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I've already looked at quite a few other threads on this forum but I still haven't been able to find one that answers this question nor gives me any clues on how to proceed.

First Thread

Second Thread

These 2 links proved to be particularly helpful. There's just a part I don't quite understand because of arildno's answer on the second thread which has messed up LaTeX formatting which was referred to in rcqldr's answer on the first thread.

I don't quite understand where u came from nor what it represents.

This is what I've come up with so far on my own:

I originally tackled this as an object approaching a planet so I only looked at it from the perspective of the object.

Finding the time taken:

[tex]v-v_{0} = a_{0}t + \frac{1}{2}jt^{2}[/tex]

[tex]\frac{1}{2}jt^{2}+a_{0}t-(v-v_{0})=0[/tex]

[tex]t=\frac{-a_{0}\pm \sqrt{a_{0}^{2}+2j(v-v_{0})}}{j}[/tex]

Finding acceleration:

[tex]a=\frac{GM}{r^{2}}[/tex]

Finding jerk. I read on another thread that the derivative of gravitational force is jerk. Is this right? What's making me wonder is the fact that it's negative. Does that mean the acceleration is decreasing?

[tex]j=\frac{\partial F}{\partial r}=\frac{\partial }{\partial r}\left(\frac{GmM}{r^{2}}\right)=-\frac{2GMm}{r^{3}}[/tex]

Finding final velocity:

[tex]KE_{i} + PE_{i} = KE_{f} + PE_{f}[/tex]

[tex]\frac{1}{2}mv_{i}^{2}-\frac{GMm}{r_{i}}=\frac{1}{2}mv_{f}^{2}-\frac{GMm}{r_{f}}[/tex]

Assuming that the other object has no initial velocity, the starting KE can be canceled out.

[tex]\frac{1}{2}v_{f}^{2}=\frac{GM}{r_{f}}-\frac{GM}{r_{i}}[/tex]

[tex]v=\sqrt{\frac{2GM(r_{i}-r_{f})}{r_{i}r_{f}}}[/tex]