1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to find the mass (or volume) of Saturn's ring?

  1. Nov 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Saturn has the largest ring system of any planet, with the inner edge about 7000 km above
    Saturn’s surface and extending to its outer edge about 80,000 km above the surface. However,
    the rings are only about 20 m thick. The rings are made of billions of small particles, most
    of which are water ice, which has a mass density of 1000 kg/m3. Approximating the ring as
    being made of solid ice, what is the mass of the ring compared to the mass of Earth’s Moon?
    Show how your got your answer.

    2. Relevant equations

    Volume = L x W x H and Mass = Density x Volume

    3. The attempt at a solution

    I have the density (1 g/m3) but I don't know how to get the volume from the other information because I've never taken the volume of a ring before, could someone please help me find the volume? I know that once I have the volume I can easily get the mass and compare it with the moon's mass
     
  2. jcsd
  3. Nov 15, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The volume is just the thickness times the area. And the area is the region between two circles. You should be able to find that just by subtracting the area of the inner circle from the area of the outer.
     
    Last edited: Nov 15, 2013
  4. Nov 16, 2013 #3
    But how am I suppose to find the areas of both circles? I tried to use the equation of finding the area of a ring, which is: ∏(R2 - r2)
    I added both 80,000 km (which is R) and then 7,000 km (which is r) by the radius of Saturn which is 58,232 km (to extend their radiuses to Saturn's core/centre, not just its surface)
    Then I got ≈ 4.6662 x 1010, I then multiplied it by 0.02 km (which is the thickness) to find the volume which I then plugged into the Mass = Density x Volume equation, and compared it to the moon's mass but still got an incredibly small number which is not one of the choices. Any more help would be greatly appreciated, thank you
     
  5. Nov 16, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are off to a good start. 4.6662 x 1010 is an ok numerical answer but the units on that are km^2. You need to keep track of the units, and show the rest of your work. I did it and I get that it's small compared to the mass of the moon. But not 'incredibly small'.
     
  6. Nov 16, 2013 #5
    Thank you for replying back but I still don't think its right because the answer I got at the end was ≈1.27 x 10-11, where all the choices are exactly the following:
    0.0001 x Earth's moon, 0.001 x Earth's moon, 0.01 x Earth's moon, 0.1 x Earth's moon, and 1 x Earth's moon.
    But the answer I have is 1.27 x 10-11 x Earth's moon is is obviously much smaller than all of these answers, thank you again.
    p.s I was aware of all the units but none of them needed to be changed
     
  7. Nov 16, 2013 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, you do need to keep track of units. You've correctly got that the area of the rings is 4.6662*10^10*km^2. That's kilometers, not meters. The units on thickness (20m) and density (1000kg/m^3) are given in meters. The answer is not that small.
     
  8. Nov 16, 2013 #7
    Actually I did say that I used 0.02 km for thickness in my previous comments, but for density I tried using both 1000 kg/m3 and 1 g/cm3 to see what answer they would give me. The 1000 kg/m3 gave me 1.27 x 10-11 (which I stated before) and the 1 g/cm3 gave me 1.27 x 10-14. So I really don't know what I'm doing here but I still get a much smaller answer than the smallest choice I have even after I tried using different units when I was unsure, thanks.
     
  9. Nov 16, 2013 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Can you please show your whole working for one of those? Your exponents are drastically off. You are NOT handling units correctly. 1000 kg/m^3 and 1 g/(cm^3) are exactly the same. How can using one or the other change the answer??
     
    Last edited: Nov 16, 2013
  10. Nov 17, 2013 #9
    That's what I'm wondering too. But here's what I did:
    I used ∏(R2 - r2) to find the area of the ring
    R being 138,232 and r being 65,232, the area I get is then = 4.6662 x 1010 km2 (which is supposedly right), then I multiply it by the thickness (0.02 km) to get a volume of 933,240,000 km3.
    Then I multiplied the volume by the density which is 1000 kg/m3, to get a mass of 933,240,000,000 (or 9.332 x 1011)
    Lastly, I divide the mass of saturn's rings by the moon's mass, therefore it looks like this: (9.332 x 1011 kg)/(7.348 x 1022 kg) to give me the final answer of something ranging from 0.0001 to 1 x the moon's mass
    But instead, I get 1.27 x 10-11 which is obviously very wrong for some reason
     
  11. Nov 17, 2013 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Now you've got the volume right as well. In terms of km^3. That's not the same as m^3. If you want to multiply by 1000kg/m^3 you had better either convert the km^3 to m^3 or do vice versa for the density. You still aren't tracking the units correctly until the end. I'll give you a big hint. There are 10^9m^3 per km^3.
     
    Last edited: Nov 17, 2013
  12. Nov 17, 2013 #11
    Oh ok, so after I multiplied 933,240,000 km3 by 109, I get a volume of 9.3324 x 1017 km2 instead, then I multiply it by the density of 1000 kg/m3 to get a mass of 9.3324 x 1020 for the rings. Then I divide it by the moon's mass and get ≈ 0.01 which is one of the answers. I hope this is right and thank you for helping.
     
  13. Nov 17, 2013 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You mean you get a volume of 9.3324 x 1017 m3, I hope. You are still messing the units up on intermediate steps. But, yes, the final answer is pretty right. See how important units are? If you had realized that you should have put a unit on the final mass.
     
    Last edited: Nov 17, 2013
  14. Nov 17, 2013 #13
    Oh yeah, I meant m3, not km2, I just got confused with all the units, and I forgot to put the kg for the mass at the end but I was just focusing more on the number for now
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: How to find the mass (or volume) of Saturn's ring?
  1. Finding the Volume (Replies: 6)

Loading...