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