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If so, that means gravity is only 9.8 m/s^2 at the surface of the earth?

Qmechanic
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    In physics, better to ask "what experiments suggest X?" rather than "How is X derived?" Nature gives us phenomena, not axioms. – John Doty Apr 01 '23 at 22:59
  • Possible duplicates: https://physics.stackexchange.com/q/286360/2451 and links therein. – Qmechanic Apr 02 '23 at 04:36
  • @JohnDoty Disagree. Sure, data is the final arbiter, but sometimes one wants to know what the data is, and sometimes one wants to understand how theory explains or predicts that data, and hence needs to see a derivation. So it's up to the OP which kind of question they ask (though maybe, if they're asking the latter, the title could be reworded to ask what explains the result in question). – J.G. Apr 03 '23 at 07:12
  • @J.G. Circular reasoning. The theory is crafted to capture the data. It cannot explain its own source. Good theory, though, provides a unified story for a variety of data, and has predictive power for new data. – John Doty Apr 03 '23 at 13:13
  • @JohnDoty But since a theory explains/predicts many claims from just a few assumptions, questions such as "Does one equation explain the other?" and "Did this equation lead us to conclude the other?" are (i) worth asking, (ii) unfortunately at risk of conflation in the OP's wording, and (iii) unaddressed by mentioning evidence each equation enjoys when considered separately. – J.G. Apr 03 '23 at 13:25
  • @J.G. A theory can't explain anything: the phenomena are what they are. A theory can have the power to connect different phenomena, and can make predictions. For a good theory within its bounds, the predictions usually conform to reality. "Could this equation lead us to conclude the other?" is a fine math question, but it's not physics. Overemphasis on the abstract mathematical story rather than the concrete physical phenomena is why we're so poorly educating students that they can't manage to light a bulb given a battery and a piece of wire. – John Doty Apr 03 '23 at 13:48
  • @JohnDoty We may have different meanings of explain in mind, but we're agreed about predictions. I maintain the battery issue results from their misunderstanding one or more theories rather than their misunderstanding the general role of theories. – J.G. Apr 03 '23 at 14:03
  • @J.G. They understand the theory just fine. Do you really imagine that MIT students can't manipulate the symbols of electromagnetic theory? That's what understanding in the world of theory is. But it's only mathematics. They cannot connect their understanding to physics. – John Doty Apr 03 '23 at 14:09
  • @JohnDoty I construe the "theory" as including how that connection works, not just the mathematical symbols. – J.G. Apr 03 '23 at 14:13
  • @J.G. That's not how it's taught. Nor is it how theorists think. Walter Lewin used to put items from his collection of physics puzzles on his desk at MIT. They weren't that hard, but theorists would sometimes be thoroughly vexed. – John Doty Apr 03 '23 at 15:23
  • @JohnDoty How it's taught and how people think vary (I for one was taught the sensible version), but I agree both sometimes go awry, and I'd be interested to learn more about these Lewin puzzles. But I think this issue is tangential to whether the OP should have asked about experiments. – J.G. Apr 03 '23 at 15:40

2 Answers2

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Yes, $F=mg$ can be derived from applying $F=\frac{G m_1 m_2}{r^2}$ to the surface of the Earth. The acceleration felt by an object of mass $m$ dropped a distance $R_{\rm Earth}$ from the Earth's center (where $R_{\rm Earth}$ is the Earth's radius) is (see, eg, Wolfram Alpha)

$$ a = \frac{F}{m} = \frac{G M_{\rm Earth}}{R_{\rm Earth}^2} = \frac{\left(6.67 \times 10^{-11} {\rm N\ m^2\ kg^{-2}}\right)\left(5.97 \times 10^{24} {\rm kg}\right)}{\left(6.37 \times 10^6 {\rm m}\right)^2} = 9.81 {\rm m\ s^{-2}} = g $$ It's an interesting exercise to calculate the acceleration due to gravity on, say, the moon and the sun, by adjusting $M$ and $R$ in the above formula appropriately.

Andrew
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    I like "can be". Historically, this isn't what happened. The proportionality of force and mass was part of the understanding of experiments from Galileo to Hooke. Extending that to orbits resulted in the inverse square law. – John Doty Apr 01 '23 at 22:58
  • It's worth noting that you can view $F = mg$ as the zeroth-order approximation to Newton's law of gravitation. You can Taylor expand $F = Gm_{1}m_{2}\cdot 1/r^{2}$ about $r = R_{\textrm{Earth}}$ and then truncate the series. – Maximal Ideal Apr 02 '23 at 00:07
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Yes, the GM/r^2 is basically g, which is 9.81 m/s^2 near surface of Earth. When further up it's drops as r increases. The m is the small mass.

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