Determining the Mass of Jupiter Using Io's Orbital Parameters

Click For Summary

Homework Help Overview

The problem involves determining the mass of Jupiter using the orbital parameters of its moon, Io, specifically its orbital period and radius. The context is rooted in celestial mechanics and Kepler's laws of planetary motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various equations related to orbital mechanics, including Kepler's Third Law and the gravitational force equation. There are attempts to relate the orbital period and radius to the mass of Jupiter, with some participants expressing confusion about the steps involved.

Discussion Status

There is an ongoing exploration of different approaches to the problem, with some participants providing guidance on relevant equations. Questions about the application of these equations and the interpretation of variables are being raised, indicating a productive dialogue without a clear consensus on the method to use.

Contextual Notes

Participants note the need to convert units, such as the orbital period from days to seconds, and question the relevance of certain equations based on their sources. There are mentions of potential errors in calculations and the importance of understanding the relationships between variables.

Husker70
Messages
89
Reaction score
0

Homework Statement


Io, a satellite of Jupiter has an orbital period of 1.77 days and an orbital
radius of 4.22 x 10^5 km. From this Data determine the mass of Jupiter


Homework Equations


4pie^2/GM
Kepler's Third Law

The Attempt at a Solution


I keep getting turned around. I know the answer but I
keep finding different ways to start
I also used T^2 = Ka^3
But that seems independ of Mass?
Thanks,
Kevin
 
Physics news on Phys.org
Husker70 said:

Homework Statement


Io, a satellite of Jupiter has an orbital period of 1.77 days and an orbital
radius of 4.22 x 10^5 km. From this Data determine the mass of Jupiter


Homework Equations


4pie^2/GM
Kepler's Third Law

The Attempt at a Solution


I keep getting turned around. I know the answer but I
keep finding different ways to start
I also used T^2 = Ka^3
But that seems independ of Mass?
Thanks,
Kevin

A generic equation for orbitting bodies that you may want to commit to memory or have handy for some quick calculations is:

T=2\pi a\sqrt{a/GM}.
 
That was nowhere to be found in my book. Thanks...
Should I next find acceleration by taking 1.77 days
and making it 152928s and then find the circumference
to be 2.65 x 10^9 and then velocity is 17338 m/s?
Is this the right direction to go.
Kevin
 
Husker70 said:
That was nowhere to be found in my book. Thanks...
Should I next find acceleration by taking 1.77 days
and making it 152928s and then find the circumference
to be 2.65 x 10^9 and then velocity is 17338 m/s?
Is this the right direction to go.
Kevin

The a given in that equation is the radius. I remember it from long ago, but, if it's not found in your book, don't bother with it. We can derive it from scratch, can't we? ;)

You want to find the mass of the planet, given by the equation \vec{F}=\vec{G}Mm/\vec{R}^2. You know that this system follows uniform circular motion, too: \vec{a}=\vec{v}^2/\vec{r}. And, that period is time, which can be related with displacement and velocity: \vec{x}/\vec{v}=T. Do you agree?
 
I agree
 
Husker70 said:
I agree

Then you can solve for M. :) Let me know what you try.
 
Except how can I use your first equation when I don't know the mass of Io?
 
Husker70 said:
Except how can I use your first equation when I don't know the mass of Io?

Because \vec{f}=m\vec{a} so the small masses cancel. ;)
 
So a = G(M/r^2)
 
  • #10
Husker70 said:
So a = G(M/r^2)

Yup, which is also equal to the quotient between the square of velocity and radius. Just relate the equations, you'll end up with M.
 
  • #11
So I get v^2 = G(M/r)
I'm not sure how to get M =
 
  • #12
Husker70 said:
So I get v^2 = G(M/r)
I'm not sure how to get M =

\vec{v}=2\pi\vec{r}/T...
 
  • #13
I'm just not seeing it sorry
 
  • #14
Husker70 said:
I'm just not seeing it sorry

\frac{4\pi^2\vec{r}^2}{T^2}=\frac{\vec{G}M}{\vec{r}}\rightarrow M=\frac{4\pi^2\vec{r}^3}{\vec{G}}.
 
  • #15
Thanks a lot for the help but using that equation I don't get the right answer for some reason.
Kevin
 
  • #16
asleight said:
\frac{4\pi^2\vec{r}^2}{T^2}=\frac{\vec{G}M}{\vec{r}}\rightarrow M=\frac{4\pi^2\vec{r}^3}{\vec{G}}.

You forgot the T^2 in the formula.
And never mind the vectors. G is not a vector and r^3 is the magnitude cubed and not the vector cubed.

M=4Pi^2 r^3/(G T^2)

I've got about 1.9 x 10^27 and it's very close to the accepted mass of Jupiter.
The period should be in seconds, right?
 
  • #17
asleight's posts are so full of errors, you're better off ignoring them.
 
  • #18
I figured that out. Thanks again. You were a great help.
Talk to you soon.
Kevin
 

Similar threads

Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
27K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
7K