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Relative speed and coin dropping

  1. Jan 30, 2016 #1
    Hello Forum,

    We are on planet earth that is rotating at an angular speed of omega = 7.27 × 10-5 rad/s. The earth atmosphere is also rotating with earth at the same speed (that is why it is at rest relative to the earth surface).

    Let's say I have a coin in my hand that is 1 meter above the earth surface. That coin has the same angular velocity as anything that is on the earth surface but a different and higher tangential speed v which is higher by omega*r = omega (1 meter). Point that are further away from the earth center must have a linearly larger tangential speed v.

    That said, If I launch the coin in the air and that coin reaches at height of 2 meter, I believe the coin will fall down and not drop exactly on my hand where it was initially. Explanation: the speed need at h=2 meter is higher than the speed at h=1 meter. When the coin reaches h=2m its speed is not sufficient to stay align with the speed of my hand at the lower height h=1 meter....

    This effect may be so negligible that we may state that the coin fall back on my hand on the same spot where it took off.... but if the height difference is significant we should notice this effect.

    The same goes it we drop a coin from a very tall tower where the tangential speed v2 is higher than the tangential speed at lower altitudes v1. The coin will fall slightly ahead of the position from which it was dropped because its speed v2 will allow the coin to cover a longer distance compared to the distance covered by object traveling at speed v1...

    A helicopter that raises straight up vertically from the surface of the earth should progressively see the point from which it took off move away from directly below...

    Is my thinking correct?

  2. jcsd
  3. Jan 30, 2016 #2
    Why don't you run the numbers and see what you get? You don't need our help for that.
  4. Jan 30, 2016 #3


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    I'd do as Chestermiller suggests and try some numbers for coins dropped at various heights. But keep in mind, that at some point, air resistance will come into play, not only with regard to the falling speed of the coin, but also its sideways motion with respect to the ground. But to get a feel for things, you can ignore it for now.

    However, with the helicopter, you cannot ignore the air. It is the air that the Helicopter is supported by when it is hovering or climbing vertically. Any incremental increase in tangential velocity the atmosphere has with altitude will be transferred to the Helicopter as it climbs. It will be carried by the atmosphere.
  5. Jan 30, 2016 #4

    omega= 7.3 × 10-5 rad/s
    h_o = 0 (earth surface)
    h1= 1 meter
    h2= 2 meters

    v1= omega*h1= 7.3x10^-5 m/s
    v2=omega*h2= 14.6x10^5 m/s

    In a time interval delta_t=1 second, an object at height h1 overs a horizontal distance of 7.3 x10^-5 m relative to an object on the ground at h_0=0 m.
    A object at height h2 travels a distance 14.6x10^-5 m relative to an object on the ground.

    Now, let's say object at height h1 is give some vertical speed v_vertical so that it reaches height h2 after 2 seconds and then it stops momentarily at that height before falling back down. My claim is that (IGNORING AIR FRICTION and only assuming the presence of gravity), the object on its way down from h2 back to h1 will not fall on the same position it was originally but will be it will be a position behind the original one. The amount of shift is:

    shift = v2*(4seconds)-v1*(4 seconds)= 29.2x10^-5 meters

    4 seconds is the roundtrip time...

    Earth as a frame of reference is not a pure inertial frame of reference. It can be consider one only approximately...

  6. Jan 30, 2016 #5


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    I haven't looked at any of the calcs yet, but I can't believe your v1 and v2. Surely tangential speed must be proportional to radius. The radius does not double going from 1m to 2m.
    But there is no vertical speed such that it takes 2 sec to travel from 1m height to 2m height. h1 and h2 must be different in this case.

    Generally I wonder whether, if something goes "backwards" on the way up, it will go "forwards"on the way down and end up where it started.
  7. Jan 30, 2016 #6
    I made a mistake. Let me tweak things up:

    Starting point:
    omega= 7.3x10^-5 rad/s
    h1=R_earth + 1 m
    v1 = omega* (R_earth+1meter)

    h2=R_earth + 1000m
    v2= omega*(R_earth+2meter)

    both v1 and v2 are horizontal speeds. At one point in time, the object at h1 is given a vertical speed upward v_up = 17,640m/s (ridicolous of course) So that it reaches the height h2 in a time of 1800 seconds. Once the object reaches h2 it will only have its original horizontal speed v1 = omega* (R_earth+1meter). The object will then start its descent toward h1. It will reach h2 1800 second later. The total round trip time is 3600 second.

    The object will fall somewhere at height h1 that is shifted from the original position by

    [v2-v1]*(3600)= omega*[(R_earth+2meter) - (R_earth+1meter)]*(3600)
  8. Jan 30, 2016 #7


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    I still don't see why the object ever moves at v2.

    The problem is that all these calculations are based on the idea that vertical and horizontal don't change during the 3600sec. They must change by 15o during that time. For that reason I think you are right to say that it will land behind the starting point: gravity will be acting to reduce the original horizontal velocity and it will land moving backwards.

    But the calculations seem more complex than those we use in calculations based on a stationary flat earth. I'm not familiar with doing that, so I'll have to leave you to the experts.
  9. Jan 31, 2016 #8
    I truly believe that as the object originally at altitude h1 reaches altitude h2, it gets there with the horizontal speed v1 it had a h1. The speed v1 is not sufficient for the object to keep the point of departure at height h1 directly underneath. To do so, it would have to travel at the higher speed v2.

    My other example: coin dropped from a tall tower. The coin, even if dropped straight down, will not drop directly below the point of release but ahead of it. Of course, these shifts are so negligeable that we can ignore them, the same way we ignore that earth is not a true inertial frame of reference....
  10. Jan 31, 2016 #9


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    I would now have to agree with all of that.
    My original concern was that the two effects you mention here would cancel out for the object thrown up and falling back down - ie. the object moving upward would become behind the launch point, and the object dropped from the tower would land ahead, and for the object thrown up then falling down, these would cancel. I still can't calculate whether the two effects cancel, but I think not.
  11. Jan 31, 2016 #10


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    First you say that h2 is 1000 m above the surface of the Earth, and then you say that the object leaves h1 with a vertical speed of 17,640m/s and reaches h2 1800 sec later, having lost all its vertical velocity, in which case h2 cannot be 1000m. Also, 17,640m/s is greater than the escape velocity from the surface of the Earth, so anything leaving H1 at that speed will never return. Here's the problem, Assuming a constant 9.8m/s/s deceleration, the object would have to climb 15876000 m before it comes to a stop and begins it fall back down. But that would put it 3.49 times further from the center of of the Earth than h1. Since gravity falls off with the square of the distance, this means that at that altitude acceleration due to gravity will have dropped to 0.8 m/s/s. Which means that by the time the object reaches this height, it will not have decelerating at 9.8 m/s/s for the whole climb and thus will not have lost all of its upward velocity and come to a stop. It will continue to climb to regions where gravity is even weaker.
    To correctly calculate just where the object lands with respect to where it was launched upwards requires one to delve into orbital mechanics. In essence, throwing the object straight up from the surface of a rotating Earth puts it into orbit with respect to the Earth's center. The orbital velocity at the moment of release will be the vector sum of the initial upward velocity and the tangential velocity due to the Earth's rotation at the point of release. From this you can derive the orbital parameters, and from these you can work out where the orbit will return to the initial launch height, and how long it will take to do so. Then you work out how much the Earth rotates in that amount of time and work out distance between launch and landing point.

    But you don't need to work out the exact numbers to determine whether it will land "behind" the launch point. Consider two objects with the same mass, one is toss upward from h1 and the other remains at h1. They both have the same angular momentum ( Since the upwards velocity of the tossed object is along the radial it has no effect of the object's angular velocity). Since the tossed object's angular momentum must remain constant, but its radial distance from the center of the Earth increases as it climbs, it's angular velocity must be lower during the time it is higher than h1 and thus the total angular distance it travels will be less than that traveled by the object that remained at h1.
  12. Jan 31, 2016 #11
    Thanks Janus.

    I guess your this part of your answer captures what I was trying to say and find validation for: "....Since the tossed object's angular momentum must remain constant, but its radial distance from the center of the Earth increases as it climbs, it's angular velocity must be lower during the time it is higher than h1 and thus the total angular distance it travels will be less than that traveled by the object that remained at h1...."

    This is small shift is therefore caused by taking into account the earth rotation and it is not discussed when introducing free fall motion in basic physics books because the books assume that earth is an inertial system for all practical purposes and low altitude situations.

    Same goes for the parabolic trajectory described in projectile motion chapters. In reality, the trajectory is a section of an ellipse (that is what Kepler orbital laws tell us) which is well approximated by a parabola very close to the earth surface...
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