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Gravity-free Fall

  1. Jul 13, 2006 #1
    Quenstion about netwonian mechanics.
    We have a small body that is moving against a very massive object with big density.
    For a given time t i want to know what can be the maximum distance of this object.
    So we have the mass of the big object M , the mass of the small m with speed Ua and the time t.
    the velocity at any moment is Ua+Uf where Uf is the speed because of the acceleration of the gravity.
    F=M*m*G/r^2
    d(dr/dt)/dt=a
    where a is the acceleration
    a=M*G/r^2
    r=R+s where R is the radius of the object and s the distance tha the object traveled at time t. I need an equation to find the s.
    Dont forget that the object begins with a speed Ua not from 0 speed.
    thanks and please quick
    Of course the gravity force is not constant but change with distance,
     
    Last edited: Jul 13, 2006
  2. jcsd
  3. Jul 13, 2006 #2

    HallsofIvy

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    If I read your question correctly, you have an object of mass m, initially at distance R from a planet of mass M and initial speed U,away from the planet.
    The simplest way to determine the distance to which the object will go is to use "conservation of energy".
    With force function
    [tex]-\frac{GmM}{r^2}[/tex]
    the potential energy function is the negative of the integral of that
    [tex]-\frac{GmM}{r}[/tex]
    where we take the constant of integration equal to 0 so that the potential energy is 0 at infinity.
    When r= R, the potential energy is
    [tex]-\frac{GmM}{R}[/tex]
    When the velocity is U, the kinetic energy is
    [tex]\frac{1}{2}mU^2[/tex]
    and so the total energy is
    [tex]\frac{1}{2}mU^2- \frac{GmM}{R}[/tex]
    the object will keep moving away from the planet until it kinetic energy is 0- when its potential energy is equal to the total energy
    [tex]\frac{GmM}{r}= \frac{1}{2}U^3- \frac{GmM}{R}[/tex]
    Solve that for r.
     
  4. Jul 14, 2006 #3
    Rephrase the question.

    my question was not exactly this, please let me rephrase it. thanks for the answer and please answer me again
    we have a massive object (neutron star) with radius Rn and mass Mn and we have a particle that is moving toward the massive object.
    The particle has initial speed Ua and we have the force of the gravity that pulls the particle.The time that needs the particle to fall on the nautron star is t (given).
    What is the distance that the particle traveled?
    so we have initial speed, to the same direction of the gravity force.
    The particle is in distance r from the surface of the massive object.

    p -->Ua---->Fn-----------------| Massive Object with Rn,Mn
    |------------r-----------------| distance
     
  5. Jul 14, 2006 #4

    HallsofIvy

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    Oh, so you are going the opposite way. You have an object at distance r= R from (the center of) a massive object (neutron star). You know that the time to impact on the object, at r= Rn, is T (I'm going to use capital T here so I can use t for the time variable). You want to know what R must be in order that the time to impact is T.

    Okay, since "conservation of energy" doesn't involve time, we will have to do this by "force= mass*acceleration". (Actually, I will now have to repeat much of what I did before and erased after I realized "conservation of energy" would answer the question I thought you had asked more simply.)

    The force, at distance r, is [itex]\frac{GmM}{r^2}[/itex]. The equation of motion (force= mass*acceleration) is
    [tex]m\frac{d^2r}{dt^2}= -\frac{GmM}{r^2}[/tex]
    We can, of course, immediately divide both sides by m to get
    [tex]\frac{d^2r}{dt^2}= -\frac{GM}{r^2}[/tex]

    That's a non-linear equation but since t does not appear explicitely in it we can use "quadrature" to simplify.

    Let v be the speed: [itex]v= \frac{dr}{dt}[/itex]. Then [itex]\frac{d^2r}{dt^2}= \frac{dv}{dt}[/itex] and, by the chain rule,
    [tex]\frac{d^2 r}{dt^2}= \frac{dv}{dt}= \frac{dr}{dt}\frac{dv}{dr}= v\frac{dv}{dr}[/tex]
    so the equation is now
    [tex]v\frac{dv}{dr}= -\frac{GM}{r^2} [/tex]
    which is a separable, first order equation
    [tex]v dv= -\frac{GM}{r^2}dr[/itex]
    Integrating,
    [tex]\frac{1}{2}v^2= \frac{GM}{r}+ C[/tex]
    Since v= U when r= R, we can evaluate that constant of integration.
    [tex]\frac{1}{2}U^2= \frac{GM}{R}+ C[/tex]
    so
    [tex]C= \frac{1}{2}U^2- \frac{GM}{R}[/tex]
    (which, if you look closely, really is the initial total energy: kinetic energy plus potential energy). For simplicity, I'm going to keep calling that "C" but we can substitute that value later.

    Now solve for v:
    [tex]v= \sqrt{2\frac{GM}{r}+ 2C[/tex]

    Since [itex]v= \frac{dr}{dt}[/itex] we now have the separable differential equation
    [tex]\frac{dr}{dt}= \sqrt{2\frac{GM}{r}+ 2C}[/tex]
    or
    [tex]dt= \frac{dr}{\sqrt{2\frac{GM}{r}+ 2C}}[/tex]
    Unfortunately, that integral is difficult. I'm going to have to think about it.
     
  6. Jul 14, 2006 #5
    Thanks alot, the answer was pretty difficult.
    the intergral is little... difficult but i found the answer (im mathematician).
    Can i ask another problem i have? i really appreciate your time.
     
  7. Jul 14, 2006 #6
    More difficult problem

    My next problem is like this. this is really tough

    we have a particle that is moving along the axe x, then we have the gravity force that is not on the axe x but the center of the gravity (the center of the massive object) has a distance h from the axe x.
    In time t what will be the distance that the object will travel?
    What are the equations of velocity,displacement with time?
    We need the r and h according to time.If i have the equation then i can put different h and r and see the result. Its really important the time, is given.
    Particle
    *---->-------------------------->x
    | Uin (initial Velocity)
    |
    |
    |
    |
    |
    h
    |
    |
    |------------r-----------------* Massive star
    _y___________________________________________________

    Its really important to find the solution , thanks for your time
     
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