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)


Is it correct to say that different acceleration of free fall at the poles and equator is due to centripetal acceleration?

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.

edit 2

please avoid using pseudo forces to answer the question as I want to understand what's going on exactly from a true inertial frame

  • $\begingroup$ Your book is talking about net acceleration, i.e. $g-a_{cent}$, free fall acceleration $g$ is same for all bodies, i.e. all bodies falling in vacuum under influence of gravity field, will change speed at the same rate. So in principle $g$ can be a measure of gravity field strength. $\endgroup$ Commented Sep 4, 2020 at 10:24
  • $\begingroup$ Net acceleration from what frame? In the intertial frame, electric field strength IS net acceleration i $\endgroup$ Commented Sep 4, 2020 at 10:27
  • $\begingroup$ In rotating Earth frame, where centrifugal pseudo force is acting, hence such difference from $g$ $\endgroup$ Commented Sep 4, 2020 at 10:31
  • 1
    $\begingroup$ Yes the value for g changes (that is net acceleration is slightly lower at the equator due to centrifugal forces produced by the planet's rotation). This difference is due only to the fact that the earth rotates. If the earth was stationary, g would have the same value everywhere (as you said, assuming earth to a perfect sphere with uniform density and $g = GM/r^2 = F$ per unit mass by definition). $\endgroup$
    – joseph h
    Commented Sep 4, 2020 at 10:32
  • 1
    $\begingroup$ @OVERWOOTCH No it is due to centrifugal force not centripetal force. $\endgroup$
    – joseph h
    Commented Sep 4, 2020 at 10:47

5 Answers 5


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.

  • $\begingroup$ Just to be clear, does this mean that If i shoot a ball vertically upward, provided that the heigh is high enough, it wont land exactly in the pervious position? Since the ball will already have a tangential velocity component that can’t be maintained due to decreasing g? $\endgroup$ Commented Sep 7, 2020 at 3:39
  • $\begingroup$ That is actually also true! You have intuited something called the Coriolis effect, where what it means to be corotating with the Earth is a faster velocity at further radius from the axis, so if you jump you should lack that velocity and land slightly west of where you started. This also affects motion along Earth’s surface at latitudes other than the equator, as you move towards the axis with Northward motion by a factor of the sine of your latitude. $\endgroup$
    – CR Drost
    Commented Sep 7, 2020 at 6:19
  • $\begingroup$ With that said you do not need g to diminish (presumably you mean by the inverse square law?) in order to observe this and indeed I do not think that would actually make a difference directly. $\endgroup$
    – CR Drost
    Commented Sep 7, 2020 at 6:25
  • $\begingroup$ Thank you for your answer. I get it now $\endgroup$ Commented Sep 7, 2020 at 6:57
  • $\begingroup$ Really nice answer. $\endgroup$ Commented Sep 11, 2020 at 16:25

Free-fall acceleration is assumed to be the constant number g whereas the field strength is the acceleration just defined for any radius. The closer you get to the surface the closer the approximation gets.


If you take a free fall at the equator and with Newton gravitation law you obtain this equation:


because the free fall height h is much smaller then the earth radius R we can take the Taylor expansion and obtain:


thus as long as the mass falling (free fall) g is depending on the free fall height, but this term is very small.

$$g_r=9.79$$ $$ g_h(10000~[m])=0.03$$


Gravitational field strength is simply the vector $\vec{E}$ at a point such that a mass $m$ at the point experiences the force $m\vec{E}$ due to gravity. No confusions there.

Acceleration of free fall as defined in your book seems to be the acceleration experienced by us with respect to the rotating Earth frame, which is an accelerated frame.

For an analogy, you can consider your accelerating measured by you when you are inside a life accelerating downward at $9.8 ms^{-2}$. Even though your acceleration is $g$ as seen from an inertial frame, you have zero relative acceleration with respect to the lift floor (as both have the same acceleration). Since both you and the lift floor are falling at the same rate, your feet are not trying to penetrate into the lift floor. If your feet don't apply any force on the lift floor, then the lift floor does not apply any force on you. In the absence of reaction force from the floor, you feel weightless inside the lift (it's the reaction force that we feel as our weight).

The rotating Earth frame is a similar scenario. The acceleration of the ground beneath you is again pointing in the downward direction (the centripetal acceleration). The difference is that this acceleration is less than $9.8ms^{-2}$, so it's not enough to make you feel completely weightless. It's enough for you to feel less weight than $mg$ though.

It's like the case of an elevator frame which is accelerating downward with an acceleration less than $9.8ms^{-2}$. Your feet do try to penetrate into the floor (as gravity pulls you in with a rate higher than the acceleration of the floor). So the floor does apply some normal force on you to keep you moving with the same rate as the floor is moving (prevent you from penetrating)

Let's solve for this normal force:

As seen from an inertial frame, you are having the same accelerating $a$ as the lift (as you both must fall together, your weight is not enough to break the lift floor). The forces on you are $mg$ and the normal reaction $N$ from the lift.

So $$ma=mg-N$$

where $N$ is the normal force, giving $N=m(g-a)$. This reaction force is what you feel as your weight.

So if you analyse a person standing on a rotating Earth from an inertial frame, you would see that the person is being pulled by gravity at a rate of $g$, while the ground beneath him is being pulled towards the center at a rate $a$ which smaller than $g$. To prevent you from penetrating into the ground, the ground must make sure that you have the same falling rate as it has. To make sure of that, it has to apply a reaction force $m(g-a)$ on you.


No, it is not.

What makes the free fall acceleration different at the two poles and at tge equator is the shape of the Earth which actually is not a perfect sphere as usually referred to. It has an ellipsoid shape with two flat surfaces on the two poles areas which makes the radius from tge Earth's centre of mass to the two poles shorter than tge radius from its centre of mass to the equator, and tge gravitational force is inverse proportional to the distance "R" between objects and the Earth's centre of mass It means that greater distance results in weaker gravitational force on the Earth's surface of the equator, abd stronger gravitational force on Earth's surface of the two poles Note that the angular velocity of the Earth's rotational motion around its axis is always the same or constant.


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