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I am studying Fowles' Analytical Mechanics. I thought I understood the concept and derivation of acceleration in a non-inertial frame. enter image description here I am not sure how to explain the highlighted question.

My try: Since the body is not moving initially in my frame $z=h$ which is rotating with Earth, it may have an initial velocity bigger than my frame origin's when someone sees it 'outside'(not related to Earth rotation). But if this is correct, shouldn't the initial velocity of the body seen in my frame when released($t=0$) exist?

I am confused. Please help me understand the matter.

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  • $\begingroup$ I'm sorry but it's not totally clear what the textbook example is asking us to explain. $\endgroup$
    – psitae
    Commented Dec 22, 2018 at 17:09
  • $\begingroup$ Think about the tangential speed of the body at height $h$ and then at height zero above the Earth's surface. $\endgroup$
    – Farcher
    Commented Dec 22, 2018 at 17:38

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The author is suggesting in the highlighted portion that a person unfamiliar with Newton's First Law (i.e. inertia) might believe that the object will fall straight downwards while the earth spins beneath it. Since the earth turns to the east, the effect would be that the ball (which is falling straight down) will appear to move westward. This is, of course, not what happens.

Because the object is initially attached to something (your arm, a platform, a geosynchronous satellite) which is rotating at the same angular speed as the earth, then the object has some tangential velocity eastward when released. However, that tangential velocity is given by $v_t=\omega r_0$, where $r_0$ is the initial height of release relative to the axis of rotation. Since the object is descending towards the axis of rotation, it finds itself among other objects with progressively slower and slower tangential velocities (due to smaller $r$). Therefore, it moves eastward more quickly than the things on the ground, and thus lands to the east of the point directly beneath the point from which it is dropped.

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  • $\begingroup$ Why smaller $'r'$? $r$ increases as you move down from pole to equator, due to the earth's shape. $\endgroup$
    – Eisenstein
    Commented Jan 27, 2021 at 7:13
  • $\begingroup$ @Eisenstein Yes, but in this example, we are not talking about that. As the object falls from a great height above the Earth's surface to the ground, its radius of rotation decreases. $\endgroup$
    – Geoffrey
    Commented Jan 27, 2021 at 16:07
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I agree that the highlighted question is weird, because that question doesn't deal with physics at all. If meaningful at all it is in the area of psychology.

The premise of the highlighted question is fundamentally flawed. It suggests thinking in terms of say, Aristolian mechanics (for lack of a better expression), but if your world perception is Aristotelian mechanics then the suggestion that the Earth might be rotating is absurd to begin with.

Apart from that, the stated problem has no particular relation to the case of non-inertial frames.

Dropping an object from a height above the ground is a particular instance of the thought demonstration known as 'Newton's cannonball'. Because the Earth is rotating: when an object is dropped with zero initial horizontal velocity relative to the Earth it actually does have an angular velocity relative to the Earth's axis of rotation.

Once released the subsequent motion is a Kepler orbit (if we treat air friction as negligable). As we know, orbital motion has the following property: as the Earth's gravitational force pulls the orbiting object closer the angular velocity of the object increases. (Vivid example: the highly elliptic orbits of comets. As the comet moves towards perihelion its angular velocity increases dramatically.)

In this case an efficient method to calculate a good approximation for the change in angular velocity is to apply conservation of angular momentum. Since angular momentum is conserved at the point of impact the object has a higher angular velocity then when it was released. Since 100 meters is a very small height it's probably a sufficient approximation to calculate it as a linear increase in angular velocity. For a more precise calculation you'd have to integrate the angular velocity over time.

Conversely, a wrong calculation would be to treat this as a case of conservation of linear momentum. That is, given the difference in height there is a difference in linear velocity. If you would use only that difference in linear velocity then the calcularion will arrive at a wrong answer.

It's not clear why you are referring to this problem as one of acceleration relative to a non-inertial frame. This problem is unrelated to that: to arrive at a correct answer there is no need to consider motion relative to some non-inertial frame. How does Fowles present this problem: as one of orbital mechanics, or as one of motion relative to a non-inertial frame?

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