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Homework Help: Confused about physics of freefalling (through sky)

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Basically, i was told to describe the physics involved with a person who just happens to be falling out the sky...However, i'm a bit overwhelmed as to where i should start off
    Human mass: 54kg
    Earth's gravity: 9.8m/s^2
    Air resistance:???perhaps negligible or would that throw my answer off?

    2. Relevant equations
    So im guessing some relevant equations would be....
    Newton's Law of Gravitation
    however i was reading that his law only applies to 3 situations and one of which is a sphere (earth in my case) and a "particle" which is much smaller (i assume that would be the human) Does that work??

    Acceleration due to gravity=9.8m/s^2

    3. The attempt at a solution
    I can find the gravitational force between the human and the centre of earth through newton's law of gravitation and explain this.

    is there anything else i'm missing/can explain? How would I incorporate air resistance into this...
  2. jcsd
  3. Apr 5, 2010 #2
    You should recite Newton's Laws every night before bed =].

    Also, I see a lot of '[tex]9.8ms^{-2}[/tex] and I assume this is just the examining board setting lower standards for some subjects, but for the last 4 years I have been using [tex]9.81ms^{-2}[/tex] and don't see the need to drop a 0.01!
    Last edited: Apr 5, 2010
  4. Apr 5, 2010 #3


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    You don't really need Newton's law of gravity. Near the surface of the Earth, the force exerted by the Earth on the person is the person's weight, mg. A sky diver reaches terminal velocity. You can find a rough number for that by doing some research on the web. Then you need to explain what "terminal velocity" means in terms of the person's acceleration and the forces acting on him/her.

    You might also wish to discuss whether the acceleration of the person is constant or varies with time and provide a rough sketch of the acceleration as a function of time and the velocity as a function of time.
  5. Apr 5, 2010 #4
    Why wouldn't Newton's law of gravity not needed, wouldn't it be a good way to describe the relationship between the human and the centre of the earth (earth is pulling him towards)?

    I'm looking up terminal velocity right now, personally haven't learned that yet, but might get some brownie points for mentioning things we haven't discussed yet in class : D

    haha, no that's just me, not a superb physics student in general..............
  6. Apr 5, 2010 #5
    Hi xchococatx
    Could you just provide a little more information.
    From what height is this unfortunate person falling? "Out of the sky" is a little vague if we are to give a detailed answer. It may not be significant but it provides a starting point.
    You are right that the most important and obvious force/acceleration is that due to gravity.
    This does vary with height above the Earth's surface, though; so depending on where the fall starts, you may need to take this into account.
    Air resistance is certainly not negligible; even if he doesn't have a parachute!
    I would suggest taking a look at "viscosity" and the concept of "terminal velocity" to include them in your analysis.
    Newton's Law is fine in this instance. It will help you to calculate the reduced value of g if the skydiver starts from a very great height.
    There may be other things to think about but that seems plenty for now. I'm sure others on here will think of more.
  7. Apr 5, 2010 #6
    Sorry for being vague, he is falling out of the sky at an altitude of about 2 kilometers...where the stratocumulus clouds are.
  8. Apr 5, 2010 #7
    To see if you need to take Newton's law of universal gravitation into account, evaluate the values of g at the point from which the person is falling and at the point he will land. If the difference between these values is small, you may ignore Newtons law of universal gravitation and assume constant g.

    To see if the difference is negligible, complute the following:
    g(s) at starting point
    g(e) at end point
    g(e)-g(s) = Δg , which is the difference between the g's.
    If g(s) + Δg = g(s) within the significant figures (2 figures in this case), you may assume constant g.
  9. Apr 5, 2010 #8
    In which case you probably don't need to consider any change in the value of g as it (the change) would be negligible at this height.
    Air resistance is by far the most important other thing to consider.
  10. Apr 5, 2010 #9

    D H

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    Ahem. g is 9.80665 meters/second2. While adding a 0.00335 m/s2 is, perhaps, acceptable (see side note), changing the exponent from 2 to 1 is not.

    Side note: Gravitation is not constant over the surface of the Earth; it varies from about 9.789 m/s2 at the equator to about 9.832 m/s2 at the poles.


    In that case you cannot ignore air resistance.
  11. Apr 5, 2010 #10
    Sorry, simple error, no need to get anal about things.

    Oh, and if that sounds hypocritical, I have reason to be anal about 0.01; just try and do every gravity based calculation you have ever done and miss out some decimals.

    While I won't argue that it was silly of me to put the wrong exponent, 0.00335 kinda DOES matter.

    And I'd rather you didn't try and disprove me with what you read on Wikipedia.

    I do think your question has been answered, hasn't it choco? ( I just don't want to feel like a spammer ;p)
  12. Apr 6, 2010 #11
    Thanks for everybody's reply. air resistance would be very important at the height falling from 3 km because we would assume that at somepoint the terminal velocity would be reached (air resistance=gravity?) and so therefore g would be constant?**clarification, by constant g, i mean that it will always be the same not increasing** Roight? Wrong?

    i can see why now that the law of gravitation isnt necessarily needed XP there are much more other things that can be discussed .......
    Last edited: Apr 6, 2010
  13. Apr 7, 2010 #12
    Of course g varies over the distance, but given the small distance travelled and the low number of significant digits, its change is negligible. This has nothing to do with air resistence. Terminal velocity inceases with increasing g, so you would need to correct for that as well is the distance was longer.
  14. Apr 7, 2010 #13

    D H

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    Terminal velocity also decreases with increasing density, and the density variation over 2 km is a much greater effect than that of the very slightly increasing gravitational acceleration. Gravity will of course change as the person falls, but this change is very, very small. The free air correction is -0.3086 mgal/meter, or 3.086×10-6 m/s2 per meter. Over the course of a 2 km fall, gravitational acceleration will change by a paltry 0.06%. Since terminal velocity is proportional to the square root of the gravitational acceleration, that 0.06% change in gravitational force represents a 0.03% change in terminal velocity.

    Density variation, in comparison, is a huge effect. The standard atmosphere model yields a density of 1.0649 kg/m^3 at 2 km compared to a density of 1.2250 kg/m^3 at sea level. This alone represents a 6.8% change in terminal velocity over the course of the fall. Deviations from standard atmospheric conditions will have a larger effect than will the tiny change in gravitational acceleration. A person falling on a hot, humid day will hit the ground quite a bit harder than if the person fell on a cold, dry day.
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