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Acceleration due to Gravity and Final Velocity

  1. Aug 3, 2014 #1


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    Two objects, both of mass "m", start out, at rest, a distance "h" apart (the radius of both objects is negligable). What speed will one of the objects have when they collide? (relative to the original "resting" speed)

    The whole point (without going in to why) is to solve the problem without using conservation of energy.

    The problem should be solvable right? Everything you could possibly need to know is known, so it's reduced to pure math.

    Despite knowing everything, I can't solve the problem.

    This is how I've expressed the problem mathematically:




    Where x is the final velocity that I'm solving for.

    My question is, have I correctly set up the equations? And if so, what 'tools' are required to solve this?
    Do I need to know about differential equations?

    I have a feeling I'm not yet prepared to solve the problem, but I mainly wanted to know if this was the correct way to set up the problem?

    And I wanted to know if there is another way of expressing the situation (mathematically) that makes it easier to solve?
  2. jcsd
  3. Aug 3, 2014 #2


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    Try calculating the acceleration of each object towards a common center of mass, then consider the acceleration of the pair of objects to be the acceleration towards each other which would just be the sum of the accelerations (since one of the accelerations would be negative). Assume the masses are point masses located at x1 and x2, with x2 to the right of x1. Using f for force and a for acceleration, and changing the distance to x:

    $$ f1 = \frac{ G \ m1\ m2}{(x2 - x1)^2} = m1 \ a1 $$
    $$ f2 = \frac{-G \ m1\ m2}{(x2 - x1)^2} = m2 \ a1 $$
    $$ a1 = \frac{G \ m2}{(x2 - x1)^2} $$
    $$ a2 = \frac{-G \ m1}{(x2 - x1)^2} $$
    $$ r = (x2 - x1) $$
    $$ \frac{d^2 r}{dt^2} = \frac{d \dot{r}}{dt} = \ddot{r} = \frac{-G (m1 + m2)}{r^2} $$

    I'm not sure if the final velocity is finite, but the time to collision is finite.
    Last edited: Aug 4, 2014
  4. Aug 3, 2014 #3


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    Thanks for the reply. Your math helped me realize a mistake in my equations, namely I ignored the increase in force from the movement of the other object. I think that [itex]a=\frac{Gm}{(h-2∫∫a.dt)^2}[/itex] is what I should have wrote.

    Though I still don't understand how such equations would be numerically evaluated. Do you need to know some sort of "differential equation techniques" or something? Maybe I should save this question for the future when I've learned more maths.

    You say this because of the fact that they are point masses?

    Very good point, I did not recognize this downfall of the assumption that the object's radius is negligable. :redface:
    I guess I would need to re-do it with a finite radius.
  5. Aug 3, 2014 #4


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    Your post shows an integral of "a dt", but acceleration is defined in terms of distance, not time a(h) not a(t). You'll have to integrate to get velocity versus distance v(h), then distance versus time (dh/dt), and it's possible that you end up with a function of time versus distance t(h) as opposed to distance versus time h(t), in which case you'd have to hope to be able to convert t(h) into h(t), then take derivatives to get v(t) and a(t).

    For the final intergral (distance versus time), it won't matter much if the objects are point masses or have finite radius, as this will just change the limits of the integral and not the integral itself.
  6. Aug 3, 2014 #5


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    My thinking was that I could write a(t) by writing h(t) and then a(h(t)) would be a function of time. Although h(t) depends on a(t) which depends on h(t) etc., so I did realize that my equations were very recursive (and this is what was challenging when I thought about how I could evaluate it)

    I guess setting up the equations in such a recursive way doesn't help at all in solving the problem?

    How does integrating acceleration versus distance yield velocity versus distance? If I integrated acceleration versus distance wouldn't I end up with units of [itex]\frac{distance^2}{time^2}[/itex]?

    I meant "r(t)" or "x(t)" instead of "h(t)"
    Last edited: Aug 4, 2014
  7. Aug 4, 2014 #6


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    h is given as the initial distance between the masses, so it is not correct to use it as variable. Let be the distance between the bodies at time t x(t)=x2-x1. Then you have the equation ##\frac{d^2 x}{dt^2} = \frac{d \dot{x}}{dt} = \frac{-G (m1 + m2)}{x^2}##

    ##\dot x=dx/dt = V## V being the relative velocity of the particles, which is twice the velocity of an individual particle. So ##d^2x/dt^2=dV/dt##. Applying chain rule in the differentiation, ##dV/dt=(dV/dx) (dx/dt) =(dV/dx) V = 0.5 d(V^2)/dx##. Choosing V^2 = z the new variable , the differential equation becomes

    ##\frac{d z}{dx} = \frac{-G (m1 + m2)}{x^2}##

    You need simply integrate both sides from x=h to the final distance xf, to get the square of the final relative velocity.

    If you are not familiar with Calculus yet, apply conservation of energy. The gravitational force is conservative.

  8. Aug 4, 2014 #7


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    Continuing with my updated previous post and changing the initial distance to be 'a' intead of 'h':

    $$ \frac{d^2 r}{dt^2} = \frac{d \dot{r}}{dt} = \ddot{r} = \frac{-G (m1 + m2)}{r^2} $$
    $$ \frac{dr}{dr} \frac{d \dot{r}}{dt} = \frac{dr}{dt} \frac{d \dot{r}}{dr}= \dot{r} \frac{d\dot{r}}{dr} $$
    $$ \dot{r} \frac{d\dot{r}}{dr} = \frac{-G (m1 + m2)}{r^2} $$
    $$ \dot{r} d\dot{r} = \frac{-G (m1 + m2)\ dr}{r^2} $$
    $$ \dot{r} = 0 \ at \ r \ = a : $$
    $$ 1/2 \dot{r}^2 = G(m1+m2)/r - G(m1+m2)/a $$
    choosing the negative root:
    $$ \dot{r} = dr/dt = -\sqrt{2G(m1+m2) \left ( \frac{1}{r} - \frac{1}{a} \right ) } = -\sqrt{2G(m1+m2) \left ( \frac{a - r}{a\ r} \right ) } $$
    If you wanted to continue, you end up with an integral that's tricky to solve (but solved in an older thread if interested):
    $$ -\sqrt{\frac{a}{2G(m1+m2)}}\ \sqrt{\frac{r}{a-r}} \ dr = dt $$
    Last edited: Aug 4, 2014
  9. Aug 5, 2014 #8
    Can we do this question with reduced mass approach.
    Here μ=m/2.
    By using conservation of energy

    Initially potential energy of system=Kinetic energy of system.

    If we find v from this,will we get the required answer?
  10. Aug 5, 2014 #9


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    Yes, but remember, in the reduced mass approach the velocity v is the difference between the velocities of the individual masses: v=v2-v1. As they are equal in the problem and of opposite direction, v1=v2=v/2.

  11. Aug 6, 2014 #10
    Yes I forgot that.I think solving this question with reduce mass approach and conservation of energy is a lot easier.
    How will we approach to the question if masses of particles would have been different?
    Let the mass of particles be m[itex]_{1}[/itex] and m[itex]_{2}[/itex]
    So μ=m[itex]_{1}[/itex]m[itex]_{2}[/itex]/m[itex]_{1}[/itex]+m[itex]_{2}[/itex].
    So by conservation of energy
    Let v be the velocity of reduce mass just before collision.
    Here in this case final velocities of both particles is different.
    So how will we find the velocities of m[itex]_{1}[/itex] and m[itex]_{2}[/itex] just before the collision.
  12. Aug 7, 2014 #11


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    There is no external force, so the CM of the system stays in rest. That means m1v1+m2v2=0 where v1 and v2 are the velocities. Again, v=v2-v1. Getting v, you can solve the system of equation for the individual velocities.

  13. Aug 7, 2014 #12
    Thank you ehild.:smile:
    This problem was easier to solve with this approach.
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