# Mass of Planet

1. Nov 9, 2008

### Husker70

1. The problem statement, all variables and given/known data
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

2. Relevant equations
4pie^2/GM
Kepler's Third Law

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 9, 2008

### asleight

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

$$T=2\pi a\sqrt{a/GM}$$.

3. Nov 9, 2008

### Husker70

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. Nov 9, 2008

### asleight

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?

5. Nov 9, 2008

I agree

6. Nov 9, 2008

### asleight

Then you can solve for M. :) Let me know what you try.

7. Nov 9, 2008

### Husker70

Except how can I use your first equation when I don't know the mass of Io?

8. Nov 9, 2008

### asleight

Because $$\vec{f}=m\vec{a}$$ so the small masses cancel. ;)

9. Nov 9, 2008

### Husker70

So a = G(M/r^2)

10. Nov 9, 2008

### asleight

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. Nov 9, 2008

### Husker70

So I get v^2 = G(M/r)
I'm not sure how to get M =

12. Nov 9, 2008

### asleight

$$\vec{v}=2\pi\vec{r}/T$$...

13. Nov 9, 2008

### Husker70

I'm just not seeing it sorry

14. Nov 9, 2008

### asleight

$$\frac{4\pi^2\vec{r}^2}{T^2}=\frac{\vec{G}M}{\vec{r}}\rightarrow M=\frac{4\pi^2\vec{r}^3}{\vec{G}}$$.

15. Nov 10, 2008

### Husker70

Thanks alot for the help but using that equation I don't get the right answer for some reason.
Kevin

16. Nov 10, 2008

### nasu

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. Nov 10, 2008

### borgwal

asleight's posts are so full of errors, you're better off ignoring them.

18. Nov 10, 2008

### Husker70

I figured that out. Thanks again. You were a great help.
Talk to you soon.
Kevin