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Relationship between Gravity and Density

  1. Jan 18, 2014 #1
    It would be greatly appreciated if someone could help me solve this problem.

    There are two objects, let's call them object A and object B. Assume that object B is the Earth and object A is some other planet. If object A has 10 times the gravity of object B (Earth) and assuming object A and object B (Earth) have the same density, how do I find the mass, radius, and volume of object A?

    Thank you.
     
  2. jcsd
  3. Jan 18, 2014 #2

    Jonathan Scott

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    Just some hints:

    If the two objects have the same density, then their masses are in proportion to their relative volumes, which is proportional to a power (work it out) of the ratio between their radius values.

    If you write out the gravitational field at the surface of each object, that also involves the mass and some other power of that ratio for the radius part.

    If you give the ratio of the radius values a name, say k, you can write down the formula for the gravitational field of one object in terms of the mass and radius of the other. If you know what the relative strength of the gravitational field is, then you can solve for the ratio and hence work out the relative radius and mass.

    The result is a lot simpler than most people would suspect.
     
  4. Jan 18, 2014 #3

    HallsofIvy

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    What, exactly, do you mean by "gravity" when you say "10 times the gravity of the earth"? Gravity is a force, depending on distance, not a fixed number. If you mean "the force of gravity at the planet's surface", then the force is proportional to the mass of the planet, which is proportional to it volume, which is proportional to the radius, cubed, but inversely proportional to the square of the radius.
     
  5. Jan 18, 2014 #4
    Thank you for your response. And, yes, by gravity I mean "the force of gravity at the planet's surface." I also understand just about everything you said. I understand that the force of gravity is proportional to mass, which is proportional to volume, which is proportional to radius cubed, etc. However, what are those proportions? I know the formula for calculating surface gravity and I know the formula for the volume of a sphere, so how do I figure out those proportions and solve the problem?

    Please understand that I'm a neophyte who has some interest in astrophysics but the last time I took a math class was algebra in college 7 years ago. I'm not looking for you to solve the problem, only point me in the right direction.
     
  6. Jan 18, 2014 #5

    Jonathan Scott

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    Just give names to the quantities that you are working with and then work through the equations.

    For example, let ##r_A## be the radius of object A, ##r_B## be the radius of object B, and let ##k## be the ratio between them, so ##r_A = k r_B##. You can then let ##m_A## be the mass of object A and ##m_B## be the mass of object B and use the volume ratio to work out what ##m_A## is in terms of ##k## and ##m_B##, and so on.
     
  7. Jan 18, 2014 #6
    Thanks for your response. However, I am not sure I understand you. Are you telling me that once I have this equation: rA=krB that I need to solve for k in order to get the ratio between rA and rB? Also, what do you mean by "use the volume ratio to work out what mA is in terms of k and mB"? What is the "volume ratio" and how do I use it as you have advised?
     
  8. Jan 18, 2014 #7

    Jonathan Scott

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    You're on the right track. I was hoping to leave you something to work out for yourself!

    For a fixed density, the mass is proportional to the volume.

    The formula for volume involves ##r^3##, so if everything else is the same and the radius is increased by factor ##k##, ##r^3## becomes ##(kr)^3## which is the same as ##k^3 r^3## so the mass is increased by a factor ##k^3##.

    The formula for the gravitational acceleration in terms of the mass and radius is ##g = Gm/r^2## so the gravitational acceleration ##g_A## for object A is given by the following:

    $$g_A = {G m_A}/{r_A}^2 = {G (k^3) m_B } / (k \, r_B)^2 = {G (k^3) m_B } / (k^2 \, r_B^2) = k \, {G m_B}/{r_B}^2 = k \, g_B$$

    That means that for constant density, the gravitational acceleration at the surface is proportional to the radius.

    I hope can you finish that off from there now!
     
  9. Jan 19, 2014 #8

    Jonathan Scott

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    Note that if the density of the objects is different, the mass is affected in the same proportion.

    For example, the Moon has about 27% of the radius of the Earth and about 60% of the density, so the strength of the gravitational field on the surface of the Moon is about 27% of 60% of that of the Earth, which is about 16%.
     
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