1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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


    User Avatar
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted