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!

Grade 12 Force Field question

  1. Jan 21, 2012 #1
    1. The meteoroid Perturbation passes near the planet Conniption which has a mass of 5 * 10^25kg, a charge of +10^6 C and has a radius of 10500km. The meteoroid has a mass of 10 tonnes and a charge of +5 C. The meteoriod has an altitude of 9500km and a velocity of 10km/s. The planets magnetic field is 0.001T perpendicular to the meteoroids velocity. Determine the instantaneous acceleration of the meteoroid.
     
  2. jcsd
  3. Jan 21, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Relevant Equations? Attempt at Solution?
     
  4. Jan 21, 2012 #3
    Unsure on where to begin!
     
  5. Jan 21, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    Begin with relevant equations. What forces are involved and from what fields?
     
  6. Jan 21, 2012 #5
    The forces that are involved in this question from what I can tell are gravitational and magnetic.

    Some possible relevant formulas are:
    Fg = GMm / r^2
    Fe = kQq / r^2
    Fb = qvBSin(theta)

    I think it would be required to substitute these into F = ma to solve for 'a' somehow, but I am not sure.
     
  7. Jan 21, 2012 #6

    gneill

    User Avatar

    Staff: Mentor

    Okay, that's progress.

    Each of those forces will have a contribution to the total force on the mass. Each force will act in a direction that depends upon the nature of the field. That is, each is a vector quantity and you will add the vectors to determine the net force. The acceleration can then be found via [itex] \vec{a} = \vec{F}/m [/itex]
     
  8. Jan 21, 2012 #7
    mPerturbation = 10 tonnes = 10000 kg
    mConniption = 5 * 10^25 kg
    charge of Conniption [Q] = +10^6 C
    charge of Perturbation [q] = +5 C
    rConniption = 10500000 m
    altitude Perturbation = 9500000 m
    r = rCon + altitPert = 20000000
    v Perturbation = 10000 m/s
    B Conniption = 0.001 T (perpendicular to meteoroid's velocity)
    a = ?

    Attempt at solution.

    Find instantaneous acceleration:

    F = ma
    a = F / m
    a = ( -Fg + Fe + Fb ) / m
    a = [(-GMm/r^2) + (KQq/r^2) + (qvBsin90)]/m
    a = [(-(6.67*10^-11)(5*10^25 kg)(10000 kg)/(20000000 m)^2) + ((9.0*10^9)(+10^6 C)(+5 C)/(20000000 m)^2) + ((+5 C)(10000 m/s)(0.001 T)(1))/(10000 kg)]
    a = -8.32125 N/kg

    Not sure if the vectors are absolutely correct though or if something is out of place.
     
  9. Jan 21, 2012 #8

    gneill

    User Avatar

    Staff: Mentor

    You're answer looks okay for the given values, although the problem makes the gravitational force so much larger than the others that it swamps their effects.

    I also note that the specification of the magnetic field is ambiguous -- it is directed perpendicular to the velocity of the meteoroid, but that only defines a plane in which the field lines lie, not the direction of the field within that plane. That makes a difference to the direction of the force that results; if the resultant magnetic force did not align with the others you'd have to add their vector components rather than simply sum their magnitudes. Even aligned the magnetic force could be negative rather than positive.
     
  10. Jan 21, 2012 #9
    Forgot that Fg = -GMm / r^2 so that the - cancel out:

    F = ma
    a = F / m
    a = ( -Fg + Fe + Fb ) / m
    a = [-(-GMm/r^2) + (KQq/r^2) + (qvBsin90)]/m
    a = [-(-(6.67*10^-11)(5*10^25 kg)(10000 kg)/(20000000 m)^2) + ((9.0*10^9)(+10^6 C)(+5 C)/(20000000 m)^2) + ((+5 C)(10000 m/s)(0.001 T)(1))/(10000 kg)]
    a = 83537.5 / 10000
    a = 8.35 N/kg

    That is probably the final answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Grade 12 Force Field question
Loading...