplanes are pressurized at that altitude" He looked around the room nervously and then told me to assume the airplane wasn't pressurized.
A related one is to add a second section to the question asking how much extra rope would be needed to go round a tennis ball the same 1ft further out from the ball.
For someone who doesn’t know, or hasn’t calculated/thought about it that is pretty astounding.
It’s likely less than the reading error in most speedometers including those which are used for autonomous vehicles unless you are writing a guidance system for bombs or attempting to land on another planet that’s pretty negligible.
Given that the ground speed was likely drawn from gps data, which should have velocity readings more accurate than 1mph (particularly for an airborne gps unit), the choice of calculation method for ground speed does make a difference.
That is also an interesting question. If they’re measuring speed as if the airplane were at the surface rather than at 30,000 feet, is it speed along the WSG84 earth model?
"Assume the Earth is a perfect sphere, and its circumference is 25,000 miles. You have a 25,000 mile long rope looped all the way around the Equator, so it is tight against the surface. Now you want to raise the rope one foot in the air, all the way around. How much more rope do you need?"
I thought about this and said, "You need a lot more rope! I don't know how much, but it will be a lot. It has to reach all the way around the Earth, only higher!"
Of course the real answer is "You only need about three more feet. To be specific, you need another 3.14159... feet."
That's all you need to raise a rope a foot in the air, all the way around the Earth.