Find the mass of Jupiter, in kg, based on given data (in "Description") for the orbit of one of its moons.

Question

Moon: Io

Average orbital radius (r) in km: 4.22*10⁵ km

Period Time (T) in years (y): 0.00485 y (or 1.77 days)

r³/T² in km³/y²: 3.19*10²¹ km³/y²

Moon: Europa

Average orbital radius (r) in km: 6.71*10⁵ km

Period Time (T) in years (y): 0.00972 y (or 3.55 days)

r³/T² in km³/y²: 3.20*10²¹ km³/y²

Moon: Ganymede

Average orbital radius (r) in km: 1.07*10⁶ km

Period Time (T) in years (y): 0.0196 y (or 7.16 days)

r³/T² in km³/y²: 3.19*10²¹ km³/y²

Moon: Callisto

Average orbital radius (r) in km: 1.88*10⁶ km

Period Time (T) in years (y): 0.0457 y (or 16.19 days)

r³/T² in km³/y²: 3.20*10²¹ km³/y²

Answer 1

From Newton’s Law of Universal Gravitation we get

If we plug in G = 6.674 × 10^–11 we get M = (5.915 × 10¹¹)(r³/T²)

To get M in kg we need to have r³/T² in m³/s²

1 y = (365)(24)(3600) = 31,536,000 s

1 y² = (31,536,000)² = 9.945 × 10¹⁴ s²

1 km = 10³ m

1 km³ = 10⁹ m³

1 km³/y² × [10⁹/(9.945 × 10¹⁴)] m³/s² = 1.0055 × 10^–6 m³/s²

r³/T² = (3.195 × 10²¹ km³/y²)(1.0055 × 10^–6) = 3.2126 × 10¹⁵ m³/s²

So, the mass of Jupiter is calculated to be

M = (5.915 × 10¹¹)(3.2126 × 10¹⁵) = 1.90 × 10²⁷ kg

Answer 2

There's a few ways to do this. You might want to consider what you are covering in class.

This problem gives you information to calculate the mass a couple different times. In cases like this you might want to use the method from the lab portion of your class, because you often measure things many times.

Kepler's Third Law gives a³/T² ≈ G M/(4 π²). Notice the right side is constant and almost constant in the data you are given. Consider making a graph where a³ is your y-axis and T² is your x-axis.

Here's the graph, (sorry these links are difficult to share)

https://www.wolframalpha.com/input?i=linear+fit+%7B%281.77%29%5E2%2C+%284.22*10%5E5%29%5E3%7D%2C%7B%283.55%29%5E2%2C+%286.71*10%5E5%29%5E3%7D%2C%7B%287.16%29%5E2%2C+%281.07*10%5E6%29%5E3%7D%2C%7B%2816.19%29%5E2%2C+%281.88*10%5E6%29%5E3%7D

with slope 2.5437×10¹⁶ km³/day².

With this slope you can solve for the mass of Jupiter by setting the slope from the graph to the relationship from the equation.

slope = G M/(4 π²) ⇒ M = (4 slope π²)/G

This gives the mass of Jupiter of 2.01×10²⁷ kg (remember to convert units to seconds and meters) which is close to the actual mass of Jupiter 1.90×10²⁷ kg.

https://www.wolframalpha.com/input?i=%284*2.5437%C3%9710%5E16*%28%281000%5E3%29%2F%2824*3600%29%5E2%29*pi%5E2%29%2F%286.674*10%5E-11%29

You also might want to set the intercept to zero before doing your best fit line. I will leave that to you.