How Do You Calculate Orbital Period with Limited Data?

[blackbyron]

Homework Statement

An Earth station receives data transmitted back in time from a future intergalactic expedition. The table summarizes the data for the moons of a planet that will be discovered in a distant galaxy.


Moon 1: only has orbital radius = 5E7 meters
Just only this information.


Question: the orbital period of Moon I is closest to:

Homework Equations

Using Kelpers Law

Force = GMeM/R^2

The Attempt at a Solution

How do I find out the orbital period if I don't know that acceleration, and the mass, and what's the difference between radius and orbital radius? I just don't understand what to do.

 

[tiny-tim]

hi blackbyron!

what are the possible answers?

my guess is that you're expected to assume that the planet's mass is somewhere between Mercury's and Jupiter's

 

[blackbyron]

Hi, tiny-tim, thanks for your reply.

No, I don't believe the moon on the mass has something to do with mercury and jupiter.
It only talks about the 3 moons.

Here is the table, that might make sense to you.

The possible answer is 1.2 *10^5s. Not sure if that asks you a question.

 

[gneill]

You have period and radius information for Moon II. You could use Kepler's 3rd law to solve for the orbital period of Moon I by setting up the appropriate ratios.

 

[tiny-tim]

ohhh …

then can't you find the orbital period of moon I from the orbital period of moon II ?

 

[blackbyron]

@tiny-tim
hmm, I probably would give it a try, but I don't know if I'm going to to do it right. haha

You have period and radius information for Moon II. You could use Kepler's 3rd law to solve for the orbital period of Moon I by setting up the appropriate ratios.

Alright, so I use the ratio of Time to solve for time 2 using the Kepler's 3rd law.

So For example,

Tm1/Tm2 = kepler's third law?
m1 = moon 1
m2 = moon 2

Is that right?

 

[blackbyron]

I think I got it. Finding out the ratio of Tm1/Tm2 gives me the right answer.

So, I think I did it right. My answer is 1.24e5. Do you guys agree? :)

 

[gneill]

The value looks fine.

 

[blackbyron]

Thank you so much guys. It's very helpful.