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I Duration of a fall

  1. Nov 10, 2017 #1
    • New member has been reminded to please be as complete and clear in their OP as possible.

    Sorry for my English...
    I have made some calculations. By Iterative calculation. With Excel.
    The time that need two objects to fall from 10 meters to 2 meters.
    The mass of the objects i choose are:
    1 and 10 kilos.
    2 and 10 kilos
    5 and 10 kilos.
    1 and 9 kilos
    3 and 7 kilos.
    5 and 5 kilos.

    What i found as duration surprise me.
    More time for 1 and 10 kilos than for 5 and 10 kilos.
    And the same time for 1 and 9 kilos than for 5 and 5 kilos.

    Can someone redo my calculations?

    Luc B.
  2. jcsd
  3. Nov 10, 2017 #2


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    WHAT calculations? I don't see any calculations. If I understand your problem correctly, you obviously have the wrong answer so what we need is to see YOUR calculations to see where you went wrong
  4. Nov 10, 2017 #3
    Thanks for your reply.
    How do i make the calculation?
    First, i calculate the distance an object travels. x.m1 = y . m2 and x+y = 10 - 2 meters.
    I calculate the cinetic energy. 1/2 . m . v² at different places along the travels.
    And i calculate the average speed for each portion of the travels.
    And i have the duration of the travel.
    With Excel, it is possible to divide the travels in thousands parts.

    What are the results?

    1 and 10 kilos. 14.4 days
    2 and 10 kilos. 13.8 days
    5 and 10 kilos. 12.3 days
    1 and 9 kilos. 15.1 days
    3 and 7 kilos. 15.1 days
    5 and 5 kilos. 15.1 days.

    Luc B.
  5. Nov 10, 2017 #4


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    DAYS ???
  6. Nov 10, 2017 #5


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    What is x? What is m1? What is y? What is m2? What is the relevance of x+y?

    Like @phinds, I still see no calculations.
  7. Nov 10, 2017 #6

    Phinds. Yes, days. But if you want, give me other mass and distance, and Excel make the calculation in a few micro secondes....
    Jbriggs444. x is the distance that travels one object. y the distance that travels the other object.
    And the total distance is x + y. here 8 meters (10 - 2)

    I have already post this in french forum. My first language.
    A lot of people say me that i am wrong.
    Noone make the calculation by himself.............

    Luc B.
  8. Nov 10, 2017 #7


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    OK, SOMETHING is weird here. You problem statement says you want the time for an object to fall from 10 meters to 2 meters. This would be a few seconds at most, yet you say DAYS. ???
  9. Nov 10, 2017 #8
    Hello phinds.

    It is not object that fall on the earth.
    It is an object of 1 kg that fall towards an object of 10 kg.
    And sorry, it takes days.....................
    But perhaps, my calculation are wrong.
    When an astronaut goes out the international station. He doesn't fall on its in a few seconds. Does he?

    Luc B.

    It is very late in my country. Sorry but i have to sleep. See you tomorrow......
  10. Nov 10, 2017 #9


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    You have two objects in free space and the only forces acting on the objects are mutual gravitational attraction . You are attempting to work out how long it takes for the objects to move towards each other by a specified distance . Initial separation distance is known and initial relative velocity between objects is zero .

    That's not too difficult a calculation if the objects are relatively small compared to the distances involved ?
    Last edited: Nov 11, 2017
  11. Nov 10, 2017 #10


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    Ah. OK, I misunderstood the problem.
  12. Nov 10, 2017 #11


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    Not your fault. The original question was lacking clarity in the extreme; we all were scratching our heads. o0)
  13. Nov 10, 2017 #12
    But maybe not very simple as the acceleration is not constant. I found this thread that has some suggested solutions (which I can't vouch for):
  14. Nov 10, 2017 #13


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    Here they give an equation to find the fall time for two masses from a distance of r to a distance of x:

    Note that if x and r remains unchanged, then the relative time only depends on

    where u = G(M+m)
    with M and m being the masses involved.
    1kg+10kg = 10kg and 5kg+10kg =15kg, and 1/10 is larger than 1/15, which agrees with your results with the 1kg-10kg pair and the 5kg-10kg pair.

    1kg+9kg, 3kg+7kg, and 5kg+5kg all equal 10kg, so you should get the same fall time for all three combinations of masses.
  15. Nov 10, 2017 #14
    And thanks you very much.

    That means:
    1/ Aristote was rigth. Heavy objects fall faster ...
    2/ Galilée was rigth. All the objects of the earth fall in the same way on earth.
    3/ Einstein was wrong. And time is not relative.
    That's all for now

    Luc B.
  16. Nov 10, 2017 #15


    Staff: Mentor

    4/ You have misunderstood what one or more of the above said

    You may want to read about the reduced mass
  17. Nov 10, 2017 #16
    Hello Dale.

    I will come back here in a few time.
    I am sure that i am wrong somewhere.
    Perhaps in the calculation.
    I am sure someone will redo them. And i will find my mistake.
    See you here later.

    Luc B.
  18. Nov 10, 2017 #17


    Staff: Mentor

    Did you even bother to read the link? What is the reduced mass of the combinations you mentioned? What is the gravitational force? How does it depend on the product and on the sum of the masses?

    The conundrum you mention here does not need numerical calculations. It can be investigated analytically very easily using the concept of the reduced mass. Which is the reason that I pointed it out to you
  19. Nov 10, 2017 #18
    Hello Dale.
    Thanks for your answer.
    I know about reduced mass. I have read a lot but in french. I promise i will read in English. Very good exercice for me.
    But i think (not sure) it is an approximation. But i am not sure.
    I have an other question. I am not sure i use the good words.
    The liberation speed of the earth is 11,2 km/s. I know how we can calculate its. How we find its.
    We take only the attractive force of the planet on object that goes away.


    If we have a small planet of 1000 kg. And a rocket of 1000 kg.
    What is the liberation speed of the rocket about this planet.

    Sorry. But i do not find its.

    Luc B.
  20. Nov 10, 2017 #19


    Staff: Mentor

    No, it is exact.
  21. Nov 10, 2017 #20
    Hello Dale

    Ok ok. I will read it.
    Did you make any analitic calculation?
    To confirm or not my numbers?
    I read it. In English....

    Luc B.
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