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1g (gravitational acceleration) Sphere of U238

  1. Jun 17, 2015 #1
    How big would a sphere of U238 have to be to reach 1g at its surface?
     
  2. jcsd
  3. Jun 17, 2015 #2

    phinds

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    Do you not know how to look up the density of U238? Do you not know how to calculate the mass of a sphere? Do you no know the equation for the force of gravity at the surface of a sphere? That is, what part of solving this problem do you not understand?
     
  4. Jun 17, 2015 #3
    Physics is not my forte, yet.

    If I understand correctly do I just solve for R (where g=1)?


    from http://nova.stanford.edu/projects/mod-x/ad-surfgrav.html:

    You have learned that the surface gravity (g)of a body depends on themass (M)and theradius (r)of the given body.

    The formula which relates these quantities is:g = G * M / r2

    whereG is called the Gravitational constant.
     
    Last edited: Jun 17, 2015
  5. Jun 17, 2015 #4

    phinds

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    right

    OK, that's a start. Now you need to find the density of U238 and write an equation for the mass of a sphere of radius r.
     
  6. Jun 17, 2015 #5
    Thank you.

    So if I understand correctly, I'd need two equations then. One to figure out the mass (with the known density of U238) given a particular radius. And another to figure out the radius given a particular mass in order to equal 1g. I'm guessing I could use some method to graph the two and where they intersected in mass and radius would be my solution?
     
  7. Jun 17, 2015 #6

    phinds

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    Just solve algebraically. No need to graph.
     
  8. Jun 17, 2015 #7
  9. Jun 17, 2015 #8

    phinds

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    Hm ... give that U238 is 3 or 4 times as dense as the Earth's average density, 5949 km seems a bit high, but I didn't do the math so maybe it's right. I'd double check it though. I don't like using those apps ... I prefer to do it myself. You should try it that way and see what you get.
     
  10. Jun 17, 2015 #9

    DaveC426913

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    Hm. Back of napkin calcs suggest your number may be a little too high.
    Density of Earth is about 5g/cm3
    Density of Uranium is about 19g/cm3.
    So, volume of U sphere should be about .26 (1 / 3.8) of Earth. (excluding any compression effects)
    Then cube root of .26 (to go from volume to linear) is .64.
    .64*6371 = 4082
    I think that leads to a radius closer to 4100km.


    [EDIT: Dagnabit you! phinds!]
     
  11. Jun 17, 2015 #10
    I was thinking that it wasn't quite small enough either, but could the cubic nature of the volume of a sphere lead to a slightly unintuitive answer?

    @Dave I don't think it'd work that was as it wouldn't be a linear comparison.
     
  12. Jun 17, 2015 #11

    DaveC426913

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    I was busy fleshing out my answer. You can see where I took the cube root to convert volume ratio (.26) to radius ratio (.64).
     
  13. Jun 17, 2015 #12
    That definitely seems reasonable. The gravity at the surface also has the inverse square law from the center of mass to consider as well though, which is why it is maybe bigger than it first would seem....or my equation is wrong.
     
  14. Jun 17, 2015 #13

    phinds

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    Now, now. Don't whine. Being a day late and a dollar short is common for Canadians :-p
     
  15. Jun 17, 2015 #14

    jbriggs444

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    You are correct. Surface gravity does not scale with the cube of radius. It scales linearly. The cube factor from volume and the square factor from the inverse square law cancel to leave a simple linear relationship of surface gravity to radius (neglecting compression effects).

    If Uranium is 3.8 times denser than Earth on average then the required radius of a 1 g planet is a factor of 3.8 times less. Roughly 1500 km.
     
  16. Jun 17, 2015 #15

    ChrisVer

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    I think the right answer should be around 1800km... (at least that's what maths tells me)
     
  17. Jun 17, 2015 #16
    What maths though? Where did the equations I found go wrong? And how does a cube factor and a square factor cancel each other out?
     
  18. Jun 17, 2015 #17

    jbriggs444

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    Here is how I work these sorts of thing in my head:

    Double the radius and you increase volume (and therefore mass) by a factor of 8. Double the radius and you reduce gravity for a given mass by a factor of 4. Factor of 8 divided by factor of 4 is a factor of 2. You've doubled radius and you've doubled surface gravity. Therefore it is a simple linear relationship.
     
  19. Jun 17, 2015 #18

    ChrisVer

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    The gravitational force an object of mass [itex]m[/itex] feels from another of mass [itex]M[/itex] in a distance [itex]r[/itex] is: [itex]F= \frac{GMm}{r^2}[/itex]
    The newton's 2nd law for the object of mass [itex]m[/itex] is [itex]F=ma[/itex]
    Equating them:
    [itex]ma =\frac{GMm}{r^2}[/itex]
    [itex]a= \frac{GM}{r^2}[/itex]
    Now the mass [itex]M[/itex] of a sphere of radius [itex]r[/itex] and density [itex]\rho[/itex] is the density*volume. [itex]M= \rho V_{sphere} = \rho \times \frac{4}{3} \pi r^3[/itex]...

    So the acceleration is:
    [itex]a= \frac{4 \pi G \rho r^3}{3 r^2}= \frac{4 \pi G \rho}{3} r[/itex]

    So the distance [itex]R[/itex] at which [itex]a=g[/itex] is given by:
    [itex]R= \frac{3g}{4 \pi G \rho}[/itex]

    For: [itex]G= 6.674 \times 10^{-11} \frac{m^3}{s^2 kg}[/itex] , [itex] \rho= 19 \frac{g}{cm^3}=19000 \frac{kg}{m^3}[/itex] and [itex]g=10 \frac{m}{s^2}[/itex]:

    [itex] R\approx 1883 km[/itex] and gets a little less if you take g=9.81 m/s2 and ρ=18900 kg/m3, around 1856 km
     
    Last edited: Jun 17, 2015
  20. Jun 17, 2015 #19
    @jbriggs444 I don't think that's right about gravity scaling linearly. Also, I think pi helps to rule out the linear relationships.

    What about my equation (linked above) is wrong?

    I took the surface gravity equation here:
    http://nova.stanford.edu/projects/mod-x/ad-surfgrav.html -- g = G * M / r2

    And substituted mass with the density of U238 (18.9 g/cm3) multiplied by the volume of a sphere, which is 4/3πr^3.

    Since the radius was in km in the surface gravity equation on converted the density one which was in cm by multiplying by 100,000.
     
  21. Jun 17, 2015 #20
    @ChrisVer That looks more correct that what I was trying to do. I think I got all flummoxed with the units. Thank you!
     
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