Mass of Jupiter using its moon Sinope

AI Thread Summary
The discussion focuses on calculating Jupiter's mass using Sinope's orbital data and modified Kepler's third law. Participants emphasize the correct formula, M = a^3/p^2, where M is the mass in solar masses, a is the average orbital distance, and p is the orbital period. There is confusion regarding the application of the formula, particularly in converting units and incorporating Sinope's mass, which is deemed unnecessary for this calculation. The thread highlights the importance of using the correct version of Kepler's law to avoid significant errors in the mass estimation. Understanding these principles is crucial for accurate astronomical calculations.
johnq1
Messages
3
Reaction score
0

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!
 
Physics news on Phys.org
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?
 
Back
Top