What does this tell you about how the moons were created?

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The discussion centers on the gravitational dynamics of two moons orbiting a planet, specifically analyzing their relative masses and densities based on their orbital radii. Using the equations of motion, participants derive that if the orbital radii differ by a factor of 2, the mass of the larger moon is four times that of the smaller moon. Furthermore, if both moons have similar radii, their differing masses imply distinct densities, suggesting varied origins, such as formation from a gas disc or collision events.

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  • Familiarity with orbital mechanics and Kepler's laws
  • Knowledge of basic algebra and equation manipulation
  • Concept of density and its calculation
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Homework Statement


Two identical spaceships are sent to explore a planet with two moons. Each spaceship orbits one of the moons.

a)The orbital radii are observed such that the spaceship have the same period. What is the relative mass of the two moons if the two radii differ by a factor of 2.

b) If the two moons have a similar radius, what does this imply about their density? What does this tell you about how the moons were created?


Homework Equations


a=v^2/r
v=2pi^2/T
ma=Gm(Mearth)/r^2
F=Gm1m2/r^2


The Attempt at a Solution


a) 4pi^2/T^2=G(Mearth)/r^2
since T is the same, T=1

b) If the two moons have similar radius, then they should have different densities (if the relative masses are different in part a)
 
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Hi physics(L)10! :smile:

(have a pi: π and try using the X2 and X2 tags just above the Reply box :wink:)
physics(L)10 said:
a=v^2/r
v=2pi^2/T
ma=Gm(Mearth)/r^2
F=Gm1m2/r^2

a) 4pi^2/T^2=G(Mearth)/r^2
since T is the same, T=1

Not really following that … can you write it out more fully? :confused:

(and v = 2πr/T)
b) If the two moons have similar radius, then they should have different densities (if the relative masses are different in part a)

So where did they come from? :wink:
 
That was just a guess because I don't really have a clue on what to do, that's why you can't understand lol. I'm guessing they came from different planets. I can't answer b without knowing the answer to a though.
 
Hi physics(L)10! :smile:
physics(L)10 said:
That was just a guess because I don't really have a clue on what to do

ok, start with your two equations for acceleration, then eliminate the acceleration to get an equation relating v T and r …

show us all the steps. :smile:
I'm guessing they came from different planets.

Well, from different places … they needn't have come "from planets", they could have been part of the same gas disc that the planets were formed from, or they could have been limps knocked off planets, or they could have been comets etc from a long way away. :wink:
 
a=v^2/r, where v=2πr/T
Therefore, a=(2πr/T)^2/r=4π^2r/T^2

Then, you plug this value for a into the equation ma=GmMearth/r^2

b) Believe it or not I was actually thinking that lol.
 
physics(L)10 said:
a=v^2/r, where v=2πr/T
Therefore, a=(2πr/T)^2/r=4π^2r/T^2

Then, you plug this value for a into the equation ma=GmMearth/r^2

(please use the X2 tag just above the Reply box :wink:)

ok, so your equation relating v T and r is … ? :smile:
b) Believe it or not I was actually thinking that lol.

We believe you … but the examiner needs you to say it. :wink:
 
The equation would be 4π2r3/T2=GMearth. I`m guessing you have to isolate for r now, but I`m confused on how to integrate that the periods are the same and the radii differ by a factor of 2.
 
Hi physics(L)10! :smile:

(just got up :zzz: …)
physics(L)10 said:
The equation would be 4π2r3/T2=GMearth.

(earth? … is one of the moons called "earth"?! :rolleyes:)

ok, this is now a dimensions question …

never mind the constants (4π2 and G), they don't matter :wink:

you're only interested in the powers of r M and T …

if you keep T the same and double r, what happens to M? :smile:
 
Lmao, I`m used to putting earth. M gets bigger if you double the radius. So would one of the moons mass just be double the other moon and that would be your answer?
 
  • #10
uhh? :confused:

r3/T2 = constant times M, so … ? :smile:
 
  • #11
Ok, this is what I thought of first. When you cross multiply the T (which is constant) multiplies with M so you get M. Then you divide M by 2 and then cube root the answer and you get a smaller answer. But then I used common sense and thought that shouldn`t the mass be larger if the radius is bigger?
 
  • #12
physics(L)10 said:
Ok, this is what I thought of first. When you cross multiply the T (which is constant) multiplies with M so you get M. Then you divide M by 2 and then cube root the answer and you get a smaller answer.

hmm … doing it with equations is easier than doing it in English …

you have T2 = r3/M …

so, if T is constant, then r3/M is constant ……

does that help? :smile:
 
  • #13
Not really lol. I was thinking more along the lines of r3=M and then cubed root M to find the radius. The T can be ignored since its a constant and is the same for both moons.
 
  • #14
physics(L)10 said:
I was thinking more along the lines of r3=M and then cubed root M to find the radius.

Yes, that's right :smile: … what is worrying you about that? :confused:
 
  • #15
So one moon would just be half the mass?
 
  • #16
physics(L)10 said:
a)The orbital radii are observed such that the spaceship have the same period. What is the relative mass of the two moons if the two radii differ by a factor of 2.
physics(L)10 said:
I was thinking more along the lines of r3=M and then cubed root M to find the radius.
physics(L)10 said:
So one moon would just be half the mass?

How did you get that? :confused:
 
  • #17
One would be r3=M and the other would be 2r3=M which is r3=M/2
 
  • #18
physics(L)10 said:
One would be r3=M and the other would be 2r3=M which is r3=M/2

Sorry, but you'll never be able to do these questions in the exam if you don't learn to tidy up your notation. You can't keep using "r" for everything without losing track. :redface:

Your basic formula is r3 = kM (for a constant k).

Now you can put values in for each of the two satellites, making …

r13 = kM1
r23 = kM2

and you are given the relationship r2 = 2r1.

Try it that way. :smile:
 
  • #19
One would be r13=kM1 and the other would be 2r23=kM2 which is r23=kM2/2
 
  • #20
physics(L)10 said:
… the other would be 2r23=kM2

Nooo!

It's still only r23=kM2

that's the point of a formula!
 
  • #21
Then where does the factor of 2 come in?? lol
 
  • #22
physics(L)10 said:
Then where does the factor of 2 come in?? lol

uhh?
tiny-tim said:
… and you are given the relationship r2 = 2r1.
 
  • #23
So the radius IS 2x larger
 

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