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  • We assume the gravitational pull of the earth to be approximately $9.81\ \mathrm{m/s^2}$.
  • If I model it as a particle and place an object $0.0254$ m away, what is the new force?
  • I tried using the inverse square law and the radius of the actual earth to figure out the force from $0$ m away, then went to use the inverse square law again to find this $F$ from $0.0254$ m away, but of course squaring this decimal will only make it smaller, making the force bigger, but this is axiomatically false.

How should the method be modified to get a coherent answer? What should be changed?

Thanks for any help.

I will point out here that I have all the correct calculations and know what I am doing in that respect, but if I am 0m away from the object, then the force is F, and if I am a distance s from it, the force is F/s^2, which will axiomatically be smaller than F as that is a fundamental property of physics. However if I am 0.254m away, then the force comes out as F/0.254^2, which is far greater than F as we are dividing by a number between 0 and 1, but this should be less than F as we have moved away. I am asking for a workaround for this, not an answer for the force as I know how to find it without this complication. I would use different units, but I'd then need to know what to multiply the output by to get a correct answer in Newtons instead of kg cm/s^2.

  • If you are at distance $0$ from a point mass, then the gravity pull would become infinite - not $g=9.81, \mathrm{m/s^2}$. You get $g$ at a distance of $R=6380,\mathrm{km}$. Because $R$ is much bigger than one inch, the pull an inch away should be immensely bigger. If you lift an object one inch from the surface of Earth, then the comparison should be between distances $R$ and $R+0.0254,\mathrm{m}$. Those two are virtually equal, because that one inch is insignificant in comparison to $R$. – Jyrki Lahtonen Nov 12 '16 at 05:36
  • How did you "figure out the force from $0$ m away"? It seems from other things you have written that there was probably a fundamental error in something you did there. If you could show what you did, step by step (edited into the question in an appropriate place), someone might be able to spot the error, and this might resolve the apparent inconsistency of your results. – David K Nov 12 '16 at 06:31

3 Answers3

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If you model the earth as a point mass so that you can place an object at $0.0254$ m away without going inside the mass distribution, then in fact the force (or rather, the gravitational acceleration) at that distance will be incredibly large.

In fact, this is only about three times the "black hole radius" (Schwartzchild radius) of an Earth mass, so if you come three times closer, a light ray would not be able to escape!

(Yes, experts, I know that is a simplisitc way of looking at it and GR should be used, but c'mon, the OP is not looking for an answer like that!)

Mark Fischler
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  • Indeed, but the problem in the maths is that by increasing the distance, the force is getting larger as we are dividing by a number far smaller than 1. Even at this small distance, the force should get smaller as the distance increases, so a different unit should be used, but which one? And how should it then be converted back to Newtons for use in further calculations? –  Nov 11 '16 at 20:45
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$ \mrm{F}\pars{R + 0} ={GMm \over R^{2}} = mg \implies g = {GM \over R^{2}} \,,\qquad\left\{\begin{array}{rl} \ds{\mrm{F}\pars{r}}: & \mbox{Force Magnitude from} \\ & \mbox{the Earth Center.} \\[2mm] \ds{r}: & \mbox{Distance from the Earth Center.} \\ \ds{G}: & \mbox{Gravitatory Constant,} \\ \ds{M}: & \mbox{Earth Mass.} \\ \ds{R}: & \mbox{Earth Radius.}\ \ds{\approx 6371}\ \mbox{Km} \\ \ds{m}: & \mbox{Particle Mass.} \\[2mm] \ds{g}: & \mbox{Earth Surface Gravity Aceleration.} \\ & \ds{\approx 9.81\ \mrm{m \over \sec^{2}}} \end{array}\right. $$

\begin{align} \\[5mm] \mrm{F}\pars{R + h} & = {GMm \over \pars{R + h}^{2}} = {GM \over R^{2}}{m \over \pars{1 + h/R}^{2}} = {mg \over \pars{1 + h/R}^{2}} \\[5mm] \implies {\mrm{F}\pars{R + h} \over \mrm{F}\pars{R}} & = {1 \over \pars{1 + h/R}^{2}} = {1 \over \bracks{1 + \pars{0.0254\ \mbox{m}}/\pars{6371000\ \mbox{m}}}^{\,\,2}} \\[5mm] & \approx \bbx{\ds{1 - 7.9736 \times 10^{-9}}} \end{align}
Felix Marin
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  • You are reading the question to be "what is the acceleration $2.54$ cm above the existing surface. I read the question to be "what would be the acceleration $2.54$ cm from a mass point of the mass of the earth. Given your reading, the answer is correct. – Ross Millikan Nov 12 '16 at 05:48
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I would read "model it as a particle" as "make the earth a mass point". Then it makes mathematical (if not physical) sense to put the test particle $0.254$ m from the point and compute the gravitational acceleration. For the second bullet, you can't compute the new force without knowing the mass of your test particle. You can compute the acceleration at that distance. Yes, the acceleration will be much larger than $9.81$ m/s$^2$ because you are so much closer to the center. If you are in a classical mechanics course you can ignore any general relativity corrections and just do Newtonian gravity calculations.

Ross Millikan
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