Introduction to significant figures - Astronomy

etween 9.05 and 9.15. Thus, to say that we know the product to the third significant figure is clearly untrue. We must limit ourselves to reporting the product as 63. It should be noted that all operations that are not addition or subtraction follow the multiplication rule. For example, 4.309 3 = 80.01, 21.7 2.6 = 3.0x10 3 , log10(34.2) = 1.53, and sin(11°) = 0.19. The results of all these operations have the same number of sig figs as the least-­‐known quantity that went into them. How do I add and subtract with significant figures? In adding and subtracting numbers, keeping track of sig figs should be done while reading a number left-­‐to-­‐right and noting where it “loses significance”, or has its last significant figure. For example, 347 loses significance after the ones place, 43.2 loses significance after the tenths place, and 5,430 loses significance after the tens place. Whichever number “loses significance” first, relative to the decimal point, limits the number of significant figures in the result. It is easiest to see this when the decimal points are aligned vertically. For example, to add 456.89 and 0.0489, you could write the operation as follows: 456.89 + 000.0489 (456.9389) = 456.93 Punching this operation into a calculator would result in the sum in parentheses, but since the top number loses significance before the other, we must cut off the sum at the hundredths place. Even though the top number has more significant figures in this case, it is still the limiting quantity. To give another example, consider 54.300 and 10.3. The former number has significance out to three places beyond the decimal, while the latter only goes out to one place beyond the decimal, and therefore limits the sum or difference of these numbers to the tenths place. Thus, 54.300 + 10.3 = 64.6 and 54.300 – 10.3 = 44.0. Now consider 1,248 and 10,200 (1.02x10 4 ). The sum or difference of these two numbers will be limited to the hundreds place, where 10,200 has its last significant figure. So, 1,248 + 10,200 = 11,400 (1.14x10 4 ) and 1,248 – 10,200 = -­‐9.0x10 3 . Note that a calculator would show the last difference as -­‐8,952, but we must round this to -­‐9,000 while keeping in mind that there are two significant figures in the result. Thus, I’ve written it in scientific notation. Can I see a more complex example? Suppose we want to find the mass of the Jupiter by observing how its moon Ganymede orbits. We know that T 2 = a 3 /M, where T is the period of Ganymede’s orbit, a is its orbital radius, and M is the mass of Jupiter. By carefully observing the position of Ganymede over several orbits, we are able to obtain averages of the orbital parameters T and a. We find that T = 6.63 days and a = 7.4 Jupiter diameters. Rearranging our equation gives M = a 3 /T 2 , but we must remember that our units are solar masses, astronomical units (AU), and Earth-­years, respectively. Thus, we must convert the orbital parameters of Ganymede into the correct units. We know that there are about 365.25 days in a year, so T = 6.63/365.25 = 3

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