Why is acceleration due to gravity a constant? [duplicate]

I just learned of Newton's law of gravitation and that distance between two bodies is a factor in the gravitational force. My question is if that's true why is the Earth's gravitational acceleration a constant $9.8\mathrm{ms^{-2}}$ as change in force mathematically should mean change in acceleration if you refer to the equation $F=ma$.

Although @JamalS's answer covers the variation of $g$ due to latitude and terrain variations, here is a simpler reason why $g$ is taken to be constant close to the Earth's surface.

If the Earth has a radius $R$, and there is a mass $m$ at a distance $r$ above the surface, the force is given by: $$F=\frac{GMm}{(R+r)^2}$$

This formula can be expanded as a series for small values of $r$, and this gives (upto three terms): $$F=\frac{GMm}{R^2}-\frac{2GMmr}{R^3}+\frac{3GMmr^2}{R^4} + \mathcal O (r^3)$$

The first term is a constant, but the second term and so on have ratios of the form $\frac{r}{R^3}$ which are very small when $r$ is small compared to the radius. (for example if $r$ is in the order of a few metres)

For most examples used in high school physics or introductory classical mechanics, $r$ usually is very small compared to the radius of the Earth, which is 6371 kilometers.

In cases like these, the other terms are as good as nothing, and the first term is the most significant contribution. This first term happens to be $mg$, where: $$g=\frac{GM}{R^2}$$

• Why the downvote? What is wrong in the answer? – Hritik Narayan Jul 1 '17 at 14:14

The acceleration due to gravity on Earth is not a constant, precisely because as you pointed out the gravitational force depends on a separation which is not constant.

Moreover, the Earth is not a perfect uniform sphere; any gravity anomaly will affect the value of $g$. At sea level, we may estimate the acceleration at latitude $\phi$ as,

$$g = 9.780327(1+0.0053024\sin^2\phi - 0.0000058\sin^2 2\phi)\, \mathrm{ms^{-2}}$$

though there are other formulas. As aforementioned, anomalies such as the free-air correction will also affect the value of $g$, to account for example for the elevation above sea level.

The Bouguer anomaly is another example which takes into account the attraction due to the terrain such as for example a mountain. More complex models also exist.

As you point out, the value of $g$ is not a constant. I will explain you how $g$ is computed, which will be useful to see why is not constant. If you take the Newton's law of gravity, you have that the gravitational force between two bodies is:

$$F = \frac{GMm}{r^2}$$

where $F$ is the modulus of the force, $G$ the universal gravity constant, $M$ and $m$ the masses of the bodies, and $r$ is the distance between them. Let me consider that $M$ is the mass of the earth and $m$ the mass of the body you want to study.

Now, it turns out that the acceleration your body is given by $F=m\cdot a$. If you look the formula above, you can carefully separe it so it looks like mass times acceleration:

$$F=m\cdot \frac{GM}{r^2}$$

The important thing to note is that $GM/r^2$ really is an acceleration, since it has units of m/s$^2$, as you can check by simple dimensional analysis. On the surface of Earth, where you are studying your object, the distance between the object and the Earth is given by the radius of the Earth. Using that $M=5.972\cdot 10^24$ kg, $r=R_E=6371$ km, we compute:

$$g = \frac{GM}{R_E^2} = 9.81970... \text{m/s}^2$$

From this simple computation you can see that $g=GM/r^2$. And this is not a constant, since it depends on the distance to the center of Earth. If you climb a mountain, then you are further from the center of the Earth, $r$ increases, and then $g$ decreases. Of course, the change is very slight, since the height of any mountain in Earth is little compared with entire radius of the planet. For example, consider that the radius of the Earth finishes at sea level, and you have an object over the Everest (~9 km over sea level), then you have $r=R_E+9$ km. If you repeat the same computation I did above, you get a value of $g = 9.79038...$ m/s$^2$. As you can see, the difference is very low. For this reason, $g$ is taken as a constant over the surface of Earth, because the variations are really negligible.

Of course, for precision computations, corrections like the one pointed out by JamalS are important and should be taken in consideration. However, for "everyday computations" the constant value is just fine.