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Size of planet with 1/50 gravity of earth

  1. Sep 14, 2011 #1
    Basically I'm wondering how big a rock in space would need to be to draw an average person to it with a gravitational force of a kilo or two. Say it was the average density of earth and it was spherical, what would its diameter be.
     
  2. jcsd
  3. Sep 14, 2011 #2

    Drakkith

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    I'm not sure I understand you. I believe mass is measured in kilograms, not force. How much gravitational force are you looking at getting?
     
  4. Sep 14, 2011 #3

    Ryan_m_b

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    Well I'm pretty sure you would need 50 times less mass. Every time you half the diameter of a sphere you reduce it's volume by 8. So a sphere 1/4 the size of Earth would have 1/65th of the gravity.
     
  5. Sep 14, 2011 #4
    But you would als get closer to the center. Every time you halve the diameter, there will be 8 times less mass, the 1/r^2 in Newton's law. The end result is that an object with 1/4 times the diameter of Earth, will have 1/4 times the gravity, so you'll need 1/50 of the diameter.
     
  6. Sep 14, 2011 #5
    basically I was wondering if a future astronaut landing on an asteroid that just had enough gravity to keep him on the surface would be able to tell he was on a small object by it's appearance.

    @ryan, that doesn't seem right because the moon is about 1/4 the diameter of the earth and is about 1/6th the gravity.

    @drakkith, my bad. I meant how much would he weigh.
     
  7. Sep 14, 2011 #6
    @willem2, ok so then that would be say a rock that was about 160 miles across. So an astronaut standing in the middle would see about 80 miles to his horizon.
     
  8. Sep 14, 2011 #7

    Ryan_m_b

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    Last edited by a moderator: Apr 26, 2017
  9. Sep 14, 2011 #8

    DaveC426913

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    Yes. He would see a horizon that's very near. Mars' horizon is something like a couple of kilometers, enough to be quite noticeable.
     
  10. Sep 14, 2011 #9
    The distance to the horizon on a perfect sphere is

    [tex] \sqrt { 2 R h } [/tex]

    where R is the radius of the sphere and h the height above the surface. This will be somewhat less than 1 km.
     
  11. Sep 14, 2011 #10
    The universal law of gravitation is:
    [tex]F_g = \frac{mMG}{r^2}[/tex]
    Where [tex]G = 6.674*10^{-11} \frac{m^3}{kg*s^2}[/tex]

    The average mass density of the earth is 5.515 g/cm^3 (According to Wikipedia),
    so the acceleration due to gravity based on a planet with Earth's density and radius r is [tex]a = \frac{\frac{4}{3}*\pi*r^3*5.515*10^6*6.674*10^{-11}}{r^2}[/tex]

    Using gravity on Earth to be 9.807 m/s, we write the equation
    [tex] 9.807/50 = 1.5418*10^{-6}*r[/tex]

    This can easily be solved to find a radius of 127 kilometers.
     
  12. Sep 14, 2011 #11

    DaveC426913

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    There has got to be a formula that, for spherical objects of a fixed density, takes only mass and radius and returns the surface gravity.
     
  13. Sep 15, 2011 #12
  14. Sep 15, 2011 #13
  15. Sep 15, 2011 #14
    about 1000 km

    ceres is the smallest one
     
  16. Sep 15, 2011 #15
    It should be dependent on the composition of the body. I doubt that there is a general answer.
     
  17. Sep 15, 2011 #16
    About 1000km given what density?

    The gravitational attraction of a planet or a moon is dependent on it's diameter and its density!
     
  18. Sep 15, 2011 #17
    NO KIDDING!!!

    See my last two posts on this thread.
     
  19. Sep 15, 2011 #18

    Redbelly98

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    This can be solved without actually knowing the density of Earth, or plugging in the value of G. We just need to know the radius of the Earth.

    The surface acceleration, for a sphere of mass M and radius r is
    [tex]a = \frac{GM}{r^2}[/tex]
    Since M = ρV, and V = 4πr 3/3, we have
    [tex]a = \frac{4 \pi G \rho r^3}{3 r^2} = \frac{4 \pi G \rho}{3}r[/tex]
    In other words, a and r are proportional for a fixed density ρ. To get 1/50th the force or acceleration as experienced at the surface of the Earth, the radius must be 1/50th that of Earth:
    [tex]r = \frac{6378 \text{ km}}{50} = 128 \text{ km}[/tex]

    This agrees with tmiddlet's calculation posted earlier, and with what others have been saying without showing a derivation.
     
  20. Sep 15, 2011 #19

    Sorry, I cannot see any other posts between your claim that you used to know the answer and my remark. Still don't see why would you yell at me.
    Your question:

    "By the way, what's the limit for mass and/or radius of a body to be gravitationally compacted into a sphere? (I used to know it, but I forgot.)"

    is not well defined as it depends on the initial state of the "body" to be (or not to be) compacted (whatever that means). I don't dispute that you may have had an answer but it should have been to a more specifically formulated question.
     
  21. Sep 15, 2011 #20

    Drakkith

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    Wouldn't it be the standard formula, but with the variable density replaced with just a number?
     
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