# Video: Calculating Jupiter’s Mass

A satellite orbits Jupiter with an average orbital radius of 963700 km and an orbital period of 3.269 days. Find the mass of Jupiter, considering a day as 864.0 × 10³ s.

02:42

### Video Transcript

A satellite orbits Jupiter with an average orbital radius of 963700 kilometers and an orbital period of 3.269 days. Find the mass of Jupiter, considering a day as 864.0 times 10 to the third seconds.

We’ll label this mass of Jupiter that we want to solve for capital 𝑀. And to start on our solution, we’ll recall Kepler’s third law of planetary motion. For a much smaller body orbiting a much larger one, Kepler’s third law tells us that the period of the smaller orbiting body squared is equal to four 𝜋 squared over 𝐺 the gravitational constant times the mass of the planet it’s orbiting all multiplied by the semimajor axis of the orbiting planet’s ellipse cubed. If we draw a sketch of this orbiting satellite, we have it in an elliptical orbit moving around a focus point of the planet Jupiter.

We’re told the average orbital radius of this satellite as it moves throughout its elliptical path. And we’re also told how long it takes for the satellite to move all the way around its orbit once, that is, its period. We can call 𝑟 sub avg that average orbital radius and capital 𝑇 the period of the satellite’s orbit given as the product of 3.269 days times the number of seconds in a day. When we apply Kepler’s third law to our scenario, where capital 𝑀 the mass of Jupiter is what we want to solve for, we see that we know the period capital 𝑇 and big 𝐺 the universal gravitational constant is a known constant.

We will let big 𝐺 be exactly 6.67 times 10 to the negative 11th cubic meters per kilograms second squared. This means that when it comes to our relationship to solve for the mass of Jupiter 𝑀, the only unknown is 𝑎, the semimajor axis of our elliptical orbit. But there’s a special relationship for elliptical path which says that the average distance from the path to the centre of the path at what we’ve drawn as an origin is equal to the semimajor axis of that orbit. In other words, 𝑎, the semimajor axis, is equal to 𝑟 sub avg. So we now know everything in this expression except the mass value we want to solve for.

We rearrange the expression so that 𝑀 is by itself on one side. When we plug in these values, we’re careful to convert our expression for 𝑎 from units of kilometers to units of meters. Entering this expression on our calculator, to four significant figures, we find that 𝑀 is equal to 6.641 times 10 to the 25th kilograms. That’s the mass of Jupiter based on these numbers.