# How to theoretically calculate the value of gravitational acceleration of my town?

We normally consider the value of gravitational acceleration $g = 9.8 m/s^2$ while solving the problem. But that is the value of $g$ at poles (if I am not wrong).

My teacher have given as homework to find the value of $g$ of my town.

I know it should be approximately 9.8, but I have no idea how to find it.

• I'm guessing this has to do with the centrifugal force at the coordinates of your town on account of the Earth's rotation. There are quite a few causes other than this for regional variations in $g$, like presence of mountains, other geological variations, but I don't think a high-school level class will involve that. – Pranav Hosangadi Dec 14 '14 at 7:20
• Did your teacher hand out an absolute gravimeter? If he didn't, then you need to look it up or calculate it from the ellipsoid shape of the planet to first order. – CuriousOne Dec 14 '14 at 7:20
• With the help of experiment or theoretically? – Paul Dec 14 '14 at 7:56
• @Paul theoretically – Freddy Dec 14 '14 at 8:03
• Tip: Most likely your teacher wants you to assume that Earth is spherically symmetric (even though this assumption is actually inconsistent with the fact that Earth is spinning). – Qmechanic Dec 14 '14 at 10:28

Gravitational acceleration at the surface of earth varies with latitude (North-South position). This is due to 1) the outward centrifugal force produced by Earth's rotation, and 2) the equatorial bulge (itself caused by Earth's rotation). Both effects cause the gravitational acceleration to decrease away from the poles. The net effect is a gravitational acceleration of about $9.832$ $m/s^2$ at the poles, and about $9.780$ $m/s^2$ (0.5 % lower) at the equator.

For any sea level position on Earth (your city, Jamnagar, at 17 m elevation can indeed be considered to be roughly at sea level), we can estimate the gravitational acceleration $g$ at any latitude $\phi$ using Helmert's equation:

$$g(\phi ) = g_0 (1 + 0.0053024 sin^2 \phi - 0.0000058 sin^2 2\phi)$$

where $g_0 = 9.780327 m/s^2$ denotes the gravitational acceleration at the equator.

Local variations in Earth's topography and geology cause local deviations from the above formula. Such local variations are known as gravitational anomalies. To include these would go way beyond the intentions of the exercise.

• Thanks for answer. I am thinking why my teacher asked to find the stuff which requires something that i haven't learned! – Freddy Dec 14 '14 at 13:55
• @Freddy the problem your teacher gave you was probably to make you search for information about the problem yourself. This would make you learn both about finding information and the subject itself. Smart teacher. – Communisty Jun 28 '17 at 13:20

You can calculate the value of acceleration due to gravity in your town by finding the latitude of your town. Once you find the latitude use this equation $g'=g-R\omega^2\cos^2\lambda$.

Here $R$ is the radius of the earth, $\omega$ the angular velocity of the earth and $\lambda$ the latitude. This equation comes from the fact that the earth is rotating and a particle thus will experience a centrifugal force.

Source

The value of acceleration due varies with altitude as well. It actually decreases with increase in altitude.

The following formula approximates the Earth's gravity variation with altitude:

$g_h = g_0\left(\frac{r_\mathrm{e}}{r_\mathrm{e}+h}\right)^2$

where

• $g_h$ is the gravitational acceleration at height $h$ above sea level.
• $r_e$ is the Earth's mean radius.
• $g_0$ is the standard gravitational acceleration.

So all you need to do is find the elevation of your town. But remember that this formula treats the Earth as a perfect sphere with a radially symmetric distribution of mass and you will only get an approximate value. However, if you take into account the fact that the earth is spinning about it's axis then it will depend on the latitude (this is because of the centrifugal force and also because of the "equatorial bulge") but this is usually very small. Moreover, it also depends on a host of other factors like the density of the ground, air density etc. You can see the Wikipedia article for more details.

• You say that the centrifugal contribution is really small, but what about that height variation? From see level to $2km$ height, it gives less than $0.1\%$ deviation. You should give more number to support your statement that the answer by @Paul is not relevant. – Bernhard Dec 14 '14 at 9:05
• Well, I plugged in the numbers into both formulas. And the 'altitude formula' gives a value of $9.79m/s^2$ and the 'latitude formula' gives $9.77m/s^2$. (I stand corrected, both of them give nearly equal values) – noir1993 Dec 14 '14 at 10:30
• Both the formulas hold for different situations. I am thinking out loud here but I think what if we applied the latitude formula first to find the effective $g$ at that point and them used that value of $g$ as $g_0$ in the altitude formula, we might get a more precise value of g at the town which takes both factors into account? I am having doubts because as Qmechanic pointed out in the comment that the assumption that the earth is spherically symmetric (as we did in deriving the altitude formula) is inconsistent with the fact that earth is spinning. – noir1993 Dec 14 '14 at 10:44