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Radius and Mass of an Unknown Planet

  1. Feb 3, 2015 #1
    About:
    As a project, I'm trying to solve for the approximate radius and mass of a approximately earth-sized planet, Mann, from Interstellar. According to what was provided,
    g = 7.84m/s2
    The solid part of the planet is composed of a honeycomb of rock and ice
    The space inside the honeycomb is filled with largely ammonia, about 20% oxygen, gaseous chlorine (can support human breathing for a short amount of time) and possibly carbon dioxide
    The "surface" temperature is between 0 and -33 celsius​
    From research, I have found a moon with compositions similar to Mann: Ganymede.
    On Ganymede,
    g = 1.42n/s2
    The planet is composed of 50% rock and 50% ice
    The planet has oxygen in its atmosphere, although it's probably too insignificant to count​
    Problem:
    I do not know how to find the mass and radius of a planet, given its density and gravitational pull.
    Can gravitational pull and radius increase without changing the density?
    Is it better to consider Mann a bigger version of Ganymede or a small, frozen gas giant?

    PS more information on Ganymede is available on Wikipedia; Dr Mann talked about presence of hydrocarbons (thus carbon dioxide) as a trap; however, there was no proof that carbon did not exist on this planet.
     
  2. jcsd
  3. Feb 3, 2015 #2
    mass = volume x density

    volume of sphere = 4pi/3 x radius3

    surface gravity = GM/R2 ~ density x radius
     
  4. Feb 4, 2015 #3
    I do know all the formulas, I just don't know how they can be combined to solve for two unknowns (radius and mass). Hints please.
    Also, I heard that when densities are equal,
    r1 g1
    --- = ---
    r2 g2
    where r1 and r2 are the radii of the planets being compared, and g1 and g2 are the gravitational forces of them​
    Is that true?
     
  5. Feb 4, 2015 #4
    First get an expression for mass (from Newton's law) $$M = \frac{gr^2}{G}$$ and an expression for radius (from the density formula) $$r = \left(\frac{3M}{4{\pi}d}\right)^{1/3} .$$ Now either plug the equation for r into the equation for M or the equation for M into the equation for r. If you do the latter, you end up with a new equation for r that doesn't contain M. Now plug that new equation for r back into the equation for M, and you'll end up with an equation for M that doesn't contain r.

    TEFLing sort of answered your last question - density is proportional to the ratio of surface gravity to radius: $$d \propto \frac{g_1}{r_1}, d \propto \frac{g_2}{r_2}$$ so $$\frac{g_1}{r_1} = \frac{g_2}{r_2}$$ which is what you have.
     
  6. Feb 4, 2015 #5
    You might find this paper useful:

    The Mass-Radius Relation for 65 Exoplanets Smaller than 4 Earth Radii [arXiv : 1312.0936]

    It describes a method for approximating the density and mass of an object that is less than four Earth radii.
     
  7. Feb 4, 2015 #6
    I can understand most of the formula, except the d: does it mean density (which should be ρ) or diameter? Distance (but wouldn't that be r)?
     
  8. Feb 4, 2015 #7
    I meant d for density, and probably should have mentioned it or at least used the conventional ρ. That equation is derived from the first two that TEFLing listed.
     
  9. Feb 4, 2015 #8

    Bandersnatch

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    A very simple way to get the radius given density and surface gravity, is to use units of Earth ρ, g and R. Then:
    ##g=ρR##
    The above follows from the earlier posts, and is just a convenient restatement.

    What it tells you, is that if for example you have twice the density (of Earth), and the same surface gravity, then you must have half the radius. Or that with 3/4 radius, and 4/3 density, the gravity will be the same.

    This assumes some unrealistic things about the internal composition (i.e., that all planets have the same internal density distribution as Earth), but it's a rather good first approximation nonetheless.

    Once you have radius and density, calculating mass is trivial.
     
  10. Feb 4, 2015 #9

    Sadly as I stated in OP, the planet I was calculating for was more similar to Ganymede than earth. Same principle though.


    I substituted the M, and got
    r = (3g)/(4Gπρ)​
    is that correct? I'm not very good at math...
     
  11. Feb 5, 2015 #10

    Bandersnatch

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    Use the units for Ganymende, then. It's radius, density and surface gravity are known.

    The maths you're doing will net you the same end result as that simple equation.
     
  12. Feb 5, 2015 #11
    What is R? Radius?
    And I'm doing the unneeded math as a part of a presentation, so I'm open to any formula that works.
     
  13. Feb 5, 2015 #12
  14. Jun 28, 2016 #13
    Some of the basic facts regarding a planetary body are its mass (M), its radius (R), its surface gravity (g), its surface escape speed (v), and its average density (ρ). If you know any two of them, you can calculate the other three. Here's how.

    G = 6.67384e-11 m³ kg⁻¹ sec⁻²
    π = 3.1415926535897932384626433832795...

    Mass (in kilograms).
    M(R,g) = gR²/G
    M(R,v) = v²R/(2G)
    M(R,ρ) = 4πρR³/3
    M(g,v) = v⁴/(4gG)
    M(g,ρ) = 9g³/(16π²ρ²G³)
    M(v,ρ) = v³ √[3/(32πρG³)]

    Radius (in meters)
    R(M,g) = √(GM/g)
    R(M,v) = 2GM/v²
    R(M,ρ) = ∛[3M/(4πρ)]
    R(g,v) = v²/(2g)
    R(g,ρ) = 3g/(4πρG)
    R(v,ρ) = v √[3/(8πρG)]

    Surface gravity (in m sec⁻²)
    g(M,R) = GM/R²
    g(M,v) = v⁴/(4GM)
    g(M,ρ) = G ∛(16π²ρ²M/9)
    g(R,v) = v²/(2R)
    g(R,ρ) = 4πGρR/3
    g(v,ρ) = v √[(2πρG)/3]

    Escape speed from surface (in m sec⁻¹).
    v(M,R) = √(2GM/R)
    v(M,g) = ∜(4GMg)
    v(M,ρ) = (32πρG³M²/3)^⅙
    v(R,g) = √(2gR)
    v(R,ρ) = R √(8πGρ/3)
    v(g,ρ) = g √[3/(2πρG)]

    Average density (in kg m⁻³).
    ρ(M,R) = 3M/(4πR³)
    ρ(M,g) = [3/(4π)] √[g³/(G³M)]
    ρ(M,v) = 3v⁶/(32πG³M²)
    ρ(R,g) = 3g/(4πGR)
    ρ(R,v) = 3v²/(8πGR²)
    ρ(g,v) = 3g²/(2πGv²)
     
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