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Free-fall from space to earth surface

  1. Feb 16, 2007 #1
    1. The problem statement
    A object is dropped into a free-fall from a distance of 3 earthradiuses from the center of the earth. Starting at a velocity of 0. How long time does it take for the object to travel the halv distance to earth (that would be the length of one EarthRadius), and how long would it take before the object hit the surface of the earth?


    Variables and given/known data
    EarthRadius = 6.37*10^6
    EarthMass = 5.97*10^24
    Gamma = 6.67*10^(-11)

    Gravity at EarthRadius = 9.8 m/s^2
    Gravity at 2*EarthRadius = 1/4 * 9.8 m/s^2
    Gravity at 3*EarthRadius = 1/9 * 9.8 m/s^2


    2. Relevant equations
    F = m*a

    m * a = (Gamma * EarthMass * m) / EarthRadius^2

    3. The attempt at a solution

    I've tried set up a differential equation, but I dont get a answer that is correct, so Im pretty stuck.

    m * a = (Gamma * EarthMass * m) / EarthRadius^2
    (divide by m on both sides and get this)
    a = (Gamma * EarthMass) / EarthRadius^2

    Then I substitute a with d^2*EarthRadius / dt^2

    and finally get:
    (Gamma * EarthMass) / EarthRadius^2 = (d^2*EarthRadius) / dt^2

    Then I set to initial conditions:
    i) the object starting speed is 0 -> Diff(EarthRadius)(0) = 0
    ii) and the object stop at EarthRadius -> EarthRadius(0) = EarthRadius

    But I cant get the answer right, so Im pretty stuck.

    I've calculated that the time it will take for the object to hit the surface of earth must be at least 1100 seconds and not more then 3400 seconds. This because the accelration at 1*EarthRadius from the centre of earth is 9.8 m/s^2 and at 3*EarthRadius from the centre of earth is 1/9*9.8 m/s^2


    ANY help and tips will be appriciated :)
     
  2. jcsd
  3. Feb 16, 2007 #2

    Dick

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    Tough one. I think you'll HAVE to set up a differential equation. Use gravitational potential energy to find the velocity v as a function of distance r. Then write v as dr/dt and do separation of variables. It's still kind of nasty. Anyone have a better idea?
     
  4. Feb 16, 2007 #3
    Two differentials? one from 0 to terminal velocity and then a straight forward differential with constant velocity? sum them? Just guessing really, your teacher must be a real bstrd.:eek: I yield the floor to the experts.:/
     
    Last edited: Feb 16, 2007
  5. Feb 17, 2007 #4

    nrqed

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    The first thing that comes to my mind is to do the following:

    Instead of [itex] F = m \frac{d^2 r}{dt^2} [/itex] I would write [itex] F = m \frac{d v}{dt} [/itex] so that you work with a first order DE instead of a second order one. Next, in order to solve this, you need a relation between v and r. This you can obtain by using conservation of energy. Then you get a first order DE for v(t). That's easy to solve. Once you have the function v(t), you are basically done. Using again conservation of energy, find the speed of the object when it is at the two points you are interested in. Then use your function v(t) to solve for the time t.

    Does that makes sense?

    Hope it helps.

    Patrick
     
  6. Feb 17, 2007 #5

    nrqed

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    I think you dropped a factor of [itex] {\sqrt 2} [/itex]. The answer should be between 1600 and 4800 seconds. (the equation is [itex] t = {\sqrt{\frac{2 \Delta x}{a_x}}}[/itex], right? And your delta x is twice the radius of the Earth.


    I did the calculation the way I described in my previous post and got
    [tex] \frac{R_E^{3/2}}{\sqrt{G M}} (\sqrt{3} + \frac{3\sqrt{3}}{\sqrt{2}} tan^{-1} (\sqrt{2}) ) \approx 4200 seconds [/tex]

    Patrick
     
  7. Feb 17, 2007 #6

    nrqed

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    It's really not that bad (see my two previous posts). I did it while taking a short break from my marking. So I think the prof deserves an apology :wink:

    It is much more challenging if the two masses are comparable (so that we must take into account the fact that both objects are in motion) but it still not that difficult.

    Regards

    Patrick
     
  8. Feb 18, 2007 #7
    Hi, thank you for your help

    Remember that the accelration isn't constant.

    We are having trouble finding v(t), this is how we have tried:

    F = m * a
    F = m * dv/dt

    Conversation of Energy:
    F = m*g + 1/2*m*v^2

    Then we set:
    m*dv/dt = m*g + 1/2*m*v^2

    Divide by m on both sides, and we get:
    dv/dt = g + 1/2*v^2

    We are trying to find v(t) using a differintial equation, but are stuck.

    Can anybody give us some hints on how to find v(t)


    Edit: We have found the speed of the object when it hit the surface of earth, we found it like this using Maple:

    restart:
    Gamma:=6.67*10^(-11);
    mj:=5.97*10^24;
    Rj:=6.37*10^6;
    W:=int((Gamma*(mj*m)/r^2),r=Rj..3*Rj);

    v:=solve((0.5*m*v^2)=W,v);

    The speed is 9129 m/s when it hit the surface of earth, and the starting speed is 0 m/s
     
    Last edited: Feb 18, 2007
  9. Feb 18, 2007 #8

    Dick

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    Check your units. The 'conservation of energy' statement is mixing forces and energies (ie it's wrong). What you want is a statement equating a change in potential energy with a change in kinetic energy. You want an equation giving v as a function of r. On the other hand, your matlab solution looks relatively ok. Do the same thing to get the value of v at any r.
     
  10. Feb 18, 2007 #9
    I give up!
     
  11. Feb 18, 2007 #10

    nrqed

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    Of course the acceleration is not constant! My derivation does not assume that at all. I only assumed one thing: that the mass of the object falling is negligible compared to the mass of the Earth (so that I could neglect the acceleration of the Earth). It's a bit more involved to take the acceleration of the Earth into account but is not too bad. If you want to take into account the acceleration of the Earth, let us know.

    By the way, I might have made a sign mistake in my calculation, I will check and let you know.
    EDIT: No, my solution looks correct
    :confused: :confused: :confused: This is not the equation for conservation of energy! This equation mixes forces and energy.
    This is wrong as well. You seem to be confusing forces and energy!


    The correct equations are

    Force equation:
    [tex] m \frac{dv}{dt} = \frac{G m M}{r^2} [/tex]
    and the conservation of energy (assuming an initial speed of zero):
    [tex] \frac{1}{2} m v_f^2 - \frac{G m M}{r_f} = - \frac{G m M}{r_i} [/tex]
    where r_f and r_i are the final and initial distances of the object from the center of the Earth. The mass "m" of the object cancels out completely.

    You don't need Maple to find the speed of the object when it strikes the Earth! Just plug [itex] r_i = 3 R_E [/itex] and [itex] r_f = R_E [/itex] in the equation for conservation of energy to find [itex] v_f [/itex].

    To solve the problem, just use conservation of energy to find the relation between v and r. Use that to write r in terms of v in the force equation. That gives a simple differential equation for v(t) that can be solved with a table of integrals or even by hand.

    Patrick
     
    Last edited: Feb 18, 2007
  12. Feb 18, 2007 #11

    Dick

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    I give up is the wrong answer! What are the expressions for kinetic energy and gravitation potential energy?
     
  13. Feb 18, 2007 #12
    Hi

    We got it finally working now, big thanks to everybody that have helped us, especially you Patrick/nrqed. We wasted a lot of hours with d^2*r/dt^2

    We will post the enitre solution (even though it's almost here already) tomorrow in case somebody else on the forum might have use for it some time later.
     
  14. Feb 18, 2007 #13

    nrqed

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    ::wink: You are very welcome. I take it that you got the same answer I got? (I did the calculation very quickly so I am glad you rederived it by yourself).
    It is noce to be thanked for helping, so I appreciate your post.

    Patrick
     
  15. Feb 19, 2007 #14

    Dick

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    If you want to check, I got 3221 and 4228 secs for the two fall times.
     
  16. Feb 19, 2007 #15
    Once again, thanks for all your help. We couldn't do it without you guys.

    Thanks for your reply Dick, we used Maple to calculate the time and we got 3217 seconds half way and 4223 seconds to hit the surface of Earth.

    Here are ours solution as promised, we post screenshots from Maple so you can see what we have done and what answers we got.

    [​IMG]
     
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