Newtons Laws of Gravity

[MrGamma]

Unfortunately I am close to what you would call math notationally illiterate and I have found that to understand physics formulas translating it to a language which I know helps.

 

I have this formula...

 

 

After translating it I have...

 

function newtons_law_of_gravitation($G,$m1,$m2,$r12){

	# ^ 			unit vector
	# r12 		vector from m1 to m2
	# m1			mass 1 (kg)
	# m2			mass 2 (kg)
	# F1			force on m1 ( Newtons )

	$F1 = ( ($G * $m1 * $m2) / pow($r12,2) ) * $r12;

	return $F1;
}

 

 

Is this essentially correct? All of the items above the line are multiplied and then r12 below the line is multiplied to the power of two and then used to divide what is above the line and is then multiplied what is next to the line?

 

 

This I'm having trouble understanding... The Big G...

 

 

I have translated it to this...

 

function gravitational_constant(){
   return 6.67 * pow(10,-11);
}

 

However I cannot understand the significance of the units...

 

m³ // is this mass cubed?

 

kg^-1 // this is kg which is also mass to my knowledge... why is it negative...

 

s^-2 // This is seconds... but it is negative as well...

 

 

How can there be three units ( two of which represent the same thing ) be lined up next to each other and some negative?

 

It makes no sense...

[CaptainPanic]

m3 is distance to the power 3: it's volume (not mass!). It is meter*meter*meter, or in quantities: length*breadth*height. Multiply volume (in m3) by density (kg/m3, or [math]kg*m^{-3}[/math]) to get mass (kg)

 

There are many quantities which contain all 3 units: m, kg, s.

Please distinguish:

-"quantity", which is for example mass.

-"unit", which is for example kilogram (kg).

The quantity mass has the units kg.

 

Also note that when there is a small -1, -2 or -3 in superscript behind a unit, it means to the negative power, which is the same as putting it under the line of division: m/s = ms-1... like in:

 

[math]3*2^{-1} = 3*\frac{1}{2}=\frac{3}{2}[/math]

 

There exist quantities that are made up of 2 units.

For example velocity (or: speed): it is in meters per second (m/s). How much distance you travel per time.

As mentioned before: density: kilograms per cubic meter (kg/m3). How much weight per volume.

But there exist also quantities that have 3 units:

Energy: kg*m2/s2 (also known as Joule, 1 Joule = 1 kgm2/s2)

This is already becoming harder to understand when you just look at the units. But when you look at some formulas for calculating energy, it makes more sense:

E = 0.5*m*v^2 = [kg * (m/2)^2 = kgm2/s2].

Also, force in Newtons: 1 Newton = 1 kgm/s2! So that one can also be broken down into basic units.

 

So why does the gravitational constant have these weird units: volume per mass per time squared (m3/kgs2)??

Well, as many constants, they are just there to relate the other quantities: force (N = kgm/s2), mass (kg) and distance (m). The formula must have the same units on both sides, or else we would be comparing apples and pears (or is that "comparing apples and oranges in the English language? Anyway, comparing two different things).

 

And why does it have the value of 6.67E-11? The world one day kind of arbitrarily chose to use the meter as the measure for distance, kilogram for weight and second for time. We could have chosen this differently. Then the value would have been different.

[Klaynos]

to add:

 

kg^-1 is the same as saying 1/kg so the - sign means that it's divided by so:

 

a^-20 = [math]\frac {1} {a^{20}}[/math]

[D H]

The formula must have the same units on both sides, or else we would be comparing apples and pears (or is that "comparing apples and oranges in the English language? Anyway, comparing two different things).

The idiom is "apples and oranges", which are actually quite comparable and even quite similar.

 

Anyhow, back on topic. What CaptainPanic is alluding to here is an extremely powerful tool called "dimensional analysis". Links: Wikipedia article on dimensional analysis, and a tutorial from the University of Guelph.

 

So, using dimensional analysis, what must the gravitational constant look like dimensionally? Ignoring the direction of the force, Newton's universal law of gravity is that

 

[math]F = \frac{G\, m_1 m_2}{r_{12}^2}[/math]

 

or

 

[math]G = \frac{F\, r_{12}^2}{m_1 m_2}[/math]

 

I'll define the units operator [math]\mathcal{U}(x)[/math] and the dimensional operator [math]\mathcal{D}(x)[/math] as follows:

• [math]\mathcal{U}(x)[/math]. The units operator strips away the specific numbers from a quantity, leaving only the units. For example, if m₁ = 5.9742×10²⁴ kilograms and m₂ = 1.2 ounces, U(m₁) = kilograms and U(m₂) = ounces.
   
   
  
• [math]\mathcal{D}(x)[/math]. The dimensional operator abstracts away the specific units, leaving only the dimensions themselves. There are only a handful of physical dimensions: mass, length, time, electric current, and temperature (some use electric charge instead of current). In the above example, while the masses have two different units, they both have the same dimension, which is of course mass.

.

 

Applying dimensional analysis to Newton's equation for gravity,

 

[math]

\mathcal{D}(G) =

\mathcal{D}\left(\frac{F\, r_{12}^2}{m_1 m_2}\right) =

\frac{\mathcal{D}(F)\, \mathcal{D}(r_{12}^2)}{\mathcal{D}(m_1)\mathcal{D}(m_2)}[/math]

 

On the right hand side,

• [math]\mathcal{D}(F)[/math]. Force is mass times acceleration, which respectively have dimensions of mass and length/time/time, so [math]\mathcal{D}(F)[/math] is mass*length/time/time, or mass*length*time^-2, or even more compactly, MLT^-2.
   
   
  
• [math]\mathcal{D}(r_{12}^2)[/math]. The distance between the two objects has units of length. The quantity in question is the square of this distance, so [math]\mathcal{D}(r_{12}^2)[/math] is length*length, or more compactly, L².
   
   
  
• [math]\mathcal{D}(m_1)\mathcal{D}(m_2)[/math]. Each term in the denominator is a mass, so the dimensions of the denominator are mass*mass, or more compactly, M².

 

Putting it all together, the left-hand side has dimensions (mass*length/time/time)*(length*length)/(mass*mass), or (MLT^-2)*(L²)/(M²). Collecting like quantities, this becomes (length*length*length)/(mass)/(time*time), or L³M^-1T^-2. These are the physical dimensions of the left-hand side of the above equation. The only thing on the right-hand side of the above equation is the gravitational constant itself, so the physical dimensions of the gravitational constant must be (length*length*length)/(mass)/(time*time), or L³M^-1T^-2.

 

Physicists almost exclusively use the international system of measurements (SI), aka the metric system. In SI, the units of length, mass, and time are meters (m), seconds (sec), and kilograms (kg). Thus in SI units, the gravitational constant has units of m³kg^-1sec^-2.

 

If you use different units you will get a different numeric value for G, but it will still have the same physical dimensions. Astronomers, for example, occasionally use astronomic units instead of meters and days instead of seconds. To them, [math]G=(1.48789\pm0.00015)\times10^{-34}\,\text{AU}^3/\text{kg}/\text{day}^2[/math].

[MrGamma]

This is excellent information. Thank you for your time.

 

acceleration L/T²

 

So... technically this translates to...

 

L / ( T * T )

 

and in the final solution the units would be represented in the SI metrics.

 

m/s²

 

Undertood. Thank you.