1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mass dropped into hole through Earth

  1. Apr 22, 2004 #1
    assumptions:
    Earth is spherical, and it's density uniform.

    Me, G, r(earth) are known, m(object) is not known

    How fast is the object travelling when it passes the centre, and what is the period of it's oscillation ??

    --------------------------------------------------
    x(t) = r cos wt

    (r is the magnitude of the harmonic motion, radius of earth)

    x(t) at centre = 0, thus (wt) = pi/2

    v(t) = -r w sin(wt) by differentiation x(t)

    sin(wt) = 1,

    v = -rw

    w = sqrt(k/m), m is the object dropped

    k = GMmr^-2, ie the Grav. force

    so we can get wr = sqrt(G * Me)

    about 19962000 m/s

    Does this look OK so far ?

    working out T (similar to above, T = 2 pi/w ) i got about 2s, WTF that can't be right.
     
  2. jcsd
  3. Apr 22, 2004 #2
    It takes about 45 minutes for any object to pass entirely through the earth, at abotu any angle, as long as it is not affected by forces other than gravity (and the mass is negligible...but if its not, that cna be added in quite simply).

    You should also add that the earth is not spinning in your setup. If it were spinning and an object were to fall in at a vector perpendicular to the earth's axis, they would bounce back and forth on the walls.
     
  4. Apr 22, 2004 #3
    OK, ignore rotation, you're right.

    OK i re did it, T = 84 minutes which i think is right.

    T = 2*pi*sqrt(R^3/ G*M) = 84 minutes, this bit is OK

    speed at the middle = rw,

    w = sqrt (GM/r^3)

    use that and you get 7904 m/s, sound right ? I don't think so



    G = 6.67 *10^-11
    M earth = 5.9742 *10^24 kg
    R earth = 6378100 m
     
    Last edited: Apr 22, 2004
  5. Apr 22, 2004 #4
    If you need to be concerned with the velocity at the center of the earth, jsut use conservation of energy:

    [tex]E = .5mv^2 - \frac{GMm}{r}[/tex]

    where [tex]M[/tex] is the mass that's under the falling object, in other words, if at some arbitrary point inside the earth, your distance from the center is R, then R is the radius of the sphere you need to worry about. The earth that you've passed through doesnt matter.
     
    Last edited: Apr 22, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?