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Usually when on Earth we can use as an approximation:

g = a = 9.18 $\frac{m}{s^2}$

s = vi + 1/2 a$t^{2}$

However, how would I determine the distance traveled in situations where the the acceleration due to gravity is changing non-negligibly because the distances are so large?

I know that the following equation has to be used somehow:

a = $Gmr^{-2}$

I am not sure exactly how though.

For example, if given the information that an object is at a certain distance and is being attracted by a larger object of a given mass, the method should calculate where the object will be at a given time.

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The Z
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  • Asking "what's the formula" begs the question "why would you want to know?" In physics, we are interested in the principles. We don't much care about the formulas that result from the principles. You might think about what you really want to know ... it won't be "the formula", but it will be "the consequence of all the physics principles that apply. – garyp Oct 26 '19 at 04:52
  • Possible duplicates: https://physics.stackexchange.com/q/286360/2451 , https://physics.stackexchange.com/q/35878/2451 and links therein. – Qmechanic Oct 26 '19 at 06:29
  • Have you learned what a differential equation is? – G. Smith Oct 26 '19 at 07:03
  • @G.Smith . Yes, I know what differentiation is. – The Z Oct 26 '19 at 08:17
  • @Qmechanic . I don't find that any of those address my problem. I know how mgh and Gmr^-2 are related, and that is not what I am looking for. I am looking for how I can use that to determine distance traveled. – The Z Oct 26 '19 at 08:19
  • Hi The Z. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. – Qmechanic Oct 26 '19 at 08:30
  • The equation you wrote is a differential equation, because $\vec{a}$ is the second time derivative of $\vec{r}$. Solving that differential equation gives the motion in the position-dependent gravitational field. – G. Smith Oct 26 '19 at 08:44

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When a small mass $m$ is moving under the gravitational influence of a large mass $M$, the differential equation that determines the motion $\vec{r}(t)$ of $m$ is

$$\frac{d^2\vec{r}}{dt^2}=-\frac{GM\vec{r}}{r^3},$$

which comes from $\vec{F}=m\vec{a}$ combined with Newton’s Law of Gravitation.

It can be solved, and the solutions for bound orbits are ellipses. (This is of course why planetary orbits are essentially elliptical.) There are also solutions for radial infall, which is just a degenerate ellipse.

You’ve probably solved problems using the often-useful approximation of uniform gravitational acceleration $\vec{g}$ — such as the motion of a cannonball — where the trajectory is a parabola. Those parabolic arcs are all just good approximations to the actual elliptical arcs under Earth’s true inverse-square gravity outside its surface.

G. Smith
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