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Falling body where acceleration is not constant

  1. Mar 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Most physics problems that involve a falling body assume constant acceleration. How does one account for the changing acceleration if a body is dropped from a very high altitude?

    2. Relevant equations
    With constant acceleration, d = (1/2)at^2 or t = sqrt(2d/a)
    but a is not really constant so How is this solved.

    3. The attempt at a solution
    My first thought was to do a piece wise solution. Start at a given altitude, assume a small time, say one second and calculate the distance traveled. Then calculate the acceleration based on the new distance between bodies and continue. Is there a closed form solutio?
     
  2. jcsd
  3. Mar 23, 2015 #2

    PeroK

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    Is this really homework?

    Anyway, the direct approach involves solving a differential equation.

    Alternatively, you could consider conservation of energy.
     
  4. Mar 23, 2015 #3
    No, this is not homework but rather a problem I would like to know the answer to. I figured that a diff equation would be required but I don't see the trick in solving it. Ultimately, I would like to get the altitude as a function of time but since the acceleration is a function of altitude I am not sure how to set this up.
     
  5. Mar 23, 2015 #4

    PeroK

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    Suppose an object of mass ##m## starts at rest from a height ##h##. How long does it take to fall, using a varying gravitational force?

    You can start with energy conservation using gravitational potential. Let ##r## be the distance from the centre of the Earth, ##M## the mass of the earth, and ##R## is the radius of the Earth.

    ##PE(r) = -\frac{GMm}{r}##

    ##KE(r) + PE(r) = PE(R+h)##

    ##\frac{1}{2}m(\frac{dr}{dt})^2 = \frac{GMm}{r} - \frac{GMm}{R+h}##

    That gives you a differential equation in ##r##.

    The trick to solving it is to let ##r = (R+h)cos^2(\theta)##
     
  6. Mar 23, 2015 #5
    I understand how you used the conservation of energy to get the differential equation.
    However, I seem to need another hint to see how to use the substitution r = (R+h)(cos A)^2
     
  7. Mar 23, 2015 #6

    PeroK

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    Just plug it in and simplify. For example:

    ##\frac{dr}{dt} = -2(R+h)cos(\theta)sin(\theta)\frac{d \theta}{dt}##

    As you can see, things get worse before they get better!
     
  8. Mar 23, 2015 #7
    Looks like you can integrate to get r = (R+h)(sin A)^2 but then using the change of variable for limits gets messy.

    Maybe my numerical approach might be better to get a velocity at a specific height but I wold have rather get a closed form solution.

    I see why this is not a Physics homework problem.
     
  9. Mar 24, 2015 #8

    PeroK

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    It's not too bad. We have:

    ##\frac{1}{2}m(\frac{dr}{dt})^2 = \frac{GMm}{r} - \frac{GMm}{R+h}##

    ##(\frac{dr}{dt})^2 = 2GM\frac{R+h-r}{r(R+h)}##

    Letting ##r = (R+h)cos^2(\theta)## gives

    ##4(R+h)^2cos^2(\theta)sin^2(\theta)(\frac{d \theta}{dt})^2 = \frac{2GM(R+h)(1 - cos^2(\theta))}{(R+h)^2 cos^2(\theta)}##

    Which simplifies to:

    ##cos^2(\theta)(\frac{d \theta}{dt}) = \sqrt{\frac{GM}{2(R+h)^3}}##

    We can then integrate from ##t = 0, \ \ r = R + h, \ \ \theta = 0## to ##t = T, \ \ r = R + z, \ \ \theta = \theta_z##

    Which gives the time ##T## taken to fall from a height ##h## to a height ##z##

    ##T = \sqrt{\frac{(R+h)^3}{8GM}}(2 \theta_z + sin(2 \theta_z)) \ \ (\theta_z = cos^{-1}(\sqrt{\frac{R+z}{R+h}}))##

    To get the time to fall to the ground, you just set ##z = 0##

    I checked this formula out and found that for a fall of 1,000m the difference is only 0.002s compared to using the constant surface gravity.
     
  10. Mar 24, 2015 #9
    Wow! I would never have solved this problem. The real reason I wanted to solve his problem is that I was reading a si fi novel where an asteroid caused the moon to stop orbiting around he earth. I wondered how long it would take for the moon to collide with the earth if it were not moving in its orbit. Then I wanted to know what an observer would see as the moon got closer and closer to the earth. Now I can figure it out.

    thanks
     
  11. Mar 24, 2015 #10

    PeroK

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    For something as far as the moon, you can get an estimate by taking ##R = 0##. In which case:

    ##T = \pi \sqrt{\frac{h^3}{8GM}}##

    Where ##h## is the distance to the moon, give or take the Earth and moon radii.
     
  12. Mar 24, 2015 #11
    I calculate 4.76 days. Now for the final question. What would be the relationship between distance or height and time, i.e. distance = f(time)?
    From this I could calculate the angular size vs time. It obviously would not be parabolic as the acceleration is continually changing.

    I recall a movie recently where some planet was going to hit the earth and the movie ended with the star, Kirsten Duntz as I recall sitting on her and watching patio as the approaching planet got slowly larger and larger.
     
  13. Mar 24, 2015 #12

    PeroK

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    You can get it from that equation. We can set ##R = 0##, then we have:

    ##T = \sqrt{\frac{h^3}{8GM}}(2 \theta_z + sin(2 \theta_z)) \ \ (\theta_z = cos^{-1}(\sqrt{\frac{z}{h}}))##

    That gives you the time at distance ##z##.

    You could use a spreadsheet to calculate ##T## for whatever set of distances you want.
     
  14. Mar 24, 2015 #13
    Will do. thanks for your help.
     
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