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Help with gravitation and free-fall acceleration

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data
    The planet Uranus has a mass about 14 times the Earth's mass, and its radius is equal to about 3.7 Earth radii. A)By setting up ratios with the corresponding Earth values, find the free-fall acceleration at the cloud tops of Uranus. B)Ignoring the rotation of the planet, find the minimum escape speed from Uranus.

    3. The attempt at a solution
    I'm kinda clueless as to where to start by i did set up the ratios that it asks to do. Mearth=5.97*10^24kg and Muranus=83.58*10^24kg. and i just set up the ratio of Mearth/Muranus with out reducing anything. and for the radius, i have Rearth=6.38*10^6m and Ruranus=23.606*10^6m with setting up the ratio the same way. i think i just need a hint to get started and a clue to which equation these ratios would be used in
  2. jcsd
  3. Jan 19, 2009 #2
    Think about what you're trying to find in part (a)--it's much like what the value [itex]g[/itex] is on earth right? Note that [itex]g[/itex] can be expressed as


    For part (b), think of the fact that escape velocity is really something that comes from conservation of energy.
  4. Jan 19, 2009 #3
    Don't know where to begin?
    ---Ignore the question and write down whatever formula that might be useful, even odd ones.
    ---Look at what you know, then at your equations
    ---Form any links
    ---Look back at the question and what it is asking.

    Ok the question is to do with gravity, so perhaps Newton's Law of Gravitation might be useful;

    F = GmAmB / r2

    Then perhaps his Second law of motion;


    Well we know the masses and radii so this looks good. First take the link that;

    mAa = GmAmE

    So that for some object A in a gravitational field arround B it will experience a force accelarating it downwards. This is g hence;

    gE = GmE / r2

    For Earth we can just plug the data in for its mass, and a radius at a given point, let's say the Earths surface. Thus we know g on Earth.

    For Uranus, we are told it's 18 times more massive, and its radius is 3.7 times bigger. Thus we can rewrite our formula for gravitational accelaration on Uranus as;

    gU = 18GmE / (3.7*rE)2

    Simple really (you can expand the denominator and remove the numerical co-efficent as a fraction of 18 and that is your ratio of Earths gravitational accelaration to Uranus's. ;)


    The next bit is somewhat more tricky unless you've done A level or A level equivilent calculus;

    Ok raising a massive in a mass in a gravitational field means doing work against the gravitational pull of the planet. Hence we can write the expression;

    w = Integ{ F } ds

    Where the work done (w) is the infintesimal sum of all the little movements 'up' in the gravitational field. Now say we want to raise this object an infinite distance in the gravitational field, well that's gonna be tricky. But that is our end limit of intergration. Assuming we are going to raise the mass from the planets surface (in this case Uranus which is 3.7 times greater than Earths radius) that makes our other limit. Hence our equation looks like;

    w = IntegInfinity3.7rE{ F } ds

    F = GmAmE / r2
    and our direction of motion for the ds is actually in the radial direction from Uranus's centre (I won't go into conservative forces here), we will denote this 'dr'. Hence;

    w = (1800/1369)*IntegInfinity3.7rE{ GmAmE / r2 } dr

    NOTE: the ratio in bold is that ratio I was talking about ealier. I have removed it from the intergration brackets because it doesn't change with the intergration varible r. You can probally appriciate that I could have also taken out G, and the product of the masses at this stage as well should I have wanted to.

    Anyhow you'll note that this intergrates to;

    w = [ -(1800/1369)*GmAmE / r ]Infinity3.7rE

    NOTE: 1 / r is the same as r-1.

    Now you can just evaluate the numerical amout of work it takes to raise the object to infinity.

    Question: What is work a quantity of? and how is this linked to this escape speed we want to find? Well work is a form of energy, hence you'll appricate that;

    Work Done = Kinetic Energy

    Kinetic energy is 1/2 mAv2 where v is its velocity. Hence you might be able to appricate that;

    w = 1/2 mv2

    sqrt{ (2w / mA) } = v

    This v is your escape velocity, because if an object had this velocity, then it has enough energy to escape the pull of gravity that the planet would provide. Preaty neato huh?

    In practice this is only a lower limit due to air resistance etc. and the rotation of the planet would also create an accelrating frame of referance for the object at the equator, but those are trival in most circumstances.

    Hope that helps!
  5. Jan 19, 2009 #4
    I think there is a simpler way to derive the escape velocity (if we're just talking about one surge of velocity at [itex]t=0[/itex] and not a rocket or anything). From conservation of energy we have

    [tex]K_i + U_i = K_f + U_f[/tex]

    Two of these go to zero; the final potential energy (we're going to infinity) and the final kinetic energy (this is the speed just enough to get it out of the g-field). So now we get


    Where [itex]m[/itex] is the mass of the object and [itex]M[/itex] is the mass of the planet and [itex]R[/itex] is the radius of it. Solving for [itex]\mathbf{v}[/itex] we have


    And I took the norm because you just need the speed. I might be missing something about this but I think this could be a simpler way to do it.
  6. Jan 19, 2009 #5
    It would be if you already knew what the gravitational potetial was;


    But if you don't know that, then you might as well show how to derive that by first principles. Really all you need to know is;


    But memorising the hundreds of equations out there I doubt is worthwhile doing...

  7. Jan 19, 2009 #6
    thanks a lot for the help guys. its so much easier when it is explained like that. my physics 2 teacher don't do too good a job teaching it so it can be tough
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