Units in gravitational constant

Question

I was reading on the internet and I found that the gravitational constant is roughly $6.674 \times 10^{-11}~\mathrm{ m^3~ kg^{-1} ~s^{-2}}.$ I also found that it is equal to $6.674\times 10^{−11}~\mathrm{ N\cdot m^2/kg^2}.$

First question: what does the first unit of measurement mean? $6.674 \times 10^{-11}$ meters cubed over kilograms over second squared? Is that referring to the acceleration per kilogram, in meters (velocity change) per second squared? If so, why meters cubed?

Second question: the second expression. I know that a newton times a meter is basically a newton exerted for one meter, but what does a newton times a meter squared mean? Does it mean that the newton of attraction is multiplied by the meter squared? What does the meter squared refer to - the distance between the objects? Why is the attraction in newton times meter squared over the kilogram squared? Please, can someone just explain the equation and why it is expressed in that way?

Also: if this is just a constant, why is it measured like this? Wouldn't a straight-out acceleration over kilogram (mass) work as well?

Answer 1

Well, the way to find the units of the constant are to consider the equation it takes part in:

$$ F = G\frac{m_1 m_2}{r^2} $$

$F$ is a force: so it's measured in newtons ($\operatorname{N}$). A newton is the force required to give a kilogram an acceleration of a metre per second per second: so, in SI units, its units are $\operatorname{kg}\operatorname{m}/\operatorname{s}^2$. $m_1$ and $m_2$ are masses: in SI units they are measured in kilograms, $\operatorname{kg}$, and $r$ is a length: it is measured in metres, $\operatorname{m}$.

So, again in SI units we can rewrite the above as something like

$$\phi \operatorname{N} = \phi \operatorname{kg} \operatorname{m}/\operatorname{s}^2 = G \frac{\mu_1 \mu_2}{\rho^2}\frac{\operatorname{kg}^2}{\operatorname{m}^2} $$

where $\phi$, $\mu_1$, $\mu_2$ and $\rho$ are pure numbers (they're the numerical values of the various quantities in SI units). So we need to get the dimensions of this to make sense, and just doing this it's immediately apparent that

$$G = \gamma \frac{\operatorname{m}^3}{\operatorname{kg} \operatorname{s}^2} $$

where $\gamma$ is a pure number, and is the numerical value of $G$ in SI units.

Alternatively if we put newtons back on the LHS we get

$$G = \gamma \frac{\operatorname{N} \operatorname{m}^2}{\operatorname{kg^2}} $$

Answer 2

The first set of units is in fact equal to the second. If you replace the Newton in the second expression by its definition in terms of kilograms, meters and seconds

$$ 1 N = 1 \frac{\mathrm{kg ~ m}}{\mathrm{s^2}} $$

you recover the first expression.

The SI system has a number of basic units (meter, kilogram, second, ampere, kelvin, mole and candela). All other units are defined based on these seven, and they are really nothing more than convenient shorthands in notation.

The meaning of the second expression, which I imagine is the one you are more familiar with, is that it is the number which you should multiply with the masses of two objects (hence the $\mathrm{kg^{-2}}$) and divide by the square of the distance between them (hence the $\mathrm{m^2}$) so that you will recover the force of gravity that the objects exert on one another.

The meaning of the first expression is exactly the same, because it is the same expression. It has just been obscured by a less familiar notation, replacing the easily-recognizable Newton by its component units. Trying to directly intuit its meaning from looking at the units is not impossible, but it is unnecessarily confusing. Once you have checked that both expressions are in fact identical, I would advise you not to worry too much about the 'meaning' of the units in the first expression.

As to your last question, no it wouldn't. This is because the equation for gravitational force needs to output a force, and take into account the masses of both objects, as well as the square of the distance between them. Thus the gravitational constant must have units to match.

I hope this helps.

Answer 3

To answer this we need to take a look at the equation $F_g=Gm_1m_2/ d^2$. So if G is measured in $\rm m^3/kg~s^2$, and mass is measured in kg and distance is measured in m, then force is measured with $\rm m^3/kg ~s^2 \cdot kg^2/m^2$, which simplifies to $\rm kg~m/s^2$

And now to define $\rm kg~m/s^2$ your instincts might be to split it into $\rm m/s^2$ and kg. If $\rm m/s^2$ is a unit of acceleration and kg is a unit of mass, then force must be mass times acceleration. This is described by Sir Issac Newton PRS' second law of motion describes:

$F=ma$

So it makes sense the gravitational constant G is measured in $\rm m^3/kg^1~s^2$.

Answer 4

Its a problem.

Constants allude to pure numbers so indeed its funny that a constant should have units of measurement.

It's a fitting problem. You find, or guess that something depends on something else, proportionally like when x goes from 3 to 4, y goes from 6 to 8, (so y=2*x where 2 is a constant) or inversely proportional (y=x/2), so when you are satisfied that you found everything that can affect that something you pretty much have your equation, like y=ax^2+bx+c the simple quadratic in one dimension or something like w=xy.

The last step is to add constants so that the numbers, the results match.

However if by your units of measurement principles the units don't match, you got a problem. You will sacrifice for this if your constant holds even though it has units, but perhaps be aware that there is more to the equation than this simplification or of course that your original idea of units of measurement has a flaw. Its more of a mess to redefine your first principles, i.e speed is not meter/seconds so lets leave that out for now.

The gravitational equation in this form is also very similar to Coulombs law, too similar in fact, both are mostly guides to say that the force is proportional to the masses of the objects and inversely proportional to the square of their distance (in gravity's case)

You do get neat squares with the gravitational force, i.e. (kg/m)2 so if the whole thing is squared then you might wonder what kg/m is.

For example: Squares appear when you are adding stuff through integration, integrals another fine mathematical concept which however, at least graphically, is an approximation.

So we say if y=x^2 then dy/dx=2x and integration being the reverse of differentiation, using notation 'Integral of x' as I(x), then I(2x)=2*(x^2)/2 + K (we always add a constant in integration for the missing part.

So perhaps the (gravitational) force is f=I(something) so that it ends up squared.

Force is a funny animal. You got things like impulses like you got things like energy, work and power all of them concepts in physics, connected. For example iirc work=power*time but that's just common sense talking so I'll stop here.

Added:

To start thinking about kg/m and what that is, one thing that popped to mind, these two are connected when something travels a distance, how does the distance depend on the mass? Well, certainly when you got friction, the mass matters. You can think about density too, which is mass/volume.

So F~volume^2 and perhaps F=volumesomething, that brings it back to kgm/s^2. something that in the perceivable local is stable, constant. Mind you if F=I(x) and it has m/s^2 in it, there is an integral relation between speed and acceleration (s=vt+at/2) where s is distance, v is speed, a is acceleration and t time. Keep in mind that integration is subjective too, you integrate over something so if w=xy and both x and y are variables you can integrate w over x and you can integrate w over y. These are/(can be) additive provided that they are independent coz if y=f(x) you can go to single variable w=xf(x) => w=g(x)

Answer 5

Since this question had 46K(!) views it may be useful to add an answer even after 4 years.

$G$ is an experimental constant required to match the Newton potential energy to experiment. The Newton potential energy is $$E_P = - \frac{GM m}{r} ~.$$ Dividing by the energy $m c^2$ you get the dimensionless potential $$V = -\frac{GM}{c^2r} ~.$$ Since $V$ is dimensionless $GM/c^2$ is a length. This length is interpreted as half the radius of a black hole with mass M, $r_M/2$. G has dimension $m^3 kg^{-1} s^{-2}$. You can therefore also write the dimensionless potential as $$V=r_M/2r$$ where the only constant is a length with a clear albeit exotic interpretation.

Answer 6

This might help you visualize $G$'s value and what its units mean in a physical sense.

Consider a system with a particle sitting in space at a unit distance from a unit mass - say a grain of dust that's a metre from a $1kg$ rock. Given: $$F = ma = G \frac{M m}{r^2} ~,$$ and with $M$ and $r$ both $1$, we can see that the particle's gravitational acceleration towards the rock, $a$, is numerically equal to $G$.

The units work too, because (cancelling out the particle mass $m$ and inserting units): $$a (ms^{-2}) = G (m^3kg^{-1}s^{-2}) \frac{1(kg)}{1^2 (m^2)}$$

So that's interesting: the particle at that moment is accelerating towards the rock at $6.67 \times 10^{-11}ms^{-2}$, and in any system of units, $a = G$, numerically.

Now, imagine the particle is orbiting the rock, and maintaining a unit distance orbital radius. What is its angular velocity? Using $a = \omega^2r$ (and keeping $r$ as one metre), we see $\omega^2 = G$, numerically.

That means that $\omega = \sqrt G$. For metric units, our particle moves about $8\mu rad/s$, or $122,000s$ per radian, and that's an orbit every 9 days.

Angular velocity has units of angles (let's stick with radians) per unit time, $\theta/s$, but angles are a unit-less ratio of $arclength/radius$, so we'd usually see $\omega$ with units of just $s^{-1}$.

We've been imagining the central mass as a rock, but it could be any co-centric sphere (up to the radius of the orbit) as long as the mass seen at the centre-of-mass was unchanged, ie we're keeping the mean density of the space inside the orbit constant. Suppose it's a $1kg$ sphere of a low-density material that's exactly one metre radius, so the particle's orbit now cruises just over the surface. And suppose we make that the defining characteristic of the orbit: it's not a $1$ metre orbit, but an orbit that's cruising the surface of a unit-radius, unit-mass sphere.

This means we've boiled gravity down to two components: for a given value of $G$ in some system of units, we relate the density of the sphere (a ratio of mass and distance$^3$), and the $\omega^2$ (that's $1/$time$^2$) of the particle that orbits just above its surface.

If in a different universe, where $G$ is bigger, then you can keep the same sphere and the particle will need to go faster, or you can reduce the density (lower the mass or increase the volume) of the sphere, and keep the same speed. So $G$ tells you, for a given universe, the ratio of $\omega^2$ to the sphere's density, in units of your choosing.

In other words, $G$ has units of $\frac{\omega^2}{mass/volume}$, which is $ 1/time^2 \times \frac{volume}{mass}$.

In metric, we call it $m^3kg^{-1}s^{-2}$. Once you have the particle orbiting that sphere in your visual register, and you imagine flexing the sphere's mass or volume, or making gravity stronger or weaker, and seeing what the particle has to do to keep orbiting, you'll see that $G$ has to have those units. It's in the nature of the structure.

Answer 7

The most direct interpretation - one that transcends the paradigm divide between Relativistic and Non-Relativistic Physics, and is connected to the Raychaudhuri Equation, is that in terms of the volume contraction.

A cloud surrounding a body of mass $M$, whose constituents are all in radial motion, has a volume that as a function of time $V(t)$ satisfies the equation $$\frac{d²V}{dt²} - \frac{2}{3V} \left(\frac{dV}{dt}\right)^2 = -4πGM.$$ If initially stationary, then the initial acceleration of the volume, under the force of gravity, is $-4πGM$, the negative indicating that it starting to contract.

So, the units for $GM$ are cubic meters per second, per second.

The generalization of this to an $n + 1$ dimensional space-time is $$\frac{d^2V}{dt^2} - \frac{n-1}{nV} \left(\frac{dV}{dt}\right)^² = -n \frac{π^{n/2}}{(n/2)!} G_n M,$$ using the convention $(-1/2)! = \sqrt{π}$, where $G_n$ is the $n$-dimensional version of the Newton coefficient; whose units would be meterⁿ/(second² kilogram).