Video Transcript
The formula 𝑀 equals four 𝜋
squared 𝑟 cubed divided by 𝐺𝑇 squared can be used to calculate the mass, 𝑀, of a
planet or star given the orbital period, 𝑇, and orbital radius, 𝑟, of an object
that is moving along a circular orbit around it. A planet is discovered orbiting a
distant star with a period of 105 days and a radius of 0.480 AU. What is the mass of the star? Use a value of 6.67 times 10 to the
negative 11 meters cubed per kilogram second squared for the universal gravitational
constant and 1.50 times 10 to the 11 meters for the length of one AU. Give your answer in scientific
notation to two decimal places.
So we have some planet in circular
orbit around a star. It’s pretty cool that given our
understanding of physics and some fairly basic math, we can use information about a
distant planet’s orbit to learn the mass of such a large and far away object as a
star. Let’s take a closer look at the
formula we’ll use. 𝑀 equals four 𝜋 squared 𝑟 cubed
divided by 𝐺𝑇 squared. Now, we have been given values for
all the terms in this formula. But before we can substitute them
in, they should all be expressed in base SI units.
𝐺, the universal gravitational
constant, is already written in meters, kilograms, and seconds. So it’s good to go. And now let’s look at orbital
radius, 𝑟, which we know equals 0.480 AU. And while the astronomical unit is
used frequently throughout astronomy, it’s not in SI unit. So let’s convert it into
meters. We’ve been told that one AU equals
1.5 times 10 to the 11 meters. Knowing this, we can multiply 𝑟 by
1.50 times 10 to the 11 meters divided by one AU, which is just equal to one. So we can cancel out the AU. And thus, we have found that 𝑟
equals 7.200 times 10 to the 10 meters.
Next, we’ll look at orbital period,
𝑇, which is equal to 105 days, and days is not the SI unit of time. For this, we’ll need to convert to
seconds. Recall that one day equals 24
hours, an hour equals 60 minutes, and a minute equals 60 seconds. We can use these three equalities
to write three conversion factors, each of which being equal to one. Now we can cancel units of days,
hours, and minutes, leaving only seconds. And now multiplying through 105
times 24 times 60 times 60 seconds gives us an orbital period value 𝑇 equals 9.072
times 10 to the six seconds.
So our values are all set to
calculate. Substituting them in the formula,
we have 𝑀 equals four 𝜋 squared times 7.200 times 10 to the 10 meters quantity
cubed divided by 6.67 times 10 to the negative 11 meters cubed per kilogram second
squared times 9.072 times 10 to the six seconds quantity squared. Now there are a lot of units here,
so let’s make sure that they’re all working out to reach a final mass value in units
of kilograms.
First, for visual clarity, let’s
group the units over here, making sure to distribute the proper exponents. Now, let’s cancel units of meters
cubed as well as seconds squared in the denominator, leaving only one over kilograms
in the denominator or plain kilograms in the numerator. Now calculating, we have 𝑀 equals
2.684 times 10 to the 30 kilograms. And finally, rounding to two
decimal places, we have found that the mass of the star is 2.68 times 10 to the 30
kilograms.
