Significant Figures - Sig Figs - s.f.

The Obvious Stuff

Circumference = 2 p R

For R= 6400 km then C=40,000 km - 2s.f. input -> 2 s.f. output

For R = 6378.1363 km then C = 40,075.012 km

Except the 8 s.f. value of R is for the mean equatorial radius. What is the meaning /value/purpose of a mean equatorial circumference calculated to 8 s.f.? It depends on the purpose. It MAY have some meaning somewhere.

Less Obvious Stuff

g = GM/R² where G = 6.6720 x 10^-11, M= 5.9742 x 10²⁴ kg and R = 6378.1363 km

When is it reasonable to quote g=9.7982429 m s^-2 and when is it not reasonable?

If you quote 5,6,... sig figs thenyou need to be sure that you have included ALL of the effects that apply at that level of accuracy. For example, if you are doing a satellite trajectory problem and you want 9 sig fig accuracy (e.g. a 4 billion dollar communications satellite or to get to Mars or Pluto) then are you taking into account atmospheric drag, higher-order gravity fields, magnetic torquing, perturbations from the Sun, Moon, etc - and other effects that contribute to the acceleration of the object. Otherwise it is not just unnecessary to quote the extra sig figs - it is dishonest.

Serious Error Stuff

What is the altitude above the surface of the Earth where is g 90% of the surface value?

Let's call 0.9 g = g90 and the radius where this is the local gravitational acceleration R90.

Method A - For earth we have g = 9.80 m s^-2 so that g90 = 8.82 m s^-2

Then R90 = (GM/g90)^1/2 = 6719.2 km

The altitude where g is 90% of the surface value is then R90 - R = 6719.2 - 6378.1 = 341 km

Method B - Alternatively, we could simplify the problem - with a little algebra.

0.90 = g90 / g = [ GM / R90² ] / [GM/R²] = [R/R90]²

R90 = R / (0.9)^1/2 = R x 1.0541 = 6723.1 km

This value of R90 does ot depend on accuracy of G, M, g. And you could apply it to any planetary object.

The altitude is then 0.0541 R = 345 km

Which answer is correct? Which is more accurate? Method A or B?

These two methods have answers that differ in the SECOND sig fig. Check the numbers - I do not think I made a calculator error. One of the answers is clearly way off.

Method B has

• limited opportunity for calculator error - punching in the wrong number by mistake
• does not depend on accuracy of M, G,
• does not involve subtracting 2 large numbers- a major source of potential errors

So, when doing such calculations I recommend the following:

1. Think about the underlying physics of the problem and pick the simplest formula that applies
2. Is there a way you can turn the problem into a ratio? Then you can cancel out terms.
3. Simplify, simplify - a little algebra goes a long way to improve accuracy and save work at the calculator
4. Check whether the number of sig figs of your answer is justified by not just the accuracy of your input parameters but also the physics of the problem you are trying to solve - what are the underlying assumptions? Are you ignoring a higher-order effect which might limit the numbers of sig figs you feel confident quoting?
5. Check your answer makes sense!