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Homework Help: Hardest and Trickiest Question I've Encountered.

  1. May 27, 2008 #1
    Hardest and Trickiest Question I've Encountered. Please Help!

    1. The problem statement, all variables and given/known data
    An alien spaceship is revolving around a spherical asteroid in a very low circular orbit. The period of revolution of the spaceship is 15 minutes. A research party of the alien students lands on the steroid and drills a tunnel through it along a diameter. They then launch a rock into the tunnel with an initial speed equal to that of the orbiting spaceship. How long would it be before the rock comes back to the launch point?

    2. Relevant equations
    None that I know of

    3. The attempt at a solution

    Here is what I think, since T= 2piR/v, I thought I could get v. However, I dont even know R and I dont even know distance if I am to use v as in d/t. There seems to be too many variables involved and too little equations to use. I believe that words, very low circular orbit, could mean something. I guess I only know that the period is 15 mintues and that the rock is launched at an initial speed equal to teh orbiting space ship. That means that the velocity of the rock is equal to 2piR/15 minutes. however, i still dont know r.
  2. jcsd
  3. May 27, 2008 #2


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    Your equations are seriously underpowered. You need to know Newton's law of gravitation and assume that the body is uniformly dense. Then you need to compute the orbital period at the surface and compare it with the time it takes on a trajectory through the center. If the only equation you have is T=2piR/v, this is way too hard for you.
  4. Jun 4, 2008 #3
    no idea if this is right but yeah...


    T=15min = 900sec

    V=(2*pi*r)/900 _____(1)

    Distance= speed*time
    t=D/S _______________(2)

    distance is the diamater which is 2r

    equation 1 into 2


    after abit of algebra

    t= 286.48 sec (2dp)
    t= 4min 46 sec

    dunno if its right but yeah there it is

  5. Jun 4, 2008 #4


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    You just computed 15min/pi. That's the time required to pass through the hole one-way ignoring gravity. That much is right. But if you ignore gravity nothing will orbit and the object shot down the hole will never come back. That's no fun!
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