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Homework Help: Time of a falling object when the force of gravity isn't constant

  1. Apr 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Distance from planet = 10^14 meters, Radious of the planet = 10^7 meters, mass of the object = 100kg
    g on the planet's surface = 10 m/s^2, g 10^14 meters away from the planets center = 2.5 m/s^2

    2. Relevant equations

    3. The attempt at a solution
    I calculated the speed of the object at the surface of the planet ( 8931.5 m/s ) using Energy Conservation. I trןed to calculate the time using r(t), r(a) graphs.
  2. jcsd
  3. Apr 11, 2015 #2


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    Are you sure this is at 1014 meters instead of the surface? That would make the "planet" a supermassive black hole.

    Can you calculate the speed at other distances, too?
    How much time does the object need to get 1 meter (or in general, dx) closer, as function of distance?
  4. Apr 11, 2015 #3
    It's 10 m/s^2 at the surface and 2.5 m/s^2 10^7 meters from the surface or 10^14 meters from the center.
    I've calculated the planet's mass is 1.5*10^25 kg so its density is 35,000 km/m^3, Whitch is close to the density of the sun's core.

    And I used Energy Conservation to find the velocity equation.
    V = √2 * √G*M/r2-G*M/r1

    I can't find any function of time.
    Last edited: Apr 11, 2015
  5. Apr 11, 2015 #4


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    10^14 is a lot more than twice 10^7.
  6. Apr 11, 2015 #5
    Oh right its not 10^14 its 2*10^14.

    But I checked my calculation and I used the
    right numbers there. So it 2.5 m/s^2 10^7 meters away from the surface and 10 m/s^2 at the surface . (This is a quote from the book).

    But the data isn't the main thing. I want to find the v(t) equation.
  7. Apr 11, 2015 #6


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    Wait, what?
    The density has another error of this type.

    Okay. v is the derivative of position with respect to time: v=dx/dt. What about 1/v? Can you see how to use that, especially if you combine it with the hint of post #2?
  8. Apr 11, 2015 #7

    Sorry its 10^7 and 2*10^7. I'm really tired.

    I think I get what you say.
  9. Apr 11, 2015 #8


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    Uhh, I'm not sure what kind of density 35,000 km/m3 is, but I'm pretty sure it's not the density at the sun's core. :rolleyes:

    That number is about 150 g / cm3, or say 150,000 kg/m3.


    Pluto orbits at an average distance of 5.85×1012 m from the sun, which is still far short of 1014 m. :oops:
  10. Apr 11, 2015 #9
    Yes, I know calculate all the data again with the correct numbers.
  11. Apr 11, 2015 #10
    OK, I calculated the planet's mass and the
    object's velocity again. I got 1.5*10^25 kg for the planet's mass. And the velocity of the object is 10,000 m/s at the surface. I know 1414<t<2828 seconds.
    I can't think of any equation that has time in it.
    I did a(r) function and I found that the average acceleration is 5 m/s^2.
    I used v(t) function for constant acceleration and I got 2,000 seconds. I'm not sure about that but I don't know anything about the time compared to something else.
  12. Apr 11, 2015 #11


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    To get the time, you will need to go back to the differential equation of motion and solve it.
  13. Apr 11, 2015 #12
    I'm not sure but:
    So I have to find the derivative of x(v).

    EDIT : I tried that and it failed.
    Last edited: Apr 12, 2015
  14. Apr 12, 2015 #13


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    I presume you know that for gravity, [itex]F= -\frac{GmM}{r^2}[/itex]. Since F= ma, [itex]a= -\frac{GM}{r^2}[/itex]. Integrate to get the speed at time t and integrate again to get the position.
  15. Apr 12, 2015 #14
    Well, I know newton law of gravtion but I don't know to integrate.
    I tried every graph possible to get the v(t) but I can't find it.
    I know the concept and basics of intgral calculus, but how I can get from the a(x) to the v(x)? In the last few hours I tried to do this with no secsuss.
    I found v(x) and x(v).
  16. Apr 12, 2015 #15
    Have you learned about conservation of energy yet in your course?

  17. Apr 12, 2015 #16
    Yes, and its just high school physics class.
  18. Apr 12, 2015 #17
    Do you know the equation for the potential energy of an object in a variable gravitational field (i.e., a gravitational field that varies with distance from a specified mass)?

  19. Apr 12, 2015 #18
    Yes .
  20. Apr 12, 2015 #19
    Excellent, except for a missing minus sign. So the sum of the potential energy and the kinetic energy of the falling body doesn't change as it falls. Let v = the velocity of the body at time t, and let r be the distance from the center of the planet at time t. What is the sum of the potential energy and the kinetic energy at time t? What is the sum at time zero? What is the velocity as a function of distance from the center?

  21. Apr 12, 2015 #20


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    Didn't we have that formula in post #3 already?
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