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Free fall formula

  1. Aug 18, 2004 #1
    I am trying to find a velocity for an object that would launch it into space from earth (straight up, NOT into orbit) and return exactly 8 days later. I cannot use x = v0t + 1/2at^2 (x=0 t=691200 a=9.8 [seconds in 8 days]) because the distance from the Earth changes in such a great way that acceleration is not realistically constant any more. I know this would require calculus, but I cannot find a formula or a method. I believe some reworking of
    escape velocity would do, but there is no reference to time.

    I know this is a very specific question, but any guidance would be a great help.
  2. jcsd
  3. Aug 18, 2004 #2
    So, you are shooting it up and waiting for it to fall back down after 8 days...that means non-constant acceleration. I'm guessing you are assuming there is no air resistance right? Air resistance causes terminal velocity. As for the acceleration...its a hard one to figure out because ive never dealt with non-constand acceleration, but if at least you knew how far up it went before coming to a stop, then you could figure out g at that height and average it with 9.8 and use that to plug into an equation. Im not sure if that would get the job done right, though.
  4. Aug 18, 2004 #3
    Try using Newtons law of Gravitation: [tex]F(r)=G\frac{Mm}{r^2}[/tex]
    with r the distance to the center of the earth, G the gravitational constant, M the mass of the earth and m the mass of the object.

    If you equal this force (by using Newtons second law) to ma (assuming constant mass) you get a (differential-)equation of motion:
    [tex] \frac{d^2r}{dt^2}=G\frac{M}{r^2}[/tex]
    Solving this with the appropriate boundry conditions leads to you answer.

    You can use this also to calculate g;
    [tex]mg=G\frac{Mm}{R^2}[/tex] so
    [tex]g=G\frac{M}{R^2}= 9,8ms^{-2}[/tex]
    with R the radius of the earth. By replacing R with your current distance to the center of the earth you get a distance dependent 'g'.

    If you have trouble solving the equation let me know...
  5. Aug 18, 2004 #4
    If you exclude air resistance (wich I think you should, to make things not too compicated :cry: ) you make the 'non-constant g fault' look like nothing :wink:.
  6. Aug 18, 2004 #5
    Air resistance should be ignored because the amount of time the object will spend passing through the atmosphere is a very very small percentage of the fall time. It spends four days rising and four days falling, and will pass through the atmosphere (say a hundred miles thick with constantly varying density, so air resistance should only be significant in the lower regions) at hypersonic speed in about five seconds. Including subsonic air resistance would be more than a little complicated. I don't even know what it'd look like at such high velocities.

    The constantly varying g is much easier to include and is far more important consideration to the final result.
  7. Aug 18, 2004 #6
    The drag force is proportional to the squared velocity of the object. At high speeds there will be a large drag force decellerating the object. This will have a large effect on the initial speed. To make a guess I think the speed spl3001 is looking for is in the some order of magnitude (maybe one lower) as the escape velocity of the earth. At a speed of 10^5 m/s the drag force is about 5*10^12 N for a (small) object of an area of 1m^2. This is a huge force. And the object has to have a large mass ( in this example >10^8kg) to not be brought to a halt, and then it has to have suffiecient speed to not return to earth. But this leads to very complicated calculations (not in the least because the density of the air depends on height also) so let's stick to the case of zero drag....

    Of course if you use some sort of propulsion mechanism during the take-off you can do the problem realistic without using air resistance.
  8. Aug 18, 2004 #7


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    Let us not forget that you cannot fire it straight up because in 8 days the earth will have moved.
  9. Aug 18, 2004 #8
    Ha, Good one. I totally forgot about that... :biggrin:

    So I guess it's more like a problem to get some practice with calculus, and not solving a very realistic problem. Or if you are getting the hang of it you can ofcourse try to solve it including the earth's motion around the sun. You also have to take into account the earth's rotation when you're firing not straight up. And while you're at it you can try to include the gravitational effect of the other planets and sun as well, and...and...

    (why isn't reality as simple as in physics problems...?) :cry:
  10. Aug 18, 2004 #9
    Now that I have a distance dependent acceleration, how do I incorporate that into a formula involving time such as [tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]?
  11. Aug 18, 2004 #10


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    How accurate do you need to be?

    You could simply approximate the path as an elliptical orbit with perigee at the Earth's radius and apogee placed such that the period of the orbit is 8 days. You'll be going fast enough that the extra distance around the planet would only contribute a few minutes to the orbit, and add a few m/s to the velocity.
  12. Aug 18, 2004 #11
    Integrate... :smile:
  13. Aug 18, 2004 #12
    spl3001, are you assuming that the earth is at rest with respect to the sun? Because you said in your first post (straight up, NOT into orbit). Also, does the projectile have to land at the place that you shot it from or does it simply have to go up and down in a straight line regardless of where it lands??
  14. Aug 18, 2004 #13
    I just need it to land on earth. A straight line up and down would be fine, even a few hundred meters away is ok. I don't really need to figure an gravitational influences from other planets/moons/the sun, and the earth can be assumed to be at rest. All i need is the initial velocity and the furthest distance from the starting point (exactly 4 days into the trip). I think the orbit idea would work, except I don't have the equations to calculate the period for an orbit, especially for such an eccentric one.
  15. Aug 19, 2004 #14
    That might be able to give me the speed. That's the orbital velocity of an object in an elliptical orbit. I need to know the semi major axis of an orbit with a period of 8 days in which its perigee point is in contact with the Earth's surface.

    I believe I'm getting pretty close now.
    Last edited: Aug 19, 2004
  16. Aug 19, 2004 #15


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    The period of an orbit is
    \frac {2 \pi a^{\frac{3}{2}}}\sqrt{(G m)}
    where a is the semimajor axis, G is the gravitational constant, m is the mass of the primary.
  17. Aug 19, 2004 #16
    I would just like to thank everyone for their suggestions and ideas. It all came together and I was able to solve for the velocity and distance.

    Results are here:
    http://www.gobean.com/sww_35.htm [Broken]
    Last edited by a moderator: May 1, 2017
  18. Aug 19, 2004 #17
    Is this really a problem?
    I think the earth and the object would be moving together, so we wouldn't have to worry about that...
    Last edited: Aug 19, 2004
  19. Aug 19, 2004 #18
    If the earth would move with a constant (linear) velocity, yes... But the earth revolves in an ellipitical orbit around the sun. So it is not true in this case..
  20. Aug 20, 2004 #19
    But the object and the earth would be at the same orbit, since the 4 days distance from earth is a small value, wouldn't they?
  21. Aug 23, 2004 #20
    The rocket would have traveled a distance [tex]2\pi R \alpha[/tex] while the earth would have moved a distance [tex]Rsin(2\pi \alpha )[/tex] in the same direction. R is 1AU and [tex]\alpha[/tex] is the fraction of the orbit traveled, so this is 4/365,25. The difference between the two is 5,43*10^-5 R. This amounts to 0,637 of the earth diameter. So with a little bit of luck it will still land on the surface of the earth. So you are right..
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