Why is he always falling at the same place even though the vehicle is moving ahead? pic.twitter.com/1My2jfynqk
— Domenico (@AvatarDomy) March 15, 2023
It is not a silly question.
The reason is that when we are on a moving vehicle, we have the same velocity as the vehicle and when we jump up, the velocity in that direction does not change and we keep moving in the same direction, with only air resistance to slow us down.
Some time ago, an elementary school teacher in Cleveland asked me a variant of this question. She asked, since the Earth rotates on its axis, instead of flying to California which takes about five hours, we could not just go up in a balloon and wait for three hours so that the Earth’s rotation would bring California to us, saving us time and all that fuel.
Yeap, the plane is expending a lot of energy to slow you down so that the earth catches up.
At least from one frame of reference.
One of the comments:
Awesome!
(The answer is obviously that it slams hard into the back of the bus.)
The balloon question isn’t even that unreasonable… if you go up high enough you need to increase your transversal velocity in order to stay above the same point, if you don’t the earth spins faster below you. It’s like the Coriolis effect but with height instead of latitude.
Unfortunately the height a balloon travels is fairly insignificant compared to the radius of the earth, so ballooners (is that the right word) don’t really notice the effect (not the mention that the air tends to move along with the earth’s surface…)
It does help in launching spacecraft though, I think/
I told some of my students that if a cannon on a train fired a ball straight up in the air, the ball would land back on the cannon no matter how fast the train was going. The father of a student complained to his son that I wasn’t taking the curvature of the Earth into consideration. The father is a creationist. Penny wise and kilogram foolish….
Oops, another story. Back when in-line skating was getting popular, I skated everywhere, including in subway stations in NYC. One day I was on a train with some friends and said “watch this”! I stood parallel to the tracks while the train was stationary. When the train started moving, I started rolling towards the back of the train. My friends thought I was playing a trick on them, but I was just obeying Newton’s first law of motion.
Matt G @5: I hope you doubly amazed them by regaining your position in the moving train and, again standing parallel to the tracks, began rolling forward when the train started braking.
robert79 @3: If the balloon hits a jet stream, it’s going to go east, pretty fast. Atmospheric dynamics can be complicated…
First thing I thought about was: they don’t say the vehicle is not accelerating, only that it’s moving ahead. So they presume a constant speed and direction.
#4 Matt
The father should instead have complained that you didn’t take friction into account.
/grump
Reminds me of a physics question:
If a jet plane is put on a giant treadmill, which moves backwards at the exact speed the jet thrusts forward, would it ever take off?
First of all, the question itself is ill formed. Jet engines do not provide speed, they provide thrust force.
But to answer it, yes, and likely with complete ease. Idealized, the wheels are free spinning and frictionless. They are completely decoupled from the jets thrust. The jet would take off identically no matter what the wheels are doing. Without friction they would provide to force on the jet, so the only things would be the engine thrust and air drag to determine its motion.
In reality of course the tires and bearings are not ideal. The treadmill spinning them would provide some drag force on the plane through friction. But you can’t just subtract the speeds. You would subtract the friction force on the landing gear from the thrust force of the engines to determine the net force on plane to determine its movement. And I’ll bet you would need a VERY fast treadmill to provide enough drag force to cancel out jet engines. Probably thousands of mph at least to hold it in place.
This isn’t even a puzzle or a paradox. This is basic physics reasoning. There’s nothing complicated or tricky.
“… we could not just go up in a balloon and wait for three hours so that the Earth’s rotation would bring California to us, saving us time and all that fuel.”
I wonder how Cleveland dealt with that constant 1000 MPH wind? Also, here in California, we have a constant struggle covering up the stories of thousands of people going up in balloons in Napa and getting swept out to the middle of the Pacific Ocean to die. The Napa tourism board is very concerned.
A fun variation on the skates-on-the-subway experiment: Try driving with a helium-filled balloon in the car. Of course it will hang out against the ceiling, and you have to be careful to keep it away from the driver’s face.
Watch what it does as you turn left and right, or speed up and slow down.
Extra credit question — What principle of general relativity can help you predict what the balloon’s behavior will be?
@MattG,5:
Watched an interesting video the other day about inline skating and how and why it died after having been so popular.
Bill burr popped up briefly to say it was killed by one homophobic joke:
“what’s the hardest thing about rollerblading? Telling your parents you’re gay”
Or you know, it’s just one of the million fads that came and went with time.
WTF?! Rollerblading’s not a thing anymore? When did this happen?
@4 Matt G
For short distances it’s close enough. If you’re talking about a howitzer or a coilgun, you have to consider that the Earth rotates under the suborbital projectile and take that into account when planning where your train will be.
beholder @17: Yeah. Suppose the howitzer is fired at the equator (for simplicity). Even though the shell will have the same instantaneous transverse velocity as the platform it’s fired from, its angular velocity will always be less than the platform’s while it’s airborne (even neglecting air effects)*. So it will land to the west of the firing point.
Quick calculation, assuming no air effects, and assuming rotation during flight is much less than a radian, gives an impact distance from the firing platform of about
2vv₀³/3Rg²
where v is equatorial rotation speed, v₀ is muzzle velocity, R is Earth radius, and g is gravitational acceleration at surface. If muzzle velocity is 500 m/s, this gives a distance of about 60 metres.
* Because angular speed ω = v/r, where v is instantaneous transverse speed, and r is distance from centre of rotation. If v is constant or decreasing with time, and r is greater than the ground value, angular speed will be less than that of a stationary observer on the ground
Further to #18: Even with a paltry muzzle velocity of 100 m/s, the cannonball would still come down about 0.5 metres from the cannon.
I think I messed up the calculation by assuming I could treat the transverse (i.e. parallel to Earth’s surface) velocity of the shell as constant, whereas it does vary slightly.
So, never mind. I’ll sort it out when I have more time.
@18:
Standard orbital mechanics. To overtake someone, slow down. To fall back, speed up.
(Ref: Osmos.)
Two people do the calculation in this thread; one analytically, one numerically. Both agree that a muzzle velocity of 500 m/s at the equator gives a “miss” of 122 meters, so twice my result.
xohjoh2n @21: That applies to two bodies orbiting a third body. In this case (in which the sun is irrelevant), the only “orbiting” body is the howitzer shell. A stationary observer on the Earth’s surface isn’t in orbit.
xohjoh2n @21:
You go faster to go up, you go up to go slower, you go slower to go down, and you go down to go faster.
Orbital mechanics is seriously non-intuitive for most people.
Regarding the howitzer shell shot straight up at the equator: Silly me, I hate working with rotating frames of reference, but sometimes they are the best approach.
In this case, it’s just the Coriolis acceleration which causes the shell to land west of its origin.
Matt G @4:
I know this is ridiculously late for this thread, but it only recently occurred to me that curvature of the Earth would make a difference if the train was moving.
The effect of the Earth’s rotation is much larger than that of a moving train (at reasonable speeds), but even if the Earth weren’t rotating, the ball wouldn’t land where it started if it was fired from a moving train. If the train had a speed of 80 kph, and the muzzle velocity of the cannon was 100 m/s, the ball would land 5 cm behind its starting point. For a bullet train travelling at 320 kph, the difference would be 20 cm. Calculations available on request.