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My eighth-grade physics teacher taught us that the equation for height ($h$) during a free fall in a vacuum is equal to the initial height ($H_0$) minus a constant ($G$, though not the $9.8$ meters per second we know and love) multiplied by time ($t$) squared, divided by $2$.
However, $dh\over dt$ is velocity ($v$), and $dv\over dt$ is acceleration ($a$). If acceleration were constant, then his formula would be fine. But acceleration is not constant. It's $G\over h^2$. So if acceleration changes, then the velocity is not a perfect quadratic equation, meaning the equation he taught us is wrong. Presumably he was dumbing it down for $13$-year-olds.

What is the actual equation? How could you arrive at it? I tried setting up a second-order differential equation, but got stuck on $$dh = G\int {1\over h^2}dt dt$$ In order to solve this problem I would need to express $h$ in terms of $t$, which is the very problem I am trying to solve. How does one get around this? I am in Calc BC, so I would love for a step-by-step solution.

William Grannis
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  • https://physics.stackexchange.com/q/184377/ – BowlOfRed May 01 '19 at 19:05
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    You might want to reconsider the meaning of your product $dt,dt$ in the integral. – ZeroTheHero May 01 '19 at 19:07
  • Possible duplicates: https://physics.stackexchange.com/q/3534/2451 and links therein. – Qmechanic May 01 '19 at 19:16
  • @Qmechanic I think you closed this as the wrong dupe. This seems to be about how the force/accceleration varies with height due to gravity's change due to height; not so much about the gravity exerted between objects. – JMac May 01 '19 at 19:26
  • Yeah, do you have a better dupe? – Qmechanic May 01 '19 at 19:28
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    Nice try. Can you fill in the gaps that leads to your integral? Something is wrong there, but I can't tell what without the steps. – garyp May 01 '19 at 19:49
  • @Qmechanic The integrating the radial freefall seems to be exactly what OP was looking for, only using the term radius instead of height. – JMac May 01 '19 at 19:52
  • @William, you are confusing $G$ and $g$. And the acceleration is neither $G/h^2$ nor $g/h^2$. – G. Smith May 01 '19 at 20:02
  • @William Yes, your teacher is “dumbing it down” but the constant-acceleration approximation is an excellent one for falling apples, projectiles, etc. An important part of doing physics is using good approximations. – G. Smith May 01 '19 at 20:08

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