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The formula for a falling object has $r^2$ in the denominator. This would mean that an object that is higher up falls more slowly than the standard $9.807\ \mathrm{m/s^2}$ that we are taught in high school.

What would happen if we took at $1$ metre pole and a $10$ metre pole up to a height of $100$ metres for the bottom of both poles, and then dropped them? Let's assume they are weighted on the bottom so both remain vertical, and that the $10$ metre pole is hollow so they both weigh the same. Would they hit the ground simultaneously or otherwise?

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    $\begingroup$ Air resistance neglected or not? $\endgroup$
    – fraxinus
    Commented Aug 31, 2021 at 16:27
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    $\begingroup$ @fraxinus No air resistance considered. Purely Newtonian Gravity. I didn't' need to put in about the weight, but I wanted ONLY the pole length to be considered. $\endgroup$
    – foolishmuse
    Commented Aug 31, 2021 at 16:30
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    $\begingroup$ The effect you are referring to is called "tide." $\endgroup$
    – Ross Presser
    Commented Sep 1, 2021 at 4:22
  • $\begingroup$ One assumes that air resistance is not relevant, as the longer object would have more air resistance over it much greater surface area? One also assumes that air buoyancy is to be neglected, as otherwise the larger object would weigh less even though massing the same? Taken to an extreme, the second object might fly like a hydrogen balloon! And your statement of "Weighted at the bottom" really muddies the situation by changing the mass distribution of the longer pole. Is its center of mass higher or lower than for the shorter pole, we don't know! $\endgroup$
    – PcMan
    Commented Sep 1, 2021 at 6:33
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    $\begingroup$ Good idea for a question, but stated using very muddy terms. You need to explicitly state the starting environment. You need to not mess with the mass distribution of the poles, or if you do want to do so, then state exactly what is being done. "Weighing one end" will do nothing to the orientation of the pole in the absence of air resistance, and is an unnecessary unquantified complication. $\endgroup$
    – PcMan
    Commented Sep 1, 2021 at 6:40

7 Answers 7

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Paul T. provides a good answer regarding the case where the height of the bottoms of the poles are the same (which is what was asked for in the question). The main difference in that case is due to the different heights of the centers of mass of the two rods.

However, you might ask, what if the centers of mass of the two rods were at the same height, would there still be a difference? It turns out that there will be, although the difference is even smaller.

Let $m$ and $l$ be the mass and length of a vertical rod of uniform density, and $r$ the height of its center from the center of the planet. The planet's mass is $M$. Consider a tiny piece of the rod, of mass $\delta m$ and distance $x$ from the rod's center (so $x$ is between $-l/2$ and $+l/2$). The force of gravity on the tiny piece is:

$$\delta F=\frac{GM\delta m}{(r+x)^2}$$

The total force on the rod is the integral of $\delta F$ over the whole mass:

$$F=\int\frac{GM}{(r+x)^2}dm$$

The mass of the small piece is proportional to its length ($m/l=\delta m/\delta x$) so we can substitute $dx$ for $dm$ with the appropriate scaling:

$$F=\int_{-l/2}^{+l/2}\frac{GM}{(r+x)^2}\left(\frac{m}{l}dx\right)$$

Doing the integral yields:

$$\begin{align} F&=-\frac{GMm}{l}\left.\frac{1}{r+x}\right|_{x=-l/2}^{+l/2} \\ &=-\frac{GMm}{l}\left(\frac{1}{r+l/2}-\frac{1}{r-l/2}\right) \\ &=-\frac{GMm}{l}\left(\frac{(r-l/2)-(r+l/2)}{(r+l/2)(r-l/2)}\right) \\ &=-\frac{GMm}{l}\left(\frac{-l}{r^2-(l/2)^2}\right) \\ &= \frac{GMm}{r^2-(l/2)^2} \end{align}$$

This is almost the same value as if the mass of the rod was concentrated at its center (in which case it would be just $GMm/r^2$). Like Paul T., let's look at the relative difference:

$$ \frac{\frac{GMm}{r^2-(l/2)^2}-\frac{GMm}{r^2}}{\frac{GMm}{r^2}} = \frac{(r^2)-(r^2-(l/2)^2)}{r^2-(l/2)^2} = \frac{(l/2)^2}{r^2-(l/2)^2} \approx \left(\frac{l}{2r}\right)^2 $$

Compare this to the case where we measure $r$ from the end of the pole, where the relative difference (between a rod and a point) was just $l/r$. If the end case had a difference of on part per million, the center case will have a difference of less than one part per trillion!

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    $\begingroup$ I was wondering when somebody was going to answer this with calculus because indeed, this is a calculus question. Well done ;) $\endgroup$
    – Kyle B
    Commented Sep 1, 2021 at 5:05
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Yes. As described in the questions, there is very small difference between the acceleration of the two poles with the shorter one accelerating faster.

The difference

The gravitational force acting on a pole is $$F = \frac{GMm}{r^2},$$ where $M$ is the mass of the Earth, $m$ is the mass of the pole, and $r$ is the separation between the center of mass (CoM) of the Earth and the CoM of the pole. Neglecting air resistance and the gravitational effect of one pole on the other, the acceleration of a pole is $$a = F/m = \frac{GM}{r^2}.$$ The masses of the poles don't matter. They could be different or the same.

If the two poles are different lengths, $L > \ell$, then their CoMs will be at different distances from the CoM of the earth. Let's define $R$ as the distance from the CoM of the Earth to the bottom of the polls, which are at the same height. Assuming the poles are uniform $$r_\ell = R + \ell/2 \quad\text{and}\quad r_L= R + L/2.$$

The shorter pole will will experience a larger acceleration.

$$ a_\ell = \frac{GM}{(R + \ell/2)^2} \quad > \quad a_L = \frac{GM}{(R + L/2)^2}$$

How big is the difference?

To get a handle on how much larger, we can do a first order expansion of the two accelerations and look at the difference. $$a_L = \frac{GM}{r^2} = \frac{GM}{(R + L/2)^2} = \frac{GM}{R^2(1+\frac{L}{2R})^2} = \frac{GM}{R^2}\left(1 + \frac{L}{2R}\right)^{-2}\approx \frac{GM}{R^2}\left(1 - 2 \frac{L}{2R}\right)$$

I personally find the fractional difference $\frac{\Delta a}{a}$ to be more illuminating than the absolute. So lets look at that by dividing out the common $GM/R^2$ bit. $$\frac{\Delta a}{a} \approx \frac{a_\ell - a_L}{GM/R^2} \approx (1- \ell/R) - (1-L/R) = \frac{L-\ell}{R}$$

The radius of the Earth is about $6\times 10^6$ m, so we're looking at parts-per-million differences in the accelerations of the two poles.

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  • $\begingroup$ I'll ask you a question I asked above. Do we use the centre of gravity of the earth and of the pole just for mathematical convenience, or is there a more precise way if we did a derivative calculation on the entire length of both? I know it's a strange question, but I'm trying to get a handle on exactly how things fall in relation to spacetime. $\endgroup$
    – foolishmuse
    Commented Aug 31, 2021 at 20:11
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    $\begingroup$ In general you must integrate over the two mass distributions to figure out the force of every part of one body on every part of the other. Spherically symmetric things act like perfect point masses at their CoMs. The Earth isn't quite a sphere, so there are small $\sim\frac{1}{R^4}$ quadrupole corrections to the force due to its shape. You might want to look up mass quadrupole or multipole expansion $\endgroup$
    – Paul T.
    Commented Aug 31, 2021 at 20:20
  • $\begingroup$ If you are thinking about falling and spacetime, you might want look into general relativity instead of Newtonian gravity. $\endgroup$
    – Paul T.
    Commented Aug 31, 2021 at 20:21
  • $\begingroup$ Would the answer to the OP under GR be any different? $\endgroup$
    – foolishmuse
    Commented Aug 31, 2021 at 20:36
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    $\begingroup$ @foolishmuse: It wouldn't be significantly different under GR. I haven't done the calculations, but I would expect there would be even smaller corrections on the order of $(L-\ell)^2/R^2$; but that would introduce a parts-per-million difference in the difference, and would be correspondingly harder to detect. $\endgroup$
    – Michael Seifert
    Commented Sep 1, 2021 at 14:09
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Assuming (reasonably) that both objects are rigid (undergo no deformation under the influence of gravity) then in the Law:

$$F=G\frac{mM}{r^2}$$

$r$ refers to the distance between the centre of gravity (CoG) of the object and the CoG of the Earth.

Then assuming the CoGs of both objects have the same mass and are at the same height above the Earth (at the start of their free fall), the resulting $F$ are equal.

Would they hit the ground simultaneously or otherwise?

In those circumstances the former would be true.

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  • $\begingroup$ It is probably worth mentioning that it won't be exactly true if their individual masses were different ( and the rods were far apart from each other ). $\endgroup$
    – Nakshatra Gangopadhay
    Commented Aug 31, 2021 at 16:40
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    $\begingroup$ @Gert If I am not wrong, the CoGs for two poles will fall together simultaneously (if they are at same height) till the bottom of the longer pole touches the ground (since bottom of the longer pole is more closer to the ground) $\endgroup$
    – KP99
    Commented Aug 31, 2021 at 18:09
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    $\begingroup$ The question says "height of 100 m for the bottom of both poles" but your answer says "assuming CoGs of both objects are at the same height". You have answered a question different from the one asked. $\endgroup$
    – Andrew Steane
    Commented Aug 31, 2021 at 18:23
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    $\begingroup$ @Nakshatra There's a good question on that topic, linked below, in the Related section: Don't heavier objects actually fall faster because they exert their own gravity? The effect for small bodies on Earth is on the order of 1 part in 10²⁴. $\endgroup$
    – PM 2Ring
    Commented Sep 1, 2021 at 7:59
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    $\begingroup$ @NakshatraGangopadhay it's really a bit pedantic to call that wrong. Not only because in practice the effect is utterly negligible for any object you could actually lift, but also in theory because it's the Earth's movement that makes the tiny difference. The objects would still reach the surface at the same time if the Earth hadn't moved closer by that time in the heavier case. $\endgroup$
    – leftaroundabout
    Commented Sep 2, 2021 at 6:45
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Yes, slightly different acceleration as others have said, leading to slightly later arrival of the longer pole. (It is the location of the centre of mass which determines the acceleration).

An interesting related effect: the poles are both drawn into tension owing to the different gravitational pull on their two ends. The extreme example of this is the 'spaghettification' predicted to happen as things fall towards the centre of a black hole.

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Ok, no air. Also I simplify out the tidal effects (poles are in no sense spherical). Mutual attraction between poles, as well as the Earth moving towards them assumed negligible (not that it will change the answer).

Other than that, 10m pole has its centre of mass 4.5m higher so it gets somewhat less gravity and less acceleration.

Since both poles have to travel 100m (equal distance), the 10m pole will arrive later.

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In non uniform gravity, discounting air resistance, any two objects having different heights for their centers of gravity will be accelerated by gravity at different rates. As you seem to understand, the increased radius to the center of gravity would give a lower rate of initial acceleration to the higher object..

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  • $\begingroup$ The centre of gravity. That makes perfect sense. Are you certain of this versus the height of the top of the pole? Or is centre of gravity just used for mathematical simplicity? $\endgroup$
    – foolishmuse
    Commented Aug 31, 2021 at 17:00
  • $\begingroup$ I guess that would also pose the question is the centre of the earth just used for mathematical simplicity? $\endgroup$
    – foolishmuse
    Commented Aug 31, 2021 at 17:09
  • $\begingroup$ The COG is the point of the objects gravitational equilibrium. In non uniform gravity the COG will be slightly different than its COM.. When considering gravitational attraction between Earth and much much smaller objects the difference in Earth's COM and its COG relative to the smaller object is normally inconsequential $\endgroup$
    – Adrian Howard
    Commented Aug 31, 2021 at 21:03
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Actually the question is a little vague. When I read it, I thought it meant like a tree, or a telegraph pole, or one of those very tall old chimneys falling over. In which case we are calculating the time for a tall thin cylinder to fall over. The taller it is, the longer it will take to fall although in real life we would have to contend with the cylinder breaking or deforming. Most of the answers here focus on dropping a vertical cylinder and have answered that thoroughly. There is also the possibility of the cylinder (pole) being held horizontally before being dropped in which case the length would not be relevant. I hope that has muddied the waters a little...

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    $\begingroup$ Please provide additional details in your answer. As it's currently written, it's hard to understand your solution. $\endgroup$
    – Community Bot
    Commented Sep 1, 2021 at 16:20

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