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Gravitation acceleration

  1. Jun 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A mass m=100[kg] is released from stand still at point A which is in a distance of 1E7[m] from the face of a planet.
    The planet's radius is also 1E7[m] and the free fall acceleration on it's face is 10[m/s2.
    What is the acceleration at point A

    2. Relevant equations
    The acceleration: ##g=\frac{GM}{r^2}##

    3. The attempt at a solution
    First i find the mass:
    $$10=\frac{6.7E-11\cdot M}{1E7^2}\Rightarrow M=1.49E18$$
    $$g=\frac{6.7E-11\cdot 1.49E18}{2E7^2}=250E-9$$
    Obviously this is wrong, the answer should be g=2.5
     
  2. jcsd
  3. Jun 28, 2014 #2

    TSny

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    You have it set up correctly. Check your calculation of M. Did you treat (1E7)^2 correctly?

    What about units, especially in your answer?
     
  4. Jun 28, 2014 #3

    adjacent

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    You have got M correct but this part is wrong.

    The radius is ##(2 \times(10^7))^2## not ##2 \times (10^7)^2##

    And yes, it's more accurate to treat ##G## as ##6.67 \times 10^7##
     
    Last edited: Jun 28, 2014
  5. Jun 28, 2014 #4
    I fixed it, i calculated wrong.
    There is a harder continuation to this problem.
    Approximate the time of fall by imposing upper and lower limits on the time.
    I took the acceleration on the face and it gave a time of 1414[sec]. the upper limit is more complicated since the result is 2828[sec], and it isn't with the acceleration at A.
    I guess i could approximate the acceleration to vary linearly with r: g=kr, but then i don't know to treat this kinematically
     
  6. Jun 28, 2014 #5

    TSny

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    If you note that g varies as the inverse square of the distance, you should be able to get the answer for g at A without pencil and paper. What happens to g if r is doubled?
     
  7. Jun 28, 2014 #6

    TSny

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    Are you sure that you don't get 2828 s if you use the acceleration at A?
     
  8. Jun 30, 2014 #7
    r = planet radius = 1 E+7 metres
    R = elevation = 1 E+7 metres
    G = 6.7 E-11 (a constant)

    Planet mass (M) from :
    M = ( r ² * 10 ) / G
    M = 1.493 E+25 kg

    Only if you disregard the mass of body 2 (100 kg) as negligible (which it is, compared to the planet) can you simply calculate the field strength (g) from the basic equation :

    g1 / g2 = ( ( r2 / r1 ) ² )
    Key :
    g1 = surface field strength ( ( m / s ) / s )
    g2 = outer field strength ( ( m / s ) / s )
    d1 = surface distance (m)
    d2 = outer distance (m)

    Transpose for g2 and calculate :
    g2 = 2.5 ( ( m / s ) / s )

    Your right with the time parameters, putting g as 2.5 and 10 gives the fall times of 2828 and 1414 seconds
     
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