Physicopoeia

φυσικοποιΐα
THINGS I WISH I'D KNOWN WHEN I STARTED TEACHING PHYSICS

What is my local value of g?

We all get so accustomed to seeing = 9.81 ms⁻² that it might be a surprise to realise how much it varies over the Earth's surface, even at sea level. For example, from this worksheet, at the North Pole, g = 9.87 ms⁻². Two further questions which might then occur to pupils are:

1) If it varies, why did we choose this value in preference to any other? Is it just an average?

2) How can I find out my local value?

To which the answers are:

1) It was set by an international committee as long ago as 1901 - google '3rd conference on weight and measures 1901' for more history. By definition, its exact value is = 9.80665 ms⁻² (so, firstly, the usual 3sf value is actually rounded up a bit) and was based on the value that would be expected at sea level at a latitude of 45°N. Since this latitude is exactly halfway between the equator and north pole, it represents an average in that sense. It's also reasonable to say that many, if not most, scientific experimental work, especially at the time, would be done not too far from that latitude. It's not a simple mean of the polar and equatorial values because does not vary linearly between the equator and the poles.

2) There are several internet sites where you can calculate a local value for g, based usually on your latitude and height above sea level. However, the density of the local rocks also will make a (small) difference, so geological survey data needs to be included as well for a really accurate answer. Wolfram Alpha has a widget that only requires you to input your location and then gives you your value of accurate to rather a lot of significant figures. If you're in the UK, especially further north, you might be surprised to see how close your local value is to 9.82 rather than 9.81!

A possibly less accurate but more computationally interesting way to get a local value of based only on latitude and height above sea level, with some corrections based on 1960s observations of artificial satellite orbits, is from Kaye & Laby (1973):

[g equation]

Where:

g = 9.7803184 ms⁻²

β = 0.0053024

β = 5.9 x 10⁻⁶

φ = Your latitude (degrees)

= height above sea level (m)

Having checked for my own location, it seems that the above formula and the Wolfram Alpha widget do not concur exactly with each other. It's difficult to know why, since I doubt the local geology would make the kind of difference seen. The widget is unclear whether or not height above sea level is taken into account, or even how accurate it is outside the US. I daresay there are other resources out there that would give similar variation in results.

 

It would seem that the take-home lesson is that varies depending on where you are due to a number of factors.