# Mass of Jupiter using its moon Sinope

• johnq1
In summary, a student is trying to solve a homework problem using Kepler's third law and Newton's version, but is having difficulty understanding where the orbital distance of Sinope comes into play. They ask for help and it is suggested that they use F = ma to solve for the mass of Sinope. However, the student finds that their answer is off by over 37%.

## Homework Statement

Using a modified version of Kepler's third law and data about Sinope calculate the mass of Jupiter.

Keplers Third Law: p^2=a^3
Newton's Version: p^2=a^3 / M + M
p^2 = a^3/M
M=a^3/p^2
Sinope's period of orbit = 2.075 years
Average orbital distance of Sinope is 0.158 AU

## Homework Equations

Keplers Third Law: p^2=a^3
Newton's Version: p^2=a^3 / M + M
p^2 = a^3/M
M=a^3/p^2

## The Attempt at a Solution

I need Jupiter in solar mass units and then in kg by multiplying it by the mass of the sun (2.00*10^30).

Here is what I have so far but its not making sense.
p=2.075
p^2=4.305625
a^3=4.305625
a=1.62684209

if m=a^3/p^2 and p^2=a^3 I am just going to come up with the same answer of 2.075 which doesn't seem to help me at all. I think the problem is worded bad and I am confused as to where the orbital distance of Sinope comes into play.

Any tips on how to proceed would be appreciated!

Let us use F = ma for the moon of Jupiter.

What is F?

Keplers Third Law: p^2=a^3

This is only valid for orbits around the sun (or a star with the same mass),
so you can't use this.

You can find an answer only using p^2 = a^3/M (with M expressed in solar masses)

Can we use:

F = ma ( for the moon of Jupiter)

$\frac{GMm}{R^{2}}$ = mR$\omega$$^{2}$

Gm = $\frac{R^{3}4\pi^{2}}{T^{2}}$

T$^{2}$ = $\frac{4\pi^{2}R^{3}}{GM}$

The mass of the Juipter moon Sinope is 8*10^16 but I looked that up, didn't figure it from what was given.

You can find an answer only using p^2 = a^3/M (with M expressed in solar masses)

so would it be...
M= .158/2.075
M= .076144578313253 AU
.076144578313253(2*10^30) = 1.52289157 × 10^29kg

That's off by over 37% so that doesn't seem right.

Last edited:
johnq1 said:
You can find an answer only using p^2 = a^3/M (with M expressed in solar masses)

so would it be...
M= .158/2.075
M= .076144578313253 AU
.076144578313253(2*10^30) = 1.52289157 × 10^29kg

That's off by over 37% so that doesn't seem right.

It's about a hundred times too heavy. You used M = a/p, and not M = a^3/p^2

johnq1 said:
The mass of the Juipter moon Sinope is 8*10^16 but I looked that up, didn't figure it from what was given.

Why would you need to know the mass of Sinope?
Which is the relevant mass?