-1

My questions are:
a) What significant forces “carry” flying objects around with the rotation of the earth,
b) How do each of those forces contribute to that “carrying”, and
c) How relatively significant is each of those forces?
d) If any of the 5 forces below are not significant contributors, please state why not.

For example, if an aircraft was to take off, thrust directly upwards and hover at say 30,000 feet for 12 hours, then come directly down again, I think we'd generally agree that it would land in the same city it took off from, not half way around the earth. The forces we have considered are:

  • Gravity
  • Centrifugal force
  • Centripetal force
  • Momentum (or is it inertia?) of the object before it left earth
  • The atmosphere's rotation with the earth
  • Any other significant contributors?

I have my opinion on which of the above are relevant, but I didn’t do much Physics at university (30 years ago), and I’d like to know what you think and why. because a friend and I disagree.

Terry
  • 103
  • Yes, @yatisagade, that was the StackExchange link I was alluding to in my original post (the paragraph that Danu has edited out for us), (and I've seen others), but it doesn't answer everything I've asked, including things like: If gravity is not significant, then why is it not?. I think I understand the physics involved, but my friend and I disagree on the answers, so I am trying to get a better understanding by asking people who should know. My friend thinks he knows the answer, but can't recall the details of the reason, and hasn't got the time to get back into it. – Terry May 31 '15 at 10:34
  • Possible duplicates: https://physics.stackexchange.com/q/1193/2451 , https://physics.stackexchange.com/q/58154/2451 and links therein. – Qmechanic May 06 '19 at 13:42

1 Answers1

2

You don't need a complicated answer. The answer is the fact that we are moving too.

enter image description here

How can this bird swoop down and catch the worm if the ground and the worm are rotating so quickly? The answer is because the ground, the air, and the tree are all moving at the same rate. The same applies to flying objects. So the forces involved are the same forces that keep everything else rotating: mainly momentum and gravity (momentum is inertia in motion).

We should remember (or learn now) that momentum only works in a straight line. Gravity forces your flying object to follow the same course that it would on earth.

enter image description here

The green arrows are gravity. The blue lines are momentum. They add to form a curved line. The same line you are following, the bird is following, and the worm is following.

Jimmy360
  • 3,942
  • A well explained answer, thank you Jimmy! Good pictures, too. Now looking at part c of my question, how would you compare the significance of gravity vs. momentum, in keeping the hovering object above the same city as the earth rotates, i.e. which one is more significant and why? I have my ideas on this (which I'd be happy to share later), but I'd like to hear yours first, please. Also, do you have a Physics degree? If so, where from? Thanks. – Terry Jun 01 '15 at 08:53
  • @Terry They are both equally significant because it would not work without the both. Both are necessary to for the curved line of motion around the earth. – Jimmy360 Jun 01 '15 at 12:25
  • @Terry I do not have a physics degree as of yet. – Jimmy360 Jun 01 '15 at 12:25
  • Thanks Jimmy. Personally, I was thinking the momentum would be much more significant than gravity in this situation. Here’s my thoughts on why: 1. Gravity is simply stopping the craft from continuing in a straight path out into space. 2. Gravity is not contributing to the “horizontal” movement of the craft. 3. For every 1 m of “horizontal” movement (from momentum), far less than 1 mm of “downward” adjustment from that straight course (from gravity) would be required, and this could be achieved with a slight downward thrust. – Terry Jun 01 '15 at 22:53
  • Take away the gravity, and the principle still basically works over relatively short distances (i.e. the object only relatively-slowly gets further from earth). Imagine your bird & worm example without gravity, for example. 5. But if you retain the gravity and have no momentum (i.e. the earth spins under an object which is not moving with it), the principle would not work at all.
    • – Terry Jun 01 '15 at 23:28
  • Take away the gravity and the rotation, and imagine a basically massless (e.g. hollow bubble) planet moving in a straight line through space. The principle would still hold, though now the shape of the earth and it’s rotation is irrelevant - it’s just about keeping up with the movement of the planet.
    • – Terry Jun 01 '15 at 23:29
  • What are your comments on my points 1-6 above, especially (but not only point 2), Jimmy? And are you currently studying Physics at university, or what? – Terry Jun 01 '15 at 23:29
  • Aside from gravity and momentum, the other significant force I’d consider is that of the atmosphere which is basically rotating with the earth. As the craft gains altitude, it would require more “horizontal” velocity for it to keep up with the earth, because its radius from the centre of the earth would have increased. The movement of the atmosphere would act as a wind against it, and help to carry it along. Agreed? – Terry Jun 01 '15 at 23:30
  • Hi again Jimmy. Did you see my comments above? What do you think? Thanks. – Terry Jun 05 '15 at 21:50
  • As for the first comment, I would say that they are equally significant because the process requires both. @Terry – Jimmy360 Jun 05 '15 at 22:02
  • In your non-rotating scenario, however, you are correct. – Jimmy360 Jun 05 '15 at 22:03
  • As for the atmosphere rotating, this is a product of momentum and gravity. Imagine a molecule in the atmosphere. We can construct the same diagram that we did for the flying object. @Terry – Jimmy360 Jun 05 '15 at 22:08
  • Thanks Jimmy. I agree the atmosphere rotating is a product of momentum and gravity, but I'd probably add friction to that. Do you agree with my points 1, 2 & 3, above? If you could number you responses for clarity, that would be helpful. – Terry Jun 05 '15 at 22:41
  • Th atmosphere wouldn't act as wind for the same reason that you don't notice thy you are going 60 mph on the highway. @Terry – Jimmy360 Jun 05 '15 at 22:49
  • But I do notice that I'm going 60 mph on the highway, especially when I put my hand out the window. And I was asking about my points numbered 1, 2 & 3, above (starting four posts under your last picture). Do you see the three points I'm talking about, Jimmy? – Terry Jun 05 '15 at 23:20
  • @Terry At a perfectly constant velocity, with the windows up, and win your eyes closed, you will not notice. This is because he car is traveling with you. This is why you can flip a coin and catch in a car: you, the coin, and the car travel together. – Jimmy360 Jun 05 '15 at 23:29
  • Upon reading 1-3 you are correct. @Terry – Jimmy360 Jun 05 '15 at 23:40
  • If that's what you're talking about, then I agree, Jimmy. But I'm not talking about 2 objects moving at (exactly) the same speed. If you look at the post above, where I wrote "the atmosphere would act as a wind against it", you'll see that I preceded that comment with "As the craft gains altitude, it would require more 'horizontal' velocity for it to keep up with the earth, because its radius from the centre of the earth would have increased." Understand what I mean, Jimmy? – Terry Jun 05 '15 at 23:48
  • The atmosphere wouldn't act as wind against it however. You also would need more horizontal velocity higher up. I will add a diagram to explain that later. @Terry – Jimmy360 Jun 05 '15 at 23:50
  • Regarding your comment that "You also would need more horizontal velocity higher up". I know you would, Jimmy. That's what I was saying when I wrote: "...it would require more 'horizontal' velocity for it to keep up with the earth, because its radius from the centre of the earth would have increased." So if that's what your diagram is going to show, I don't need to see it because that is what I've been telling you. My point is, the atmosphere would already have that extra velocity, so it would be going faster than the aircraft, so it would act as a wind against the aircraft, right? – Terry Jun 06 '15 at 01:24
  • @Terry That was a typo. I meant wouldn't. – Jimmy360 Jun 06 '15 at 02:08
  • Why wouldn't you need more horizontal velocity higher up, Jimmy? As I've mentioned, the radius from the centre of the earth would be higher than when sitting on the Earth, so that means you've got a longer distance to cover, in the same time, which would need more speed, right? In other words, a big circle has a bigger circumference than a small circle. Please explain why you disagree. Thanks. – Terry Jun 06 '15 at 06:18
  • @Terry I'm having trouble finding the diagram that I want, so I'll try to explain it in words: Imagine how fast you need to go to orbit the earth. Now go higher up. You have more distance to travel, but you also have more time before you fall in to the earth. These effects combine to make the orbital velocity be constant. – Jimmy360 Jun 06 '15 at 06:39
  • What has the time that it takes to fall to earth got to do with the velocity required to keep up with the Earth's rotation, Jimmy? Let's imagine an extreme example to make the point: The aircraft flys out as far as the moon, and then orbits earth - still 24 hours per 360 degree rotation. Are you saying that its speed would still be the same as before it took off from Earth? How - it's got thousands of times more distance to cover per day? – Terry Jun 06 '15 at 09:31
  • @Terry Look up Newton's cannonball. After you have done that tell me. It will aide in the discussion. – Jimmy360 Jun 07 '15 at 02:20
  • OK Jimmy, I've read this: en.wikipedia.org/wiki/Newton%27s_cannonball It doesn't seem relevant, because I don't see how it takes away from the fact that the circumference of a big circle (e.g. a flight path) is bigger than that of a small one (e.g. Earth's surface), so to cover say 360 degrees in the same time (24 hours), you need to go faster on the outer circle than the inner one. I can't imagine Newton (or anyone) disproving that, but if you can, I'm all ears. – Terry Jun 07 '15 at 03:22
  • Okay, on the example of Newton's cannonball, do you understand how the cannonball falls into the earth (without orbital velocity)? @Terry – Jimmy360 Jun 07 '15 at 03:25
  • I think so, Jimmy. – Terry Jun 07 '15 at 03:28
  • So (without orbital velocity), would it take longer for the cannonball to hit the ground? @Terry – Jimmy360 Jun 07 '15 at 03:29
  • Longer than what? – Terry Jun 07 '15 at 03:31
  • @Terry Then releasing it closer to the ground. – Jimmy360 Jun 07 '15 at 03:37
  • Do you mean: Dropping it onto the ground, Jimmy? I don't see an answer to your question on the web page I linked to. If you do see one, please copy, paste & "quote" it here so I can see it. – Terry Jun 07 '15 at 03:42
  • I'm asking "which hits the ground first, a ball dropping up high or a ball dropped lower?" @Terry – Jimmy360 Jun 07 '15 at 03:43
  • OK, but I still don't see an answer to your question on the web page I linked to. If you do see one, please copy, paste & "quote" it here so I can see it. – Terry Jun 07 '15 at 03:44
  • @Terry I don't think you understand my question. It is very intuitive. – Jimmy360 Jun 07 '15 at 03:45
  • Do you want me to work it out from the description of Newton's cannonball on the web page I linked to? Or do you want me to use my common sense? – Terry Jun 07 '15 at 03:48
  • @Terry Common sense – Jimmy360 Jun 07 '15 at 04:17
  • If common sense, I'd say a ball dropped from a low altitude will hit the ground sooner than one dropped from a high altitude. (But I see nothing about those timings on the web page I linked to, above. None of the 5 scenarios (A-E) involved a ball just being dropped - all are fired from the canon which is at the same altitude for all balls.) – Terry Jun 07 '15 at 04:19
  • @Terry It is the same for Newton's cannonball (if you don't give it enough speed to put it into orbit). – Jimmy360 Jun 07 '15 at 04:34
  • Newton's cannonballs are all fired from the same altitude (on the page I linked to above). Does the first sentence of my previous post answer your question, Jimmy? – Terry Jun 07 '15 at 04:40
  • @Terry Yes, and the answer is the same for Newton's cannonball. Do you remember from your reading that to get into orbit, the cannonball just has to miss the surface of the earth? – Jimmy360 Jun 07 '15 at 04:43
  • Yes, Jimmy. How does that relate to my point? – Terry Jun 07 '15 at 04:45
  • @Terry An object up higher would, as we said, have more time before it hit the earth. That correlates to more time to move out of the way. – Jimmy360 Jun 07 '15 at 04:48
  • More time to move out of the way of what? – Terry Jun 07 '15 at 04:50
  • @Terry The earth – Jimmy360 Jun 07 '15 at 04:51
  • It would have more time to move out of the way of the earth??? The earth is not moving towards it. It is moving towards the earth, right? – Terry Jun 07 '15 at 04:57
  • @Terry Yes, but it needs enough horizontal velocity to move around to earth as opposed to into the earth. – Jimmy360 Jun 07 '15 at 05:00
  • OK. What's next? – Terry Jun 07 '15 at 05:05
  • @Terry As you noted, it has a longer path to travel. This combines with the fact that it has more time to keep orbital velocity constant. – Jimmy360 Jun 07 '15 at 05:07
  • Now let's relate that to our web page (http://en.wikipedia.org/wiki/Newton%27s_cannonball). Does that mean that the velocities in scenarios C, D & E on that webpage are the same? If not, what is your point? – Terry Jun 07 '15 at 05:11
  • @Terry No, they are not. Even if they all were orbiting, they might not. All that I'm trying to show is that a larger orbit does not always entail higher orbital velocity (in fact, closer orbits often have to be faster). – Jimmy360 Jun 07 '15 at 05:17
  • So you are talking about things which are: a) at orbit velocity, which b) may take different periods to go around the Earth. And I was talking about: a) aircraft (see my original question, where I wrote "For example, if an aircraft was to take off...") which are not at orbit velocity, which b) need to keep in sync with the rotation of the Earth. So, how can we expect the rules to be the same? – Terry Jun 07 '15 at 05:57
  • Sorry - deleted my last post then re-added it after you posted. – Terry Jun 07 '15 at 05:57
  • Agreed Jimmy, but these orbit scenarios are not having to get around the earth in the same time period, are they? I'm talking about 2 things traveling different distances, in the same time period, right? Intuitively, they need to travel at different speeds, right Jimmy? – Terry Jun 07 '15 at 06:01
  • @Terry My point is that they usually will, but not always. – Jimmy360 Jun 07 '15 at 20:10
  • My web link about Newton's cannonball says nothing about relative orbit periods (i.e. comparing how long it takes to do a full 360 degree orbit). I agree that an object in a long orbit could travel slower than one in a short orbit (though that web link don't confirm that either). Are you trying to say that it is ever possible for 2 objects to travel different distances at the same speed? If so, please: a) confirm that you are saying that, and b) link me to a webpage that clearly backs that claim up. – Terry Jun 07 '15 at 22:08
  • Or to explain it more simply, Jimmy, do you agree that: Distance/Time = Speed? For example, for things on the surface of the earth: 24,000miles/24hours = 1000miles/hour? (Where 24,000miles is the (approximate) circumference of the Earth.) – Terry Jun 08 '15 at 23:51
  • No, of course you annoy travel different distances at different speeds at the same time. You do not understand what I'm trying to say. If I were there, I would be able to show you what I mean with the drawing, but since I can not, I do not know how I would be able to explain it to you. @Terry – Jimmy360 Jun 09 '15 at 01:20
  • That sounds like a change of mind, because I thought your answer above to that kind of question was "My point is that they usually will, but not always." Answer this for me please Jimmy. Do you believe that an aircraft hovering 30,000 feet above the earth needs to travel the same speed as OR faster than the surface of the Earth (i.e. about 1000mph), to keep up with it as it rotates? Same speed OR faster? – Terry Jun 09 '15 at 04:41
  • It will usually travel faster, but in some cases, it can go at the same speed or even slower. I think I know a better way to explain my point. Imagine a satellite. orbital velocity at a radius of 50000 m is 500 m/s. This satellite is orbiting at 1000 m/s. That is faster than it needs to go. At a radius of 100000 m, the orbital velocity is 1000 m/s. That is the speed of at different satellite at that radius. The 2 satellite is at a lower seed at a higher radius. An orbiting satellite doesn't need to be at exactly orbital speed. @Terry – Jimmy360 Jun 09 '15 at 05:13
  • Is this right: Satelite #1: Speed = 500m/s, orbit radius = 50,000m, orbit period = ???. Satelite #2: Speed = 1,000m/s, orbit radius = 100,000m, orbit period = ???. (The “orbit period” is the time to complete a 360 degree orbit.) I assume both are orbiting in circles (let's make it circles because then it makes calculations easier and will be close to my example with the Earth's surface and aircraft). Is that all correct so far, Jimmy? If not, please copy and paste back here with your corrections. What orbit periods are you suggesting (fill in the ??? marks, please). – Terry Jun 09 '15 at 22:44
  • The above is not correct. If you read it again, you will see that they were not going at minimum orbital velocity. @Terry – Jimmy360 Jun 09 '15 at 23:06
  • Hi Jimmy. To make it clear for me, could you please give me the Speed, Orbit Radius and Orbit Period for your 2 example satelites so I can compare them. A nice simple format like this would be good: "Satelite #1: Speed = ???m/s, orbit radius = ???m, orbit period = ???, Satelite #2: Speed = ???m/s, ...etc...". Thanks. – Terry Jun 10 '15 at 07:33
  • Hi again Jimmy. Do you understand what I am asking for? – Terry Jun 12 '15 at 04:09
  • @Terry I gave you all of those things in the above. – Jimmy360 Jun 13 '15 at 01:09
  • Comments are not for extended discussion; this conversation has been moved to chat. – Manishearth Jun 13 '15 at 20:35