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Let's say there's an object of mass $m$ in space, $h$ meters away from the surface of the Earth. $h$ is large enough that $g$ cannot be assumed to be constant. The acceleration varies according to Newton's law of universal gravitation. How would we find the amount of time it takes for this object to fall to earth? (This isn't a homework problem to be clear. I was just wondering.)

My attempts at a solution:

1 - If the acceleration is expressed as a function of time, it's possible to integrate twice to get the equation of motion. This isn't possible here as $g = GM/(\text{Earth's radius } + h)^2$. In other words, acceleration is a function of time, and there's no obvious way to replace $h$ with $t$ (none that I can think of anyway.)

It's at least possible to take the height integral to get an expression for the velocity as a function of height though.

2 - I tried to approximate the amount of time by dividing the distance between the object and the earth into equal segments like a Riemann sum. For example, if there are two sections, you find the $g$ value and assume it's constant. Then you use kinematics to get the time.

Then you repeat the process, finding another $g$ value for the second part. Then you add these times together to get an approximation. As n (the number of segments) increases, the approximation gets more and more accurate, so this operation should be expressible as an integral?

It's possible to let Python find the limit of this approximation by brute force at this point given the mass of the object and the height, but I'm trying to get a neat algebraic expression. This sum is quite difficult to express as the limit of a Riemann sum, so I got stuck!

Thank you in advance for the ideas!

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  • $\begingroup$ I'm not very advanced at Calculus by the way. Take all of my suggestions with a grain of salt. The problem is probably beyond my level! $\endgroup$ Commented Jul 4 at 12:03

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If the body has no angular momentum (no tangential speed with respect to Earth), what you want to solve is Newton's second law in this form:

$$m \frac{\mathrm d^2h(t)}{\mathrm dt^2}=-\frac{GMm}{(R_{\oplus}+h(t))^2}$$

where $R_\oplus$ is the radius of Earth. There are many ways to solve this.

The usual trick is to manipulate the lefthand-side as such: $$\frac{\mathrm d^2 h(t)}{\mathrm dt^2}=\frac{\mathrm dv}{\mathrm dt}=\frac{dv}{dh}\frac{\mathrm dh}{\mathrm dt}=v\frac{\mathrm dv}{\mathrm dh}$$ where $v=\mathrm dh/\mathrm dt.$ Then integrate a first time to get $v$ as function of $h$, and finally try to integrate a second time to get $t$ as a function of $h$.

For the case where the body has a tangential speed then you can check solutions for the Kepler problem (Wikipedia).

See also: The Time That 2 Masses Will Collide Due To Newtonian Gravity

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The solution can be found on Wikipedia, the time $\rm t$ to fall from $\rm r_0$ to $\rm r_1$ is

$$\rm t=\left(\sqrt{\frac{{r_1} \left(1-\frac{{r_1}}{{r_0}}\right)}{{r_0}}}+arccos\left(\sqrt{\frac{{r_1}}{{r_0}}}\right)\right) \sqrt{\frac{{r_0}^3}{2 G ({M_1}+{M_2})}}$$

They don't say how they derived the analytical solution though, there must have been some tricks involved since my computer doesn't find a closed solution right away, but I compared it to a numerical integration of the $\rm d^2r/dt^2=-G (M_1+M_2)/r^2$ and it fits.

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  • $\begingroup$ Just out of curiosity why do you add \rm to all your variables? $\endgroup$
    – Mauricio
    Commented Jul 4 at 12:33
  • $\begingroup$ for aesthetic reasons since otherwise it looks like crap. i know that italic is the norm to avoid confusion with plain text, but in this case i don't think anyone would confuse my variables with text. in your answer at least the differential d should not be italic though, since that is not a variable. $\endgroup$
    – Yukterez
    Commented Jul 4 at 12:35
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Assuming the fall is purely radial and neglecting any fancy rotation and coriolis effects, you would obtain a single differential equation:

$$\frac{GM_e}{(R+h)^2} = -\frac{\mathrm{d}^2 h}{\mathrm{d}t^2}, $$

straight form Newton's law. This is a second order differential equation, which should give you the height $h$ as a function of time $t$ on solving. You will also need two initial conditions, which are your initial height and your initial velocity. The full solution can be found here. The integral cannot be done in temrs of elementary functions. It can, of course, be dome numerically (in much more advanced ways than a simple Riemannian sum too).

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  • $\begingroup$ Which integral cannot be done? The other answer has a full analytical solution. $\endgroup$
    – Mauricio
    Commented Jul 4 at 16:23

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