Introduction to significant figures - Astronomy

0.0182 years, where we keep only three significant figures. We also know that there are 1050 (1.05x10 3 ) Jupiter diameters in an AU, so a = 7.4/1050 = 0.0071 AU, where we are now limited to two sig figs. Next, we compute M = a 3 /T 2 = (0.0071) 3 /(0.0182) 2 = 0.0011 solar masses. Finally, we use our knowledge of the mass of the sun (1.992x10 33 g) to find that M = 2.1x10 30 g. Note that the final answer does not contain more sig figs than anything that went into its calculation. On a final note, if we were to treat the above work as one long calculation, we would only have to truncate the final answer. In other words, if the converted T and a values were not to be reported and were only calculated for use in the equation M = a 3 /T 2 , we wouldn’t need to chop them off at 0.0182 years and 0.0071 AU. We could keep all the precision that our calculator gave us until we finally arrived at M, which would still be equal to 2.1x10 30 g. We’d still have to limit our answer to the number of sig figs in the least-­‐precise input quantity (a = 7.4 Jupiter diameters), but we’d save rounding for the end of the calculation. Whether or not an intermediate quantity such as T or a is to be reported determines when you do your rounding. In the lab You Can Weigh Jupiter, for example, you are specifically asked for the converted values of T and a, so our first method of finding M is appropriate. The reason I mentioned this is that you may come up with a slightly different answer when you save all the rounding in your calculations for the end of the problem. 4