Units of force and acceleration in the context of gravity

What is the unit of gravity?

 

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Sorry, that too is gibberish.

 

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Gravity is usually described in a unit called newton. The newton describes gravitational acceleration in terms of meters per second squared (m/s²). Truth told, Wikipedia↗ can tell you plenty about newtons, whereas I can't. I can tell you that we humans experience Earth's gravity at 9.8 m/s², for instance. Other than that, I"ll leave the more complicated explanations to people who know more and can explain better.

 

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A Newton is a unit of force, not a unit of acceleration.

 

Can acceleration be written as m^-2/sec?

 

Oh, yes, sir! Thank you, sir! Can you please assist my humble ignorance and tell me what this unit of force describes?

 

Not if you're going to cop an attitude, no -- though reading the wiki article you linked would probably help.

 

No.

 

Part of the problem arises from the fact that we cannot directly detect gravity. We can only detect what things do in its presence. It is this force that we can apply units to.

Example, we can directly detect a magnetic field and assign units of flux density at every given point in space.

We cannot directly detect a gravitational field or assign units to it in space. All we can do is place things (such as an apple) in its field and see what the apple does. We assign values to what the apple does. We can also measure against some distant clock and assign values to time dilation. but again, that is not detecting gravity.

 

Tragic Fallacy and the Necessity of Learning to Read

That's what I thought.

Self-superior, hair-splitting hack pedantry is exactly useless when invested in illiteracy.

 

A member posts a simple 12 word statement of fact, and the thread goes down in flames?

Could we stay on topic please?

 

Mod Hat ― Skitt's Law

You do realize a variation of Skitt's Law is in effect by that post? To the one, I can certainly construct a limited context by which I see your point, but, to the other, that's just another off-topic post.

If it matters to you at all, the problem I'm dealing with in black ink is simply a matter of a useless post form:

(1) Straw man fallacy ...

(2) ... as basis for ...

(3) ... otherwise context-free ...

(4) ... extraneous correction ...

(5) ... in a low-effort post ...

(6) ... when the simple fact would have sufficed ...

(7) ... is annoying for its bad faith.​

He's exactly correct: A Newton is a unit of force.

But he also knows the answer to the question of what that unit of force describes, and that's why he doesn't want to answer. He went about it wrongly, for his own reasons, and he knows it.

And the thing is that this just happens to be a day when I am not in any mood to put up with it. He needn't fear my green ink about this; nor should you, by the way―I'm only using it in this post to make a particular point:

• This manner of low-effort, shitty discourse leads to a shameful number of compaints in our report queue, and has the effect of limiting sympathies for the "smart people" who just can't be bothered to put in the basic effort to justify that label in this community. To wit: Great, rah-rah, stand for science and accuracy and all that, but since it's more about self than anything else it's just not constructive. And when someone who has built himself a reputation for clumsy rhetorical sleight goes and does it again, yeah, it stands out. At some point, smart people need to act like they're smart. That is, what about being smart means people owe the respect of taking shit they won't put up with from idiots?​

Look at me, I'm smart, fuck you: This attitude is worse than useless.

As long as we've all got a handle on that, I'm happy enough to leave the hair split as is.

 

A newton is a unit of force equal to the weight near the surface of Earth of about 102 grams or 3.6 ounces. Force, \(\vec{F}\), and mass, \(m\), are related by acceleration, \(\vec{a}\), in the Newtonian formula \(\vec{F} = m \vec{a}\). In SI units, the standard units of force and mass have common names newton (N) and kilogram (kg), respectively, while acceleration is usually given as meters per second-squared, \(\textrm{m} \cdot \textrm{s}^{-2}\), m/s², or (rarely) newtons per kilogram.

A non-SI unit of acceleration is called the standard acceleration of gravity, \(g_0\), which is quantity adopted to substitute as a placeholder for average gravity at the surface of Earth. \( 1 \; g_0 = 9.80665 \; \textrm{m} \cdot \textrm{s}^{-2} = 9.80665 \; \textrm{N} \cdot \textrm{kg}^{-1} = 32 \, \frac{1061}{6096} \; \textrm{ft} \cdot \textrm{s}^{-2}\). When a fighter pilot talks about pulling "gees" this is the unit he is using to describe the acceleration the rapidly turning airframe is forcing his body components to endure. Near 18 gees (~175 m/s²) capillaries rupture as the body is bruising itself under its own “weight.”

https://en.wikipedia.org/wiki/International_System_of_Units#Derived_units
https://en.wikipedia.org/wiki/Metre_per_second_squared
https://en.wikipedia.org/wiki/Standard_gravity
https://en.wikipedia.org/wiki/Orders_of_magnitude_(acceleration)

The OP is gibberish because physical quantities are expressed as pure numbers times choices of physical quantities called units, and gravity is not a physical quantity but a phenomenon. The gravitational acceleration of all objects near the surface of the Earth is a physical quantity with units of acceleration which in the SI system has no nice name, but m/s² is very commonly used.

The old CGS system did have a standard unit called a gaileo (Gal). 1 Gal = 0.01 m/s² for the same reason 1 cm = 0.01 m.

In Newton's theory of universal gravitation, the formula is:
\( \vec{F} = - G \frac{m_1 m_2}{r^2} \frac{\vec{r}}{| \vec{r} | }\)
so in SI units, G, the constant of proportionality has units of meters-squared times newtons per kilogram-squared or meters-cubed per second-squared per kilogram when you expand what newtons are. G is also called the Gravitational constant because it shows up in Einstein's theory of gravity as well.

https://en.wikipedia.org/wiki/Gravitational_constant

 

Since G has units metres-cubed (sorry about the Brit spelling) per second-squared per kilogram, Gm where m has units of kilograms, therefore has units metres-cubed per second-squared.

Which can be interpreted as an accelerated volume change, or as an area (metres-squared) times an acceleration, i.e. a surface which is accelerating. Which is not to say an interpretation need have any physical meaning as such, but there it is.

 

Wow. It's not hair-splitting (nor is it a typo, since you said it twice, so no, no grammar nazi here): Your answer may have been well-intentioned, but it was wrong. You should learn to be more accepting of being corrected. And a moderator should have more professionalism than to get abusive over a simple mistake: learn and move on.

 

The question remains.

 

If you will recall Kepler's laws of planetary motion,
1) the orbit of a planet is an ellipse with the sun at a focus with semi major axis, a, and semi minor axis, b and area πab.
2) a line between the sun and planet sweeps out equal areas in equal times, ½ r² dθ/dt = π a b / P, where P is the period of the orbit.
3) the cube of the semi-major axis is proportional to the square of the period, a³/P² = K which is related to the units and even the value of GM where G is Newton's constant and M is the mass of the sun.

dθ/dt has the average value 2 π / P, and we have 4 π² a³/P² for the Mercury, Earth, Mars, Jupiter, Pluto, etc are all approximately \(1.33 \times 10^{20} \, \textrm{m}^3 \cdot \textrm{s}^{-2}\) where departures mostly indicate Kepler was not exactly correct, particularly for Jupiter and outer planets where Jupiter alters calculations in the fourth decimal place.

GM for the sun is about \(1.3271244 \times 10^{20} \, \textrm{m}^3 \cdot \textrm{s}^{-2}\)

 

Ok. One thing is, without units of mass, gravity doesn't have a physical basis let alone "units". Kepler's laws require orbital motion, which requires bodies with mass. Then there is the equivalence between gravity and acceleration. I can accept the existence of units of acceleration, so I must then also be accepting the existence of equivalent units of gravity (?)

 

Does it?

A newton is an amount of force that will accelerate a 1 kg body by 1 meter per second per second in the direction of the force.
A newton is an amount of force that will accelerate a 10 kg body by 0.1 meters per second per second in the direction of the force.
A newton is an amount of force that will accelerate a 0.25 kg body by 4 meters per second per second in the direction of the force.
A newton is an amount of force that will accelerate a 0.102 kg body by 9.81 meters per second per second in the direction of the force.

So during the time a constant force, \(\vec{F}\) is being applied, the dynamics of an object of mass m are given by:
\(\vec{s}(t) = \vec{s}(t_0) + \vec{v}(t_0) \times \left( t - t_0 \right) + \frac{1}{2 m} \vec{F} \times \left( t - t_0 \right)^2 \\ \vec{v}(t) = \vec{v}(t_0) + \frac{1}{m} \vec{F} \times \left( t - t_0 \right) = \frac{d \; \;}{dt} \vec{s}(t) \\ \vec{a}(t) = \frac{1}{m} \vec{F} = \frac{d \;\;}{dt} \vec{v}(t) = \frac{d^2 \; \;}{dt^2} \vec{s}(t)\)

That's all encapsulated in \(\vec{F} = m \vec{a}\). In Newtonian mechanics, no acceleration implies zero force while zero force implies zero acceleration. Thus we have Newton's first law of motion: An object has the same state of motion or non-motion unless acted upon by a non-zero net force.

A more modern definition (and in Newtonian theory, equivalent) is that force is the rate of change of momentum per unit time, but since momentum (like acceleration) has no "nice" name in SI units, I'm not sure explaining that 1 newton is a rate of change of (1 kilogram-meter-per-second) per second would help explain what a newton is.

Gravity isn't a physical quantity, but a phenomenon, so your sentence doesn't parse.

Actually, Newtonian predictions reduce to Kepler's in the limit of zero planetary mass. And the masses would of course still exist even if humans didn't invent units for them. So your point is unclear.

There is a local equivalence between describing falling in coordinates regarding the surface of the Earth as a stationary reference and bodies far from any large masses in accelerating coordinates where the notion of stationary is not compatible with Newton's first law.

True if you mean units of gravitational acceleration. But gravity is a class of phenomena which covers more than just the surface acceleration of gravity in various places. In places with no preferred surfaces, such as a globular cluster, you would be cut off from any standard of rest such as the one afforded by the Earth's approximate spherical symmetry and would not have a good reference for zero acceleration.

 

Ok, so I get that Kepler's laws apply to planetary motions, but also to say, grains of dust orbiting the sun. So you could replace a planet with a grain of dust and it would obey the same Keplerian laws of motion (?)
As long as the grain was large enough to 'experience' the sun's gravity, but how large is that? ( a question Roger Penrose, I recall, wasn't too sure about).

And, gravitational acceleration, as 'experienced' by some grain of dust large enough and in the sun's locality, is not gravity but a specific physical effect.