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B Simplistic question about mass and gravity

  1. Mar 20, 2017 #1
  2. jcsd
  3. Mar 20, 2017 #2
    I will assume you are speaking of the acceleration of gravity. Let's compare...
    Gravity on Earth-- 9.8 m/s^2
    Gravity on Jupiter-- 24.9 m/s^2
    EDIT: forgot to put in, Jupiter is over 300x more massive than Earth, so I don't think there is a simple proportion
     
  4. Mar 20, 2017 #3
    So, the mythical fifth planet would have a gravity that's much closer to Earth's. I figured that. But what would the "weight" of a 100 kg. person be on that planet?
     
  5. Mar 20, 2017 #4
    The mass of that person (100 kg) X the acceleration of gravity would give you their weight.
     
  6. Mar 20, 2017 #5
    But I'm mathgnostic.
     
  7. Mar 20, 2017 #6

    Chalnoth

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    Surface gravity is a function of two things:
    1. How much mass is contained within the object.
    2. How far the surface is from the center.

    If you know these, the surface gravity is simply ##Gm/r^2## (feel free to calculate this for the Earth, and you'll get roughly ##9.8m/s^2##). So if you have a small mass, you can still have a high surface gravity if the planet is very dense, so that it has a small radius.

    But I don't think the numbers work out here: the asteroid belt is tiny, at only about 0.05% the mass of the Earth today. It would need to be something like 4,000 times as dense as the Earth to have a four times the surface gravity. That's just not happening.

    More realistically, what they might have meant is that if Jupiter had not formed first, then what is today the asteroid belt might have formed into a rocky planet significantly larger than the Earth, with quite a bit more gravity. But Jupiter did form further out, which disrupted the orbits of most of the dust and rocks that would have made up that planet, so that their orbits destabilized and they ended up striking other planets.
     
  8. Mar 20, 2017 #7
    Okay, so it wouldn't have four times the gravity of Earth.
     
  9. Mar 20, 2017 #8
    Correct.

    And the weight equation you are looking for (mathgnostically) is simply F=ma, which is really W=ma where W is weight, and the a is just the acceleration of gravity on that planet.
     
  10. Mar 20, 2017 #9
    I'm not actual looking for an equation.
     
  11. Mar 20, 2017 #10
    Then I guess we're good here! :smile:
     
  12. Mar 20, 2017 #11
    Well, I noted that I was mathgnostic. Maybe I'm doing a poor job of communicating. Let's try again. If a 100 kg man was on a planet with four times the mass of Earth, what would his weight be there?
     
  13. Mar 20, 2017 #12
    F = GMm/r2
    ma = GMm/r2
    a = GM/r2

    W = ma
    W = mGM/r2

    W... weight
    m... mass of body (100 kg in this case)
    G... gravitational constant
    M... mass of planet (4 Earth masses)
    r... distance separating the two masses

    You can calculate the weight with this.
     
  14. Mar 20, 2017 #13

    phyzguy

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    I'm not quite sure where you got this, but it is waaay off. According to Wikipedia, "The mass of all the objects of the asteroid belt, lying between the orbits of Mars and Jupiter, is estimated to be about 2.8–3.2×10^21 kg, or about 4% of the mass of the Moon. "
     
  15. Mar 20, 2017 #14
    I wish you fair winds and folllowing seas.
     
  16. Mar 20, 2017 #15

    PeterDonis

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    2016 Award

    Staff: Mentor

    There's not enough information here to answer the question. You would also have to know the radius of the planet; just saying it has four times the mass of the Earth does not determine the radius.
     
  17. Mar 20, 2017 #16

    Chalnoth

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    Depends upon the density.

    If the density were the same, then the radius of a planet with four times the mass would be the cube root of four times Earth's radius, or about 1.6 times the Earth's radius. Since the equation for surface gravity is proportional to ##m/r^2##, the surface gravity would be 4 / (1.6)^2 = 1.6 times Earth's gravity.

    If the density were different, the surface gravity could be very different.
     
    Last edited: Mar 21, 2017
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