Why free-fall acceleration is considered constant rather than increasing? [duplicate]

The force acting on a body of mass $$m$$ is $$mg$$, where $$g$$ is acceleration of free fall ! But why should there be a #uniform acceleration of free-fall in the first place? As per Newton's universal law of gravitation there's an inverse square law of force acting with respect to separation ; Thus as object falls towards Earth separation decreases and so the force, and hence acceleration, should likewise increase! How to reconcile this argument with a universally accepted constant acceleration $$g$$?

marked as duplicate by user191954, Aaron Stevens, Alfred Centauri, Qmechanic♦Oct 4 '18 at 13:07

• Becuse problems with which we deal in mechanics are usually at the surface of earth and if there is any change in height it will be so less that it would not significantly change thhe value of g – Sourabh Oct 4 '18 at 11:08
• Possible duplicate of Why is acceleration due to gravity a constant? – user191954 Oct 4 '18 at 11:56

You're correct. The further away you are from Earth, the lesser the gravitational acceleration.

The way to reconcile that with the use of a constant $$g$$ for gravitational acceleration in most contexts is to look at how much your height is changing in the situation, and how much that would change the value of the acceleration.

You're likely to find that in most situations, the change in this acceleration is lower than the accuracy of the gravitational acceleration that you use. For example, even at 400 km above the Earths surface where satellites orbit, the acceleration due to gravity is still about 90% of the gravity on the surface.

People who design and analyze things that are far from the Earths surface do have to take the non-constant gravitational acceleration into effect. On the small scale it is usually applied to by the layman, the change is so small that it gets overshadowed by the lack of precision in all the values you use.

• Gotcha ! Strictly speaking free-fall acceleration g indeed increases, but for all practical purposes it hardly changes at all :g actually varies as 1/r², where r is the separation between the two centres of mass. Thus, even if dg/g = 2dr/r, the latter is barely of the order of 10⁻⁴ for the first 100 m height – Jean Luc Baudoin Oct 5 '18 at 16:39

You can reconcile this by the fact that Newton's gravitational law is nearly constant on such a small length scale.

We start with the inverse square law: $$\vec F=-\frac{GMm}{r^2}\hat r$$ where $$M$$ is the mass of the Earth, and $$m$$ is the mass of an object on the Earth's surface.

Now let's pick a point on the surface of the Earth a distance $$r_0$$ from the Earth's center (i.e., $$r_0$$ is the radius of the Earth, assuming a spherical Earth). The value of $$g$$ (the constant acceleration due to gravity at the Earth's surface) is then given by $$g=\frac{GM}{r_0^2}$$

Now, let's look at what the gravitational force is at a distance $$\Delta r$$ from the surface:

$$\vec F(r_0+\Delta r)\approx \vec F(r_0)+\Delta r\left.\frac{d\vec F}{dr}\right|_{r=r_0}=-\frac{GMm}{r_0^2}\hat r+\frac{2GMm}{r_0^3}\Delta r\ \hat r=-gm\ \hat r+\frac{2gm}{r_0}\Delta r\ \hat r$$

Now, typically when we are talking about accelerations on Earth's surface, we are looking at $$\Delta r\ll r_0$$ , or $$\frac{\Delta r}{r_0}\ll 1$$. (Or if you don't replace variables with $$g$$, you have $$\frac{\Delta r}{r_0^3}\ll \frac{1}{r_0^2}$$, or $$\Delta r \ll r_0$$). Therefore, it is a pretty good approximation that if you are looking at distances close to the Earth's surface, we can treat the gravitational force as constant. $$\vec F=-gm\ \hat r$$

From here, we can use Newton's second law for any object of mass $$m$$ accelerating near Earth's surface only due to the force of gravity $$\vec a = \frac{\vec F}{m}= -g\ \hat r$$ giving us our constant acceleration.