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Acceleration part 2

  1. Mar 21, 2007 #1
    1. The problem statement, all variables and given/known data

    density, ρ of the rock is 4000 kg m−3 and that this density is uniform throughout. You have equipment that enables you to measure the time it takes for a small object to fall from rest through a distance of 2.00 m. Your measurement is 8.0 s.

    Obtain an expression for ga in terms of the gravitational constant G, the
    radius of the asteroid Ra and the density ρ of the asteroid. Hence determine
    Ra.

    2. Relevant equations

    F= Gm/ r2

    F=mg

    volume of sphere= 4/3 * pi * r^3

    density= m/v

    3. The attempt at a solution

    substituting volume of sphere= 4/3 * pi * r^3 into density= m/v gives-

    density= 3m/ ( 4* pi * r^3 )

    4/3 * pi * r^3 * density = m

    F= Gm/ r2

    F= (G*4/3 * pi * r^3 * density) / r^2

    F=G*4/3 * pi * r * density

    F=mg

    mg= G*4/3 * pi * r * density

    4/3 * pi * r^3 * density = m

    4/3 * pi * r^3 * density* g = G*4/3 * pi * r * density

    g= Gr^-2

    or g= G / r^2


    (is this correct)???
     
  2. jcsd
  3. Mar 21, 2007 #2

    Mentz114

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    F= Gm/ r2 ?

    No.
    F = G*m1*m2/r^2
     
  4. Mar 22, 2007 #3
    if i use F = G*m1*m2/r^2, i am only given 1 object, 1 mass...

    so how would i go about doing this by using

    F = G*m1*m2/r^2
     
  5. Mar 22, 2007 #4

    hage567

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    Homework Helper

    You don't need to know the other mass, but you must include it in your equations. You'll see why.
     
  6. Mar 22, 2007 #5

    nrqed

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    Gold Member

    As others pointed out, you are not using the correct equation for the force of gravity!

    be very careful, there are two masses involved here, th etotal mass of the "rock and the mass of the falling object. You should really use two different symbols and be careful about using the correct one in each eqaution you are using.

    As for your final answer, there is an obvious simple check to make...does it have the dimensions of an acceleration??
     
  7. Mar 26, 2007 #6
    F = G*m1*m2/r^2

    substituting volume of sphere= 4/3 * pi * r^3 into density= m/v gives-

    density= 3m/ ( 4* pi * r^3 )

    4/3 * pi * r^3 * density = m

    im stuck here....

    shall i definte 4/3 * pi * r^3 * density = m = m1

    and leave m2 as m2
     
  8. Mar 26, 2007 #7

    hage567

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    Sure. Just keep them consistent.
     
  9. May 8, 2007 #8
    F = G*m1*m2/r^2

    V= (4/3)*PI*r^3

    p= m /v

    p= m1/ (4/3)*PI*r^3

    m1= p*4*PI*r^3/ 3

    now we substiute this value of m1 to this equation F = G*m1*m2/r^2

    F = G*m2*p*4*PI*r/ 3

    is final answer...can someone tell me if this correct..to Obtain an expression for ga in terms of the gravitational constant G, the
    radius of the asteroid Ra and the density ρ of the asteroid. Hence determine
    Ra.
     
  10. May 8, 2007 #9
    F = G*m2*p*4*PI*r/ 3

    F=mg

    mg= G*m2*p*4*PI*r/ 3

    g= G*p*4*PI*r/ 3

    as final answer.....can anyone..confirm if this is correct formula

    ga in terms of the gravitational constant G, the
    radius of the asteroid Ra and the density ρ of the asteroid.
     
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