Wikipedia:Reference desk/Archives/Mathematics/2015 April 28

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April 28

Steps to calculate G?

What are the steps to calculate G? I know that first I would do 6.673 × 10⁻¹¹ but what then? What does (m/kg)² mean? And finally, N = Newton right? -SGA314 (talk) 14:40, 28 April 2015 (UTC)

Yes, N stands for Newton which is the unit of force; and N·(m/kg)² is the unit of the Gravitational constant in the SI system (the value of the any dimensional quantity always depends upon the units of measurement).

The Gravitational constant is a measured quantity, whose value can be derived from experiments measuring the gravitational force between two known masses kept a known distance apart. For early experiments in the area, which are commonly repeated in high-school labs nowadays, see Cavendish experiment. For a more esoteric experimental set-up, see this. Abecedare (talk) 14:59, 28 April 2015 (UTC)

So, (m/kg)² means distance between the two objects in meters / the first objects weight squared. Right? -SGA314 (talk) 15:14, 28 April 2015 (UTC)

No, (m/kg)² is a SI derived unit and not something to calculate. See Newton's law of universal gravitation#Modern form for the formula used to make calculations. You said "first I would do 6.673 × 10⁻¹¹". That indicates you are assuming the value of G is already known so you are probably trying to solve an exercise which is not about calculating G but about using the known value of G to calculate one of the other values in the formula $F=G{\frac {m_{1}m_{2}}{r^{2}}}\$. If you for example want the gravitational force F in Newton (N) and you know the two masses in kg and their distance in meter (m) then just insert the numbers with G = 6.673 × 10⁻¹¹. Then you are using SI units in all cases so F becomes in Newton. Note you have to multiply the two masses and not square one of them. PrimeHunter (talk) 15:35, 28 April 2015 (UTC)

Oh! The value of G is 6.673 × 10⁻¹¹. What I wanted to do is I wanted to create a simulation using the Blender Game Engine that would display the attraction between the sun and the earth using G. So instead of trying to calculate G I need to use this formula:

$F=G{\frac {m_{1}m_{2}}{r^{2}}}\$

where:

$G$ = 6.673 × 10⁻¹¹

$m_{1}$ = Object 1's mass

$m_{2}$ = Object 2's mass

$r^{2}$ = The distance between the two objects squared

to calculate the attractive force between the planet and the sun. Thank you for the explanation! -SGA314 (talk) 15:49, 28 April 2015 (UTC)

Yes, that's right. However, physicists and teachers don't approve of formulations like "The value of G is 6.673 × 10⁻¹¹". The value of G is 6.673 × 10⁻¹¹ N·(m/kg)², where N·(m/kg)² is the unit the numerical value is given in. Without the unit it's like only saying "The distance to the Sun is 1.5 × 10¹¹". PrimeHunter (talk) 16:11, 28 April 2015 (UTC)

@SGA314: Your method sounds fine, but want to reemphasize that the value of G is not 6.673 × 10⁻¹¹; it is 6.673 × 10⁻¹¹ N (m/kg)². The units are not optional, and leaving them out is akin to saying "Mary's weight is 80" without specifying whether that is in kgs, pounds or some other system. And you have to keep track of the units in your calculations too. For example if you use the above formula with the distance measured in kilometers or mass specified in tons, you'll get the wrong results! Abecedare (talk) 16:14, 28 April 2015 (UTC) (Silent ec with PrimeHunter above, who makes the same important point.)

Ok. So, my distance ($r^{2}$) is in meters and my mass values are in kgs. So, how would I do the last part of the equation (m/kg)²? could you substitute some values for m and kg and explain what those values stand for. I am so sorry I am still not understanding this. If I understand what PrimeHunter said, then m would equal 149597870700 meters(the value of $r^{2}$) and kg would be equal to either the weight of the sun or the earth in kgs. The kg variable seems a little confusing sense there is two variables that correspond to mass in the equation. Yet again I am soooo sorry for any frustration I am causing because of my lack of understanding.

Edit: I was just reading this and the formula states:

$g=G{\frac {m_{1}}{r^{2}}}=(6.6742\times 10^{-11}){\frac {5.9736\times 10^{24}}{(6.37101\times 10^{6})^{2}}}=9.822{\mbox{m}}\cdot {\mbox{s}}^{-2}$

that $G$ is equal to $(6.6742\times 10^{-11})$ how can that be based on what you have explained to me? -SGA314 (talk) 17:42, 28 April 2015 (UTC)

The g formula is written sloppily by omitting the units but at least it only does it during a calculation and not when a single value is written by itself. (m/kg)² is not something to calculate. m and kg are not variables representing values. m simply means the word meter. m does not represent a number of meter. And kg means the word kilogram. As in "John is 1.80 meter tall" or "Mary weighs 60 kilogram". Sometimes units are combined, often because there is no name for the combination. For example, we may say Peter runs 5 m/s. That means he will run 5 meter in 1 second if he maintains the speed. There is nothing to calculate. "5 m/s" already states his speed, and "m/s" is a required part of stating the speed. We don't say "Peter runs 5". It may be harder to see the physical "meaning" of (m/kg)² than of m/s, but it's exactly the same principle of combining units. If you dont understand how to calculate with units then here is a shortcut: If all values in a formula are given in SI units such as s (second), kg (kilogram) and N (Newton), then you can ignore the units during the calculation, and the result will be in SI units. Example calculation of gravity between two objects weighing 5 kg and 8 kg, with 2 m between them: $F=G{\frac {m_{1}m_{2}}{r^{2}}}\$ "=" (6.673 × 10⁻¹¹) × (5 × 8) / 2² = 6.673 × 10⁻¹¹ × 10 = 6.673 × 10⁻¹⁰. I wrote "=" in quotation marks the first time because it isn't a real equality when one side omits the units. We are computing a force F and the SI unit for force is N (Newton) so the result must be 6.673 × 10⁻¹⁰ N, since we only used SI units in the calculation. The result could also have been written 6.673 × 10⁻¹⁰ kg⋅m/s², because 1 N = 1 kg⋅m/s². PrimeHunter (talk) 18:58, 28 April 2015 (UTC)

Oh I get it. the (m/kg)² is just a unit of measurement. Since I will be using all SI units(more specifically I will be using kilograms and meters) then I can omit the last part. So what is the name of the unit that (m/kg)² represents? —SGA314 (talk) 19:18, 28 April 2015 (UTC)

I'm not aware of a name for (m/kg)² or of properties which have that as a unit of measurement. PrimeHunter (talk) 19:56, 28 April 2015 (UTC)

Ok. Thanks for all of your help. I know understand at least how to impliment the equation with SI units. Thank you soooo much! —SGA314 (talk) 20:00, 28 April 2015 (UTC)

• (m/kg)² itself doesn't mean much of anything; and I think it's poor style, if not misleading, to write it that way rather than m²/kg². The units N·m²/kg², sometimes written (m³/s²)/kg, are what you need to multiply in so that the result is in the right units.

Newton wrote the formula F = G m1 m2 / r², or something equivalent to it, without knowing G, because he did not know the mass of any gravitationally significant body. But if you know all the other numbers in that equation for a given system, you can derive G = F r² / (m1 m2). The right side of this equation is (number) newtons · ((number) meters)² / ((number) kilograms)². Bring the numeric values together and you have (new number) newton · meter² / kilogram². What does N·m²/kg² mean as a whole? It's the unit in which we measure the gravitational constant, nothing else!

(You could measure the experimental setup in pounds [force], cubits and atomic mass units, if that's what floats your boat, but then don't forget that your result is (number) pound · cubit² / atomic mass unit².) —Tamfang (talk) 23:33, 30 April 2015 (UTC)

Diametric Drive: x & y equals what?

I have found a formula for a diametric drive from here.

In the equation for the diametric drive, what does x and y equal?

$V=(-m){\begin{bmatrix}{\frac {-G}{\sqrt {(x+d)^{2}+y^{2}}}}\end{bmatrix}}$ $+(+m){\begin{bmatrix}{\frac {-G}{\sqrt {(x+d)^{2}+y^{2}}}}\end{bmatrix}}$

where :

$x$ = ?

$y$ = ?

Once I get the answer I will add the meaning of the x and y variables to the formula section. -SGA314 (talk) 16:57, 28 April 2015 (UTC)

If one was to draw a 2-D diagram showing the positive and negative masses, these would be the x and y axes, with the masses aligned along the x axis. It is the gravitational analog of an electric dipole, as in in the section Electric dipole moment#Potential and field of an electric dipole. --Mark viking (talk) 17:55, 28 April 2015 (UTC)

So what, x and y are equal to the x and y locations of the two masses in relation to each other? also if I understand correctly, x is always = to 0 right? I don't quite understand. Where does a 2-D grid come into play? Also I am going to be using this formula in 3-D space not 2-D. -SGA314 (talk) 18:11, 28 April 2015 (UTC)

Well, that is my best guess based on just the formula. If there is a third dimension, then there should be a z-coordinate as well. Perhaps the source assumed everything was in the z plane? In this instance, it would be best to look up the original source and and go with their set up of the equations and semantics for the variables. To do much else would be original research. --Mark viking (talk) 19:49, 28 April 2015 (UTC)

Ok i will do. Thank you for your help. —SGA314 (talk) 20:01, 28 April 2015 (UTC)

Well, I get $V\equiv 0$. Probably you want $\sum _{\pm }\mp m{\frac {-G}{\sqrt {(x\pm d)^{2}+y^{2}}}}$ or $\sum _{\pm }\mp m{\frac {-G}{\sqrt {(x\pm d/2)^{2}+y^{2}}}}$ to give the masses a separation of d. y is just the distance from the axis connecting the two masses, regardless of azimuth; the problem is axisymmetric. --Tardis (talk) 06:11, 30 April 2015 (UTC)