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A falling object

  1. Jan 30, 2009 #1
    Hey everyone,

    THIS IS NOT A HOMEWORK QUESTION. This is a question I think I know the answer to but I want to see what you get to check my answer. Here is the setting. You take an object up into the atmosphere. Far enough away so that the acceleration do to gravity of the earth is 1 m/s^2. From that point, you throw the object down with an initial velocity of 10 m/s. I want to know the final velocity of the object when it touches the earth (when a=g=9.8m/s^2) and how long it took for the object to fall that distance. Since the acceleration is changing you will need to use somehow incorporate that into your equation because acceleration isn't constant. This throws out conventional kinematics like d=vt+.5at^2. Assume there is no air friction.

    a_0 = 1m/s^2

    r_0= 19954924.2m/s

    r_f= 6380000m

    v_0= 10m/s

    M=M_E= 5.97E24 kg

    This is what I get for answers
    v_f= 9215.1m/s
    t= 2943.1 seconds

  2. jcsd
  3. Jan 31, 2009 #2


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    I got a somewhat longer time (~ 4528,5 seconds) but approximately the same velocity (9223,6 m/s).

    How did you calculate it?
  4. Jan 31, 2009 #3
    I integrated [itex]F=\frac{GMm}{r^2} [/itex] to get this

    [tex]E=\int \frac{GMm}{r}[/tex]
    and this can be written as

    [itex]E=\frac{GMm}{r_f} - \frac{GMm}{r_0}[/itex]

    [itex]r_0[/itex] being the initial distance from the center of the earth and [itex]r_f[/itex] being the final.I noticed that this integral has units of energy. So I should be able to solve the final velocity using energy concepts.

    [itex] E_i+W_N_C=E_f[/itex]

    [itex] E_i [/itex] is the initial energy in the system
    [itex] W_N_C [/itex] is the energy or work that is added or taken away from the system (like air friction)
    [itex] E_f [/itex] is the final energy in the system

    We know that in this particular case we have gravitational potential energy and an initial velocity. We know there are no frictional forces like air so we discard that. Just before the object touches the earth it will have a final velocity. So the equation must look like this-

    [itex]\frac{GMm}{r_f} - \frac{GMm}{r_0} + \frac{1}{2}mv_0^2 =\frac{1}{2}mv_f^2[/itex]

    Now we want to solve for [itex]v_f[/itex]

    [itex]\frac{GMm}{r_f} - \frac{GMm}{r_0} + \frac{1}{2}mv_0^2 =\frac{1}{2}mv_f^2[/itex]

    [itex]\frac{2GMm}{r_f} - \frac{2GMm}{r_0} + mv_0^2 = mv_f^2[/itex]

    [itex]\frac{2GM}{r_f} - \frac{2GM}{r_0} + v_0^2 = v_f^2[/itex]

    [itex]\sqrt{\frac{2GM}{r_f} - \frac{2GM}{r_0} + v_0^2 }= v_f[/itex]

    That is the equation to give me a final velocity. So to find it I just plug it in.

    [itex]\sqrt{\frac{2GM_E}{6380000} - \frac{2GM_E}{19954924.2} + 10^2 }= v_f[/itex]


    So now that I have an initial and final velocity. I should be able to find an average velocity by using-



    [itex]v_a= 4612.55m/s[/itex]

    Now we have the equation [itex] \frac{d}{t}= v_a [/itex] and solve for t

    [itex] \frac{d}{t}= v_a [/itex]

    [itex] d= v_at [/itex]

    [itex] \frac{d}{v_a}= t [/itex]

    Now we solve for t

    [itex] \frac{19954924.2-6380000}{4612.55}= 2943.1 seconds[/itex]

    That's how I got my answers. For time, one could merge all the equations together to get this-

    [itex]t = \frac {2d}{\sqrt{\frac{2GM}{r_f} - \frac{2GM}{r_0} + v_0^2 }+v_0}[/itex]
  5. Jan 31, 2009 #4


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    Ah, I took the other approach. From
    [tex]m y''(t) = - \frac{G M m}{(R + y(t))^2}[/tex]
    with M and R the mass and (average) radius of the earth, and y(t) the distance above the earths surface, I first found that
    [tex]y''(0) = -1[/tex] at [tex]y(0) = \sqrt{ G M } - R [/tex]

    Then I (unfortunately, numerically) solved the differential equation
    [tex]y''(t) = - \frac{G M m}{(R + y(t))^2}[/tex]
    with the above boundary condition and y'(0) = -10. Setting the result equal to zero gave t = t0 = 4528 seconds, and the derivative of y(t) at that time is y'(t0) = -9223 m/s.

    But actually I'm not sure our values for [itex]r_0[/itex] (my y(0)) agree, to start with :confused:
  6. Jan 31, 2009 #5
    Hmm... so our final velocity's are about right. It's just the time it takes we don't agree with. I really don't know if I'm right or you are right. I'm hoping that someone will make a post backing up either your work or my work.
  7. Feb 1, 2009 #6


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    I get 4379 seconds, 9.22 km/s solving it numerically.
  8. Feb 1, 2009 #7


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    [tex]E=\int \frac{GMm}{r}[/tex]

    What are you using for little m ? Treating it as a massless test particle, m=0 and drives this whole equation to 0.
  9. Feb 1, 2009 #8
    Look at how I derived my equation, specifically when I was searching for the final velocity. When simplifying the equation to solve for the final velocity, m dropped out. I don't understand why I wasn't able to find the same exact time if we got the same final velocities.
  10. Feb 1, 2009 #9


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    Well, at least we all agree about the velocity (in two decimals, at least :smile:).

    I found a way to find the velocity analytically using the force method, as follows (same notation as before).
    We still have [itex]r_0 = \sqrt{G M} - R[/itex]. Since over an infinitesimal interval* the change in velocity is given by
    [tex]dv = a dt = - \frac{G M}{(R + y)^2}[/tex]
    and since v = dy/dt, dt = dy / v:
    [tex]dv = - \frac{G M}{(R + y)^2} \frac{dy}{v}[/tex].
    So basically I'm using a separation of variables here:
    [tex]v \, dv = - \frac{G M}{(R + y)^2} dy[/tex]
    hence (using the boundary conditions)
    [tex]\int_{-10}^{v_f} v \, dv = - G M \int_{r_0}^0 \frac{1}{(R + y)^2} \, dy.[/tex]
    The first integration is easy, changing variables to Y = R + y (dY = dy) gives
    [tex]\left. \frac12 v^2 \right|_{v = -10}^{v = v_f} = - G M \int_{R + r_0}^{R} \frac{1}{Y^2} \, dY = G M \left. \frac{1}{Y} \right|_{Y = R + r_0}^{Y = R}.[/tex]
    Note that [itex]R + r_0 = R + (\sqrt{G M} - R) = \sqrt{GM}[/itex], so
    [tex]\frac12 v_f^2 - 50 = \frac{G M}{R} - \frac{G M}{\sqrt{G M}}[/tex]
    [tex]\frac12 v_f^2 = \frac{G M}{R} - \sqrt{G M} + 50[/tex]
    and we can plug in the numbers to get
    [tex]v_f \approx 9223,6\ldots[/tex] [m/s].

    The downside of this method is that it is very hard (if not impossible) to get from v(y) to v(t) here. Although it is probably possible to calculate the average velocity by solving v(y) from
    [tex]\frac12 v(y)^2 - 50 = - G M \int_{r_0}^y \frac{1}{(R + y)^2} \, dy[/tex]
    and then evaluating
    [tex]\langle v \rangle = \left( \int_{r_0}^0 v(y) \, dy \right) / \left( \int_{r_0}^0 dy \right)[/tex]
    and using
    [tex]t = \frac{r_0}{\langle v \rangle}[/tex]

    *) I'm doing it the physicists' way here, if you insist you can make it mathematically sound :smile:
  11. Feb 1, 2009 #10
    All of our work, till the amount of time it takes, is equivalent up to this point. That makes me happy :smile:. I think I see where and why we have different amounts of time. Its literally in the very last equation you put down.

    t = \frac{r_0}{\langle v \rangle}

    You took [tex] r_0 [/tex] and divided it by [tex] v_a [/tex], or how you said it [tex] \langle v \rangle [/tex]. What you are saying in this equation is the object fell the full initial distance (19954924.2meters) to the center of the earth. I wanted the object to stop once it hit the earth's surface, which relative to the center of the earth is 6380000m. Perhaps it was the way I stated my question when I said

    "I want to know the final velocity of the object when it touches the earth (when a=g=9.8m/s^2) and how long it took for the object to fall that distance."

    I didn't make myself clear enough. That's my fault. I apologize :redface:.
  12. Feb 1, 2009 #11


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    You probably have realized this by now but you can't find the time this way since the acceleration is not constant!
  13. Feb 2, 2009 #12


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    No it is my fault - I didn't make myself clear enough. My variable y which is running between [itex]r_0[/itex] and 0, is the height above the surface of the earth. For example, in the integral, you will see that I have written (R + y) with R the radius of the earth, so y = 0 corresponds to being on the surface. My initial value is
    [tex]r_0 = \sqrt{G M} - R[/tex]
    which indeed corresponds to
    [tex]a(y = \sqrt{G M} - R) = - \frac{G M}{(R + y)^2} = - \frac{G M}{(R + \sqrt{G M} - R}^2 = - \frac{G M}{(\sqrt{GM})^2} = -1.[/tex]

    But probably there is some minor, stupid, avoidable mistake which I just don't see.
  14. May 26, 2009 #13
    Lets see now...
    I recently got an idea of how to solve (or at least try to solve) this problem. First lets forget about gravity for a moment and think about circular motion. Lets say we have a hollow cylinder. As the cylinder spins, any object inside the cylinder will "fall" or "accelerate" to the sides of the cylinder. This is relative to the frame of reference inside the cylinder. To calculate the time it takes for an object to hit the sides of the cylinder will be difficult. This is because the "magical force" that causes the object to fall increases as it gets closer to the sides of the cylinder. One would have to use more than just basic kinematics to solve the amount of time it takes. However, If we look at the spinning cylinder from outside the spinning reference frame there are no forces acting on any objects. When an object is released from inside the reference frame of the cylinder, the people looking at the the object from outside the cylinder will see the object travel in a straight line, with constant velocity until it hits the sides of the cylinder. From this standpoint it is possible to use regular kinematics to solve the problem. In particular by using

    [tex] \frac{d}{v}=t [/tex]

    Where d is displacement v is velocity and t is time. The distance will equal half the chord length relative to where it was released so

    [tex] \frac{\sqrt{r^2-L^2}}{v}=t [/tex]

    Than the velocity will simply be v=rw so

    [tex] \frac{\sqrt{r^2-L^2}}{r\omega}=t [/tex]

    r is the radius of the cylinder, L is how far away the object was released from the cylinder.

    My idea is to put my question not in a gravitational field but put it in a reference frame that is rotating and mimic the exact conditions as if it were in a gravitational field. First I need a way to define acceleration inside the cylinder. So

    [tex] a=r\omega^2 [/tex]

    where a is acceleration due to rotation r is the radius of the cylinder and w is the rate at which the cylinder rotates. Just multiply by the mass of the object and you will find out its force. Now lets look at the energy concepts from inside the cylinder. This equation will yield us the potential energy from within the cylinder.

    [tex] E=\int mr\omega^2dr [/tex]

    The final velocity of the object inside the cylinder will look like this.

    [tex] \sqrt{\int 2r\omega^2dr +v_i^2}=v_f[/tex]

    I want to find what w is suppose to be so I set this equation equal to (the now correct equation)

    [tex] \sqrt{\int \frac{2GM}{R^2} +v_i^2}=v_f[/tex]

    (in earlier posts I butched the equation)

    Set them equal, and omega will equal this

    [tex] \omega= \sqrt{\frac{2GM}{r_f^2-r_i^2}(\frac{1}{R_i}-\frac{1}{R_i})} [/tex]

    r_f is the final radius of from the axis of rotation, r_i is the initial radius from the axis of rotation. R_i is the distance from the mass of the object to planet and R_f is the final distance to the mass of the object to the planet.

    So rewording this equation
    [tex] \frac{\sqrt{r^2-L^2}}{r\omega}=t [/tex]

    [tex] \frac{\sqrt{r_f^2-r_i^2}}{r_i\omega}=t [/tex]

    So this equation should give me the amount of time it takes for an object to fall a certain distance and it works great for small distances but I get a different answer to my question than you do..... I get 6080.66 for an answer.... Where did I go wrong??

    To help save time, I created a spread sheet of my work so anyone can easily alter information to get answers without crunching too many numbers.

    I should have got the same answer as you guys.... Where did I go wrong??

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