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I’m not sure all of the details in the experiment but basically it went as follows. Someone marked their location of take off in a hot air balloon and went straight up in the air. They stayed suspended for some time before coming straight back down. Apparently the result was that they had not drifted from the original take-off point underneath them. Maybe I got all of that incorrect, but I’m sure it’s close.

But if in that experiment the Earth had no effect on the landing point, why does the “rotation of the Earth” need to be factored into making the shot with a long range rifle for example? Now I know it isn’t the best comparison and that there are many different factors in both scenarios. Such as “the bullet is traveling and in the experiment the balloon wasn't”, but I’d just like to hear different theories and opinions as to why this is.

P.S. I know a popular answer is going to be that the Earth is too massive for us to feel its spin, but I’m thinking deeper than that!

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    $\begingroup$ It's a good question, but I'm not sure that an astronomy site is the best place for it - I think this is pure physics. Have you checked on our Physics site to see whether there might already be a related question there? $\endgroup$ – Chappo Says Reinstate Monica Nov 24 '18 at 3:49
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    $\begingroup$ Related, but not duplicate: Why don't we feel the speed of earth moving $\endgroup$ – uhoh Nov 24 '18 at 3:50
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    $\begingroup$ I marked my seat while sitting in a train. I then climbed straight up into the overhead luggage compartment, and stayed there for some time before climbing straight back down. The result was that I ended up back in my marked seat; apparently the train wasn't moving at all! $\endgroup$ – user24977 Nov 25 '18 at 12:55
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    $\begingroup$ On the other hand, if I throw a ball down the length of the aisle of the train, where it lands depends on whether the train is moving in a straight line, or whether it is making a sharp turn. $\endgroup$ – user24977 Nov 25 '18 at 12:58
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There are two different things going on, and they aren't too related (from what I can see). For the hot air balloon situation, you would think that if you hovered above where you are for 6 hours, the Earth could spin under you, and you would land back in a completely different place. Unfortunately, because the hot air balloon was on the Earth to begin with, it was already moving with the Earth. There are many reference frames in play, even though we feel like we are standing still. Everyone on Earth is standing still on the surface. However, the surface is rotating around the Earth's axis. The Earth's axis (and the Earth itself) orbits the Sun. The Sun is goes around our galaxy and our galaxy is travelling through intergalactic space.

So how does this relate to the hot air balloon situation? Well because the balloon was on the surface, it was already moving with the surface of the Earth. Remember how I said that the surface of the Earth rotates around the Earth's axis? Well, since the balloon was on the surface to begin with, it will also rotate with Earth's axis, just like the surface! What if we wanted to achieve that hover effect? I said earlier that the Earth orbits the Sun. In order to achieve that hover effect, we would need our hot air balloon to orbit the Sun without rotating around the Earth. You can't do this with a hot air balloon, since the atmosphere also happens to move with the Earth, and hot air balloons can't go into space. To achieve that hover effect, we would need some sort of spaceship with lots of fuel. If I had a hovering spaceship that didn't orbit with the Earth (this would, again, cost lots and lots of fuel), then yes, I could hover in the same spot and have the Earth rotate under me.

Now, you're probably wondering how I could achieve this with a spacecraft that was originally on the surface of the Earth. There wouldn't be much point in this hovering effect since you may as well just fly to your destination (like the theoretical Big Falcon Rocket), but if you wanted to do it, you would need to go above the Earth's atmosphere, use your boosters to go opposite of the rotation of the Earth (to cancel out your velocity), hover for some time, then use your boosters to go along with the rotation of the Earth (to return to rotation velocity), and land on Earth. Of course, you could skip the 2nd part by using heat shields and ramming into Earth's atmosphere like every other spacecraft we have, and no rocket would ever do this as it's just much more practical to orbit around the Earth (hovering costs fuel, orbits don't).

Ok, what about the sniper situation? The Earth rotate on its axis. Since a day is 24 hours long, a location at the equator goes around once in the same time it takes for someone in the arctic to go around once. However, the further north or south you are from the equator, the slower that part of the Earth has to rotate to complete 1 rotation in 24 hours. Think about spinning a ball. The equator of the ball whips around, but the top and bottom move much slower. It's the same thing. Lets say my sniper is at the equator. When the sniper shoots east or west, they do not need to correct for the Earth's rotation because everywhere along that latitude, the Earth is rotating at the same speed. However, if the sniper shoots north, the bullet will go to the east. That's because when the bullet was shot at the latitude closer to the equator (shot from the south), that spot of the Earth was moving faster than the spot of the Earth that the target was standing on. It's like saying that the spot I'm standing rotates at a speed of say... 1000 mph, but my target's spot rotates at say...995 mph. Since my bullet came from the 1000 mph part, it will obviously go out of the barrel at high speeds because of the gun, but it will also be rotating around the Earth's axis at 1000 mph. However, once my bullet starts approaching my target's latitude, since the Earth rotates slower, it will appear to deviate in the direction of the rotation (which is east, or to the right). Why? Well since my target's speed is 995 mph from rotation, you do 1000 - 995 = 5. That means that my bullet will have a net velocity of 5 mph to the right relative to my target. That means my bullet will miss if the target is far away enough. If you shoot from the equator to the south, the bullet will also go east, so as a result, it will deviate to the left. It's the same effect, but "upside down". This effect is called the Coriolis effect, and it's what gives hurricanes their power.

Finally, you asked "why don't we feel the Earth rotating beneath us?" This is because we are moving with the Earth. When you are on a train that is going at a constant speed of 50 mph, you do not feel it move (you may feel some bumps due to bumpy tracks). It is only when you accelerate or decelerate do you feel things move. When you are on the train, your speed is constant, so you do not feel anything. It's the same for the Earth, except you are rotating around the Earth's axis at a constant 1000 mph. Nothing has changed, except the speed is faster.

I probably explained this horribly so feel free to ask anything.

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    $\begingroup$ You explained it very well and in great detail, thank you! $\endgroup$ – Begons18 Nov 23 '18 at 21:51
  • $\begingroup$ But, I do have some questions. Bare with me here. You said in the experiment he didn’t move because he was on the surface to start with so he was moving at the same rate as the earth beneath him. Just like when a car is moving, the person in the care is going that same speed. But let’s say you’re standing at the equator, and you fire a bullet in the opposite direction of the rotation of the earth. The bullet would still leave the barrel and travel at its intended speed wouldn’t it? Why is that? $\endgroup$ – Begons18 Nov 23 '18 at 21:55
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    $\begingroup$ @Begons18 Let say I go bowling on a train (it's a big train). The train goes 50 mph to the right. There are 2 bowling lanes, one that faces left and one that faces right. If I always bowl at 10 mph, then when I bowl to the left, it will go 10 mph left relative to the train, but 40 mph to the right relative to the ground. (50 - 10 = 40) If I bowl to the right, it will go 10 mph right relative to the train, and 60 mph relative to the ground. (50+10=60) Change the train with the Earth spinning 1000 mph east at the equator and a gun that fires 1500 mph. The situation stays the same. $\endgroup$ – User24373 Nov 23 '18 at 22:03
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    $\begingroup$ @Andreas Hurricanes cannot cross the equator. They would disappate before they got there because there isn't any supporting coriolis force to keep it spinning. A few years ago typhoon vamei developed only 1 degree north of the equator, setting a new record. Apparently, though not particularly powerful, it was so rare that it was estimated to be a 400 year storm. $\endgroup$ – User24373 Nov 24 '18 at 16:29
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    $\begingroup$ A rocket does not need to go to space for this to work. Go straight up and straight down without leaving the atmosphere and you still don't come down at the same point. it only works with the balloon because the balloon stays in the very dense parts of the atmosphere and he balloon is very light compared to the surrounding air and gets pushed along easily. $\endgroup$ – Polygnome Nov 25 '18 at 9:25
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It's hard to feel the Earth's rotation personally, but only because we are not very sensitive to very gradual changes and very weak "forces". However, there are everyday objects that can show us the rotation of the Earth. The balloon may have been a bad choice because even a slight wind over time will dominate the problem. (also see]

A pendulum

One object is a Foucault pendulum.

Foucault pendulum Source

A gyroscope

Another is an (expensive) camera with inertial image stabilization. Some cameras have small gyroscope chips that sense very tiny rotations of the camera body and compensate with image processing.

enter image description here Source

See the following questions and their answers for more on this.

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  • $\begingroup$ This is a GREAT answer. $\endgroup$ – Fattie Nov 25 '18 at 14:31
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That's three questions in one.

Why don’t we feel the earth rotating beneath us?

If you're asking what one feels when standing still on the surface of the Earth, it's because

  • Static friction and the normal force are constraint forces,
  • The difference between what you would feel on a nonrotating planet with the same mass and size of the Earth and our rotating Earth us very small, and
  • You've never experienced anything but standing on our rotating Earth.

Constraint forces take on whatever value is needed to satisfy the constraint if doing so doesn't violate some other condition. For example, the normal force pushes normal to the surface so as to keep the constrained object on the surface (the constraint). The "other condition" in the case of the normal force is that the normal force can only be directed outward. Friction comes into play if the surface on which one is standing isn't level. The other condition in the case of static friction is that it can't exceed the coefficient of static friction times the normal force.

So, what values do these forces need to take on to keep you standing still on the surface of the Earth? From the perspective of an inertial frame, an object at rest on the Earth's surface is undergoing uniform rotation about the Earth's axis. This means the net force on the object must be inward, toward the Earth's axis of rotation (note well: not toward the center of the Earth, except at the equator), and have a magnitude of $F_{\text{net}}= m r \Omega^2$, where $m$ is the object's mass, $r$ is the distance to the axis of rotation, and $\Omega$ is the Earth's sidereal rotation rate. The other forces acting on the object are gravitation and buoyancy, which is very small. Ignoring buoyancy, we must have $\vec F_\text{surface} + \vec F_\text{gravity} = \vec F_\text{net}$. The net force is largest at the equator, where it's very small compared to gravity, about 0.3% of gravitational force. In other words, the upward force exerted by the Earth's surface and the downward force exerted by the Earth as a whole are nearly equal to one another.

The same result arises if one looks at things from the perspective of a rotating frame. From this perspective, the apparent net force is zero. Rotating frames involve fictitious forces such as the fictitious centrifugal force. This has the exact same magnitude as but the opposite direction of the net force calculated above. The end result is the same.

Someone marked their location of take off in a hot air balloon and went straight up in the air. They stayed suspended for some time before coming straight back down. Apparently the result was that they had not drifted from the original take off point underneath them. Maybe I got all of that incorrect but I’m sure it’s close.

That's the second part of the question. Unless tethered, hot air balloons typically do not return to the spot from which they take off. Hot air balloons go wherever the wind carries them. That's why hot air balloons need chase teams. A hot air balloon will return to its take off point if the wind is nonexistent or if it reverses direction during the course of the balloon's flight. The buoyancy that I ignored above is anything but negligible for a hot air balloon. Here it's the normal force that is negligible (it's nonexistent). For a balloon to remain stationary, the wind needs to be nonexistent and the buoyancy force must take on the same form as the normal force takes for an object sitting still on the surface of the Earth.

Why does the “rotation of the earth” need to be factored into making the shot in a long range shot from a rifle for example?

Because the bullet is moving with respect to the surface of the Earth. Note well: It's only the very long range shots, more than a kilometer, where the Earth's rotation needs to be taken into account. The motion of the bullet brings another fictitious force into account, the Coriolis effect. The Coriolis effect is zero for a stationary person and for a stationary balloon.

The Coriolis effect has two primary effects on the flight of a bullet, a horizontal deflection and a vertical deflection. The horizontal deflection depends on latitude, the effect being strongest at the poles and zero at the equator. The Coriolis effect makes moving objects turn to the right in the northern hemisphere but to the left in the southern hemisphere. The vertical deflection depends on latitude and direction. This vertical deflection is strongest at the equator, zero at the poles or if the direction of motion is to the north or south. The deflections are small, even for a bullet fired toward a target a kilometer away. Centimeters however can be important.

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    $\begingroup$ Although centimeters can be important, for sniper shots (about 2 km or less), Coriolis effects are swamped by other error effects, so it's not normally compensated for. Long-range artillery, on the other hand, can feel the need to compensate, and rocket artillery, with even longer path lengths (and more importantly, longer flight times), will correct for the effect. $\endgroup$ – WhatRoughBeast Nov 24 '18 at 15:24
  • $\begingroup$ "The Coriolis effect makes moving objects turn to the right in the northern hemisphere but to the left in the southern hemisphere. ... This vertical deflection is strongest at the equator": this can't be right. If it were correct, the strongest rightward deflection would be felt by a bullet fired parallel to and infinitesimally north of the equator, while the strongest leftward deflection would be felt by a bullet fired on a parallel track infinitesimally to the south. In fact, both will be deflected towards or away from the origin, depending on whether they were fired eastward or westward. $\endgroup$ – phoog Nov 25 '18 at 17:49
  • $\begingroup$ @phoog - What I wrote is correct. Perhaps not best written, but correct. The eastward, northward, and upward Coriolis accelerations are $v_n\sin\phi-v_u\cos\phi$, $-v_e\sin\phi$, and $v_e \cos\phi$ where $\phi$ is latitude and $v_e$, $v_n$, and $v_u$ are the eastward, northward, and upward components of the object's velocity. Bullets fly more or less horizontally (the vertical component of a bullet's velocity is small compared to the horizontal component), so the $v_u\cos\phi$ portion of the eastward Coriolis acceleration can be ignored. ... $\endgroup$ – David Hammen Nov 25 '18 at 20:27
  • $\begingroup$ This means that the horizontal deflection is proportional to $\sin \phi$, which is zero at the equator and one (or minus one) at the poles, while the vertical deflection is proportional to $\cos \phi$, which is one at the equator and zero at the poles. $\endgroup$ – David Hammen Nov 25 '18 at 20:29
  • $\begingroup$ I see. I missed the change of gears in your answer from horizontal deflection (moving to the left or right) to vertical deflection (which is strongest at the equator). The horizontal deflection is weakest at the equator, which makes sense if that's where its direction reverses, but I misunderstood that you were asserting to the contrary that it is strongest at the point where its direction reverses. $\endgroup$ – phoog Nov 25 '18 at 20:49

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