Is Earth's surface gravity termed $g$ derived from Newton's universal law of gravity?

Question

Trying to apply the basic Newton law $F = m a$ to a falling apple and a paper clip I encountered a problem I consider a beginner's: The fact that all objects are subject to the same acceleration termed $g$, and falling off a balcony arrive on the floor the very same time cannot be derived from Newtons's first three laws. Is it correct to assume that the proportionality of earth's gravitation exerted on objects of different mass or weight needs some other law that is not implied in Newton's three basic laws of motion?

If the above is correct, is $g$ ("9.x m/s2") a derivation of Newton's general law of gravity? How exactly does this derivation look like?

Closely related:

How did Isaac Newton derive the laws of gravity?

score 1 · Answer 1 · answered Nov 25 '22 at 15:10

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You have it backwards. The observation that objects near the Earth's surface accelerate at $g$ was one of the motivations of Newton's theory that gravity is a force proportional to mass. The known motions of the Moon and planets motivated proportionality to $1/r^2$.

In physics, we don't derive the phenomena from the theory. We find theories that capture the phenomena.

answered Nov 25 '22 at 15:10

John Doty

20,827

So it's very theoretical to consider gravitational and inertial mass as one and the same thing? I mean, there is nothing "proportional to mass" if one inertial mass hits others of different sizes. Thank you! – Peter Bernhard Nov 25 '22 at 15:20
@PeterBernhard https://en.wikipedia.org/wiki/Equivalence_principle – John Doty Nov 25 '22 at 17:22

score 1 · Answer 2 · answered Nov 25 '22 at 15:14

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Is it correct to assume that the proportionality of earth's gravitation exerted on objects of different mass or weight needs some other law that is not implied in Newton's three basic laws of motion?

Yes, that is correct. The additional law required is Newton’s law of universal gravitation. It is usually written as $$F_g=-G\frac{m_1 m_2}{r^2}\hat r$$ This additional law is not derived from the others.

answered Nov 25 '22 at 15:14

Dale

99,825

But the additional law is the result of the observation that objects are subject to the same acceleration. – John Doty Nov 25 '22 at 15:24
If found out it was not Leonardo's but Galileos "leaning tower of pisa" experiment that lay the ground. – Peter Bernhard Nov 25 '22 at 15:28
@JohnDoty agreed. All laws of physics are tied to observations. – Dale Nov 25 '22 at 15:30

score 1 · Answer 3 · answered Nov 25 '22 at 15:30

Newton's three laws are general properties of all forces. Newton's Law of Gravitation and Coulomb's law of electrostatics are specific properties of the gravitational and Coulomb force respectively.

Is it correct to assume that the proportionality of earth's gravitation exerted on objects of different mass or weight needs some other law that is not implied in Newton's three basic laws of motion?

Yes. This is a specific property of gravity. So it's not derive-able from the three laws. The Coulomb force does not have this property

If the above is correct, is g ("9.x m/s2") a derivation of Newton's general law of gravity? How exactly does this derivation look like?

You can't derive laws, because laws are not theorems. Laws are axioms. They have to be postulated, not derived.

You have to deduce laws by observing nature. Observing that things fall at the same rate regardless of their mass helps in the deduction.

By observing that objects of different masses have the same acceleration, you deduce that force exerted by Earth on the object $F_{EO}$ is proportional to the Object's mass $m_O$.

By symmetry, the force exerted by an object on Earth, $F_{OE}$ is proportional to $m_E$.

By Newton's third law, $|F_{OE}|=|F_{EO}|=F$. So, $F$ is proportional to both $m_E$ and $m_O$, i. e. $F$ is proportional to $m_E m_O$

The dependence on $\frac{1}{r^2}$ is not observable on falling objects, because $g$ does not vary much near the surface. You have to deduce the proportionality of $F$ to $\frac{1}{r^2}$ from planetary motion observations.

Is Earth's surface gravity termed $g$ derived from Newton's universal law of gravity?

3 Answers3