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Ball falling through hole in the earth.

  1. Apr 19, 2004 #1
    I was just trying to figure out what would its period be and see where I went wrong (not homework, more of a thought experiment).

    So there is a hole down through the centre of the earth, spherical earth nothing complicated like air resistance etc.

    So basic stuff...

    [tex]a = \frac{GM}{r^2}[/tex]

    Acceleration of the body due to gravity, where M is the mass of the earth.


    [tex]a=-\omega ^2 r[/tex]
    because its displacement is equal to the radius
    and the period of a body undergoing SHM

    [tex]T= \frac{ 2 \pi }{\omega}[/tex]

    So then

    [tex]T^2 = \frac {4\pi^2}{\omega ^2} [/tex]


    [tex]\omega ^2 = \frac {GM}{r^3}[/tex]

    Now I'm confused as to why I got the same result as keplers 3rd law

    [tex] T^2= \frac {4\pi^2 r^3}{GM}[/tex]

    although I probably did something terribly wrong
  2. jcsd
  3. Apr 19, 2004 #2


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    In the beggining there is mistake.The mass M in the formula above isn't constant.The body is accelerated as it falls trough the earth only by spherical part of Earth's mass below it (without any contribution of a "kernel" mass above it has already passed throgh-Assuming, of course, that Earth is a perfect homogenous sphere.)Find the expression how Gravitational force drops down inside the homogenous sphere ,than ask some PF mathematician to "solve" gravity oscillator problem for you (or try to do that yourself).I promise you a fun.Enjoy!I'm not going to give you more on this,just this hint.

  4. Apr 19, 2004 #3
    Oh, didn't know that.

    so it would have to be more like
    [tex]a = \frac {G(M-A)}{r^2}[/tex]

    Where A is the mass of the "kernel" as you called it.

    so R = original radius
    r = new radius

    [tex] A = \frac { g (R-r)^2}{G} [/tex]

    I am unsure what value to put as 'g' though, would it be the original value or the value due to the decrease in radius?

    But then does the radius in both change too?
    Last edited: Apr 19, 2004
  5. Apr 19, 2004 #4


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    Keppler's 3rd Law

    Regarding your original post, it should not be particularly surprising that you came up with Keppler's 3rd law.
    You can think of objects that are in orbit as falling, but missing the planet. Since you assumed that the entire mass of the planet was acting on the falling object the entire way, it's hardly surprising that you ended up with orbital mechanics.
  6. Apr 19, 2004 #5
    Yeah I just realised that now, silly of me to assume that
  7. Apr 19, 2004 #6


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    I was wondering about something similar to your problem, and I think it requires calculus and integrals (which I'm currently punching myself in the stomach for not learning in school). By the looks of your expression, you're calculating the gravitational force exerted by each kernel by its total mass, there's a problem with that. The majority of the mass of the kernel doesn't contribute much to the net force, since it cancels itself out. Well, it's really too complicated for my level of math.
    Last edited: Apr 19, 2004
  8. Apr 19, 2004 #7
    Hehe, this problem is so fun.

    The math is actually works out pretty nicely.

    Here's a hint: Express the effective mass as a fraction of the total mass.

    The real problem is figuring out the minimum time to get from San Francisco to New York via the inside of the Earth.... I gotta work on that one a bit more.

  9. Apr 19, 2004 #8
    It works out nicely? Oh man this is going to be a real treat if I get it!

    So if

    N = mass planet - kernel
    A = mass of kernel

    [tex] a = \frac {G (\frac {N}{N + A})}{r^2}[/tex]

    I honestly don't know where to go from there. Am I even on the right track with
    [tex] T= \frac {2\pi}{\omega}[/tex] and [tex] a= - \omega ^2 r[/tex] ?
    Last edited: Apr 19, 2004
  10. Apr 19, 2004 #9
    Well I think I got it but only because of something I stumbled across.


    [tex] a = \frac {GM^_Er}{r^3_E}[/tex]

    when [tex]r = r^_E[/tex] a = g

    [tex] g = \frac {GM^_E}{r^2_E}[/tex]


    [tex] a = g \frac {r}{r^_E}[/tex]

    [tex] \omega^2 = \frac {g}{r^_E}[/tex]


    [tex] T^2 = \frac {4 \pi^2 r^_E}{g}[/tex]

    Please tell me that I'm right. And if anyone has the time could you explain Gauss's Law in this context? I don't think I could have done this without that, or could I?

    Edit: My radius of the earth seems to be wonking up sorry.

    Ha! Just noticed that

    [tex] T= 2 \pi \sqrt \frac {r^_E}{g}[/tex]

    is exactly the same as the one for a simple pendulum.
    Last edited: Apr 19, 2004
  11. Apr 19, 2004 #10
    Well, look at what the force is.

    [tex]F = -G\frac{Mm}{r^2}[/tex]


    [tex]M_\textrm{eff} = \frac{4}{3}\pi r^3 \rho[/tex]

    So F can be expressed as

    [tex]F = -G \frac{M_\textrm{tot} m}{R^3} r[/tex]

    if we let M_eff = M_tot/M_tot*M_eff = M_tot*(r^3/R^3)

    So we essentially have a force of the form

    [tex]F = -k r[/tex]

    Look familiar?

    I told you the math worked out pretty nicely. =]

  12. Apr 19, 2004 #11
    ... And the fun hasn't ended with this!

    Consider what happens if you smash the ball with racket in the center of the earth. If you smash it really hard (so it escapes the earths gravitational pull), it can get velocities that are actually much larger than if you had smashed it in free space. This is amazing. You are sitting in a potential well, but actually you can use the well to fly up faster than you could without it (at least I think so, but I have been wrong so many times before...).

    Look at http://www.dd.chalmers.se/~f99krgu/main/index.html?rocket for more information.
  13. Apr 20, 2004 #12


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    I'll give hints one more time (just hints, not a solution since I want make people think).
    Inside a homogeneous sphere force on the test particle decreases linearly (from max value on the surface to the 0 in the center).This driving force, which is linear, and produces linear differential equation of harmonic oscillator .Its' solution is trivial to find (the only little more difficult part of the job is to prove that the force is linear).
    However,generally speaking, outside the sphere driving force is not a linear one.
    Radius of Earth is Ro~6400 km.Consider the case when the body is dropped from say altitute h=3600 km (it falls and it reaches the tunnel enterance and continues to fall).
    How does the driving force between r=10000 km and r=6400 km affects the motion?This part of the problem includes a classical problem of gravitational oscillator mentioned earlier.Ask mathematicians (in PF math subforums ) for advice about nonlinear different equation involved in the problem (if you want add more fun to it..)

  14. Apr 20, 2004 #13
    Because the equations are non-linear outside the earth I use Matlab ode45-solver to solve them and get an oscillator that are somewhat modified from the harmonic if you drop the ball from 10000 km altitude.

    What I'm pointing at is that if you apply a large extra force to the ball as it passes the center of the earth it can eventually end up, far from the gravitational potential, with a larger (? or maybe equal, I guess I will have to solve more careful when I get time) velocity than if you had applied the force in free space. This is because you take a certain advantage of the attractive gravitational force (before the center) and don't have to pay as much in the non-attractive part (after the center), since your velocity, after applying the force, is so high you don't spend as much time in it.
  15. Apr 20, 2004 #14


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    Yeah Kolekrille,I understood what you said in your post before.I was using your post to address to Blistering Peanut since he didn't specify "altitude" in his post ("start up" -boundary conditions are important for DE as we know).

    Utilazing energy of gravitational field you indicate in your post is nothing new.In order to get final sattelite velocity gain along its trajectory,NASA used this trick to subsequently accelerate Voyager (It was Voyager wasn't it?) on its route.

  16. Apr 20, 2004 #15
    This is the same principal Star Trek uses to travel back in time for some whale-watching! I forgot which one that was... was it 4?
  17. Apr 20, 2004 #16


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    Don't pull the legs,or (even worse) try to impute some cranky argument in the discussion if you missinterpret what is word about.
    Certainly,this is classical physics,and total energy of the system is conserved.
    Its's also that little thingy that you cannont "fix" big object in gravitational interaction with very small one no matter how large M/m ratio,not mentioned yet-the problem shifts to new hights to if you manipulate speeds (and kollekrile sniffs close).
    NASA used 2 body gravitational system effects very dramatically during one satelite mission -I don't remember which one.
    Velocity gain when satellite with velocity V1 passes close to a moving planet
    is possible.If the planet moves at velocity V2,the satellite may lose or gain the energy.Dependicy is on the angle between velocity vectors and angle of trajectory deflection,which in turn depends how close to the planet the satellite passes by.The greatest gain of speed and the energy of the satellite happens when the two objects move in the oposite direction and when the satelite gets as close as possible to the planet while passing behind it so that its' direction almost reverses.The satellite emerges with the final velocity
    V1'~V1 + 2V2 ,from the standpoint of stationary observer reference frame.
    In this case the moving planet itself provides the energy.
    So much about "star trek" principle :cool:
  18. Apr 20, 2004 #17
    I fail to see the connection. You seem to assert that my words are based on some sort of lack of understanding. I was merely creating an association to something people may have a slight grasp of but didn't really understand. I find that such associations help one understand not only the principle discussed, but as an added bonus, also makes the movie that much cooler (setting aside the whole faster-than-light travelling through time aspect).

    Your points are valid and not out of place, but your remarks about my post are entirely out of place.
  19. Apr 21, 2004 #18


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    Pergatory,sorry, it sounded like you were mocking.If that isn't the case,you than realize than that gravity field serves just to provide force neccessary to exchange kinetic energy between two objects.It is the analogy of simple elastic collision .But something else is intersting in kolekrille's saying:
    These words may be just little changed,to enable one to show the example how velocity gain observation in the SAME direction is possible from the stanpoint of observer in free fall in a gravity field of disscussed type compared to the situation where the same observer moves uniformly in free space with same inital velocity .Pretty elementary,without need to determine "solution" of the motion.Just based on Energy considerations and gravity potential curve shape.
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