Before I ask my question, I want to make it clear that I am aware that the field strength is the property of the field while the acceleration of free fall is a property of a mass falling in a vacuum under gravity.
I originally thought that the acceleration of free fall is defined as the (net) acceleration of an object due to gravitation i.e $g=F_g/M$ as measured from an intertial frame of reference, and is thus numerically equal to the field strength.
however, my book says that the "acceleration" of free fall varies from the equator to the poles due to rotation of the earth (assuming perfecltly spherical earth with uniform mass distribution) whereas the field strength obviously does not. Does this mean that the "acceleration" of free fall is the rate of change of speed, as measured from the rotating non inertial frame rather than the total acceleration? If so, how would we even separate this component from the true acceleration? ( I suspect this involves pseudo forces)
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I have found another answer on the site but I have having a lot of trouble interpreting it.
Towards the centre of the Earth aEquator which is the free fall component and again towards the centre of the Earth REarthω2 (centripetal acceleration) which makes the mass go round in a circle of radius REarth which is the radius of the Earth.
Now the equation of motion is Fmass,Earth=maEquator+mREarthω2 from which you can see that aEquator<aPole.
This user has divided the total acceleration into 2 components; free fall(whatever that means) and centripetal; which doesn't make sense to me at all. Divining the acceleration into 2 perpendicular components like X & Y or centripetal & tangential makes sense but the centripetal and free fall components would be directed in the same direction.
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please avoid using pseudo forces to answer the question as I want to understand what's going on exactly from a true inertial frame
Best Answer
So here is the basic question I need you to answer in your head: is orbit a special case of free-fall? I think that will cut right to the essence of the question!
We grew up on this planet. We like to think about things happening on this planet. It's very easy to forget that space exists, that the Earth is round, that gravity eventually diminishes. If you do that, there is an acceleration g which varies depending on where you are, $g = 9.80\pm0.03 \text{ m/s}^2,$ and in the flat approximation free-fall from some origin is described by the equations $$\begin{align} x(t) &= u_0~t,\\ y(t) &= -\frac12 g t^2 + v_0 t \end{align}$$ for some parameters $u_0$, $v_0$. Since it sounds kind of weird to say that something is falling when it is traveling upwards, we often stipulate that $v_0 = 0$ and the physical trajectory is $y(x) =- \frac{g}{2 u_0^2} x^2$.
A comparison to a circle with radius $r$, $$y=\sqrt{r^2 -x^2} = r\sqrt{1 -(x/r)^2}\approx r - \frac1{2r} x^2,$$ lets us say that at $t=0$ the radius of curvature of this trajectory is $r = u_0^2/g.$ And so our assumption that the Earth is flat is really an assumption about $r\ll R$ which is really an assumption about how fast things are going sideways, $u_0\ll \sqrt{g R}.$ Just to give you an intuition this speed is at Earth's surface 7.9 km/s or 28,000 km/hr or 18,000 mph or Mach 23. So in terms of your everyday experience with things moving less than half the speed of sound, this is a reasonable approximation. On the flip side I have heard people say (possibly inspired by this XKCD? Uncertain source) that “space isn’t far, space is fast,” referencing these same speeds.
But at your most extreme, a high enough sideways velocity causes orbit, you “fall” over a short time exactly as much as the curved surface of the Earth is falling away under you due to its own curvature, and thus you stay at a fixed distance from the surface. What does this look like in our flat-Earth approximation, where we pretend the surface does not curve away? As the surface actually accelerates downwards away from you, if you pretend that it is flat then there is an apparent (i.e. not actual) force drawing you upwards with magnitude $m v_x^2/R$. It scales proportional to $m$ so it appears to be an attenuation of $g$, it hits all masses in your jet airplane (or rocket or whatever) proportional to their mass just like gravity does. In the flat approximation, “orbit” becomes moving at Mach 23 so that this upwards drift acceleration cancels $g$ out completely. You might say that this is not “free-fall” because of this upwards drift force, maybe.
Generally physicists still call these orbits a special case of “free-fall,” but we are acknowledging that the Earth is round and thinking of it from an inertial point-of-view, looking down from outer space. So in this context all we see is your acceleration towards the surface which is due entirely to gravity. But if you’re looking at the same thing from the Earth’s surface it looks like the acceleration is zero, so something must be canceling out the direct effects of gravity.
Here is the last fact that you need to know about this situation. You probably thought that you were traveling at $v_x = 0$ right now as you “sit still” and read this. You are actually traveling at around Mach 1 in the direction that you instantaneously call East. The actual amount is more like Mach 1.35 times the cosine of your latitude, and it causes an upwards drift force that has the magnitude of $\cos^2\theta~0.034\text{ m/s}^2$ which has already been factored into that above $9.80 \text{ m/s}^2$ number for $g$. This is because every day you complete an orbit around the center of the Earth, traveling a distance of about $2\pi R \cos\theta$ per day. So this affects you, but only a little bit, attenuating $g$ by 0.3%. To first approximation, we can still speak about freefall parabolas on Earth’s flat surface, just attenuating $g$ appropriately. (Control question: this attenuation by 0.3% seems awfully small, no? But why—what do you think would happen if it were 80% or 100%?)
So what we are left with is that unfortunately both are right, in different contexts. If you are in the inertial context, staring from space at a rotating Earth, then yes, you will define free fall purely by the gravitational field which diminishes like the inverse square of the distance from Earth’s center. But if you are in the surface context, watching planes and trains and automobiles moving around nearby, then you are very likely to want to describe free-fall with an acceleration that (a) includes rotation and mass discrepancies in Earth and anything else that affects the local falling of a ball, and (b) either does not diminish at all with height, or for very high flying planes or balloons, maybe one that diminishes linearly with height. So your definition of free-fall acceleration needs to include these terms because your surface is accelerating but you want to conceptualize it as fixed.