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Finding mass of a planet when density varies with radial distance

  1. Dec 15, 2011 #1
    1. The problem statement, all variables and given/known data

    The density of a certain planet varies with radial distance as:
    ρ (r)= ρo(1- αr/Ro)
    where Ro= 3.98 x 106 m is the radius of the planet, ρo= 4980 kg/m3 is its central density, and α = 0.17. What is the total mass of this planet ?

    Calculate the weight of a one kilogram mass located on the surface of this planet.

    2. Relevant equations

    ρ = m/v

    3. The attempt at a solution

    since ρ= m/v

    then

    m = ρv
    v of a planet = 4/3∏ R^3
    so i substitute the given info on the equations and got

    m = ρo(1- αr/Ro) * 4/3∏ R^3
    calculate all of those and I realize that i don't know what r is.

    so I assume that r = Ro but I got the wrong answer

    any help?
     
  2. jcsd
  3. Dec 15, 2011 #2

    gneill

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    Staff: Mentor

    Since the density changes with radius (r), you will need to set up an integration over the radius r and sum the mass shells from the center (r = 0) to the surface (r = Ro).
     
  4. Dec 15, 2011 #3
    i integrate the ρ as a function of r

    3.98x10^6
    ∫ ρo(1- αr/Ro)dr
    0

    and got 3641700 as my answer

    since ρv = m
    and v = 4/3 ∏ Ro^3
    m = 6.07x10^13

    but its wrong, what did i do wrong?
     
  5. Dec 15, 2011 #4

    gneill

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    Staff: Mentor

    The ρ(r) function only gives you the density at a given radius. It does not address the geometry of the object. You need to determine the mass of each shell of matter of differential thickness dr, and sum them all up (integrate). In this case the shells will be spherical, and of thickness dr. So first find an expression for the mass of a spherical shell of thickness dr.
     
  6. Dec 15, 2011 #5
    m = ρv
    ∫m = ∫ρv ???
    which results to same exact answer as before?
     
  7. Dec 15, 2011 #6

    SammyS

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Since the density only varies as the radius, and not as the latitude and/or longitude, the volume element is
    dV=4πr2dr
    After all, the total volume of a sphere of radius R0 can be obtained from [itex]\displaystyle V=\int_0^{R_0}4\pi r^2\,dr\,.[/itex]

    So the total mass of the planet is [itex]\displaystyle M=\int_0^{R_0}(ρ(r))4\pi r^2\,dr\,.[/itex]
     
  8. Dec 15, 2011 #7
    i did that integration and got 1.147x10^24 and its still wrong...
     
  9. Dec 16, 2011 #8
    Can you show us how you did the integration? I got [itex]\approx 4.1 \times 10^{24} [/itex]
     
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