Determining the Mass of Jupiter Using Io's Orbital Parameters

In summary, the satellite Io 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.
  • #1
Husker70
90
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
 
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  • #2
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:

[tex]T=2\pi a\sqrt{a/GM}[/tex].
 
  • #3
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
 
  • #4
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 [tex]\vec{F}=\vec{G}Mm/\vec{R}^2[/tex]. You know that this system follows uniform circular motion, too: [tex]\vec{a}=\vec{v}^2/\vec{r}[/tex]. And, that period is time, which can be related with displacement and velocity: [tex]\vec{x}/\vec{v}=T[/tex]. Do you agree?
 
  • #5
I agree
 
  • #6
Husker70 said:
I agree

Then you can solve for M. :) Let me know what you try.
 
  • #7
Except how can I use your first equation when I don't know the mass of Io?
 
  • #8
Husker70 said:
Except how can I use your first equation when I don't know the mass of Io?

Because [tex]\vec{f}=m\vec{a}[/tex] so the small masses cancel. ;)
 
  • #9
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 =

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

[tex]\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}}[/tex].
 
  • #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:
[tex]\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}}[/tex].

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
 

1. How is Io's orbital parameters used to determine the mass of Jupiter?

Io's orbital parameters, such as its orbital period and distance from Jupiter, are used in Kepler's Third Law of Planetary Motion to calculate the mass of Jupiter. This law states that the square of a planet's orbital period is directly proportional to the cube of its semi-major axis, which is the average distance between the planet and its orbiting satellite.

2. Why is Io's orbital period important for determining Jupiter's mass?

Io's orbital period is important because it directly affects the strength of the gravitational force between Jupiter and Io. A shorter orbital period means Io is orbiting closer to Jupiter and experiencing a stronger gravitational force, while a longer orbital period means Io is orbiting farther away and experiencing a weaker gravitational force. By measuring Io's orbital period, we can calculate the strength of the gravitational force and use it to determine Jupiter's mass.

3. What other factors are taken into account when determining Jupiter's mass using Io's orbital parameters?

In addition to Io's orbital period, factors such as Io's distance from Jupiter, the eccentricity of its orbit, and the mass of other nearby objects also play a role in determining Jupiter's mass. These factors can affect the strength of the gravitational force and must be considered when using Io's orbital parameters to calculate Jupiter's mass.

4. How accurate is this method of determining Jupiter's mass?

This method is considered to be fairly accurate, with estimates typically falling within a small range of uncertainty. However, it should be noted that there are other methods for determining Jupiter's mass, such as measuring the planet's gravitational pull on other objects in its vicinity, which may provide more precise results.

5. Can this method be used to determine the mass of other planets?

Yes, this method can be applied to calculate the mass of any planet with one or more orbiting satellites. However, the accuracy of the results may vary depending on the orbital parameters and other factors of the planet and its satellites.

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