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Can someone explain to me how to solve $6.6738410\times 10^{-11}\ \text{m}^3\ \text{kg}^{-1}\ \text{s}^{-2}$, the formula to find the gravitational constant?

I'm writing a bunch of reports about the scientific perspective about the video game Minecraft, and I'm trying to find the gravitational constant of Minecraft. I'm having a little bit of trouble solving $6.6738410\times10^{-11}\ \text{m}^3\ \text{kg}^{-1}\ \text{s}^{-2}$, as I don't know what many of the variables are. Some I get, like $\text{m}$ is the mass of something, $\text{kg}$ is something's weight in kilograms, and $\text{s}$ [for seconds] is the time it takes an object to fall in seconds.

It would be really helpful if someone could explain to me what some of these variables mean, and help guide me through the equation.

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4 Answers 4

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You're confused. The number you're referring to, $6.6738410\times 10^{-11}\ \text{m}^3\ \text{kg}^{-1}\ \text{s}^{-2}$, is Newton's gravitational constant, $G$. This is not a formula or equation. The 'variables' are actually units: meters (m), kilograms (kg), and seconds (s).

The formula where this constant most commonly appears is for Newtonian gravity: $$ F = G\frac{Mm}{r^2} $$

Where $F$ is the attractive force on each of the objects, $M$ and $m$ are the masses involved, and $r$ is the distance between their centers of mass.

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Thank you, I'm going into 6th grade so this is my first experience with it. I will use this information to further my own knowledge, and write a better report for the internet. –  Potato_Lover Jul 18 at 21:40
1  
Great that you're learning! Keep up the good work :) –  Fenzik Jul 18 at 21:43

The symbols $\text{m}$, $\text{kg}$, $\text{s}$ aren't variables that depend on a particular situation. They are the units. You can write $G = 6.67384\times10^{-11} \frac{\text{m}^3}{\text{kg}\cdot\text{s}^2}$, which means that $G$ is measured in cubic meters per kilogram-squared second, exactly the same way that you can say the Earth's radius is $6370 \text{km}$; $6370$ is the numeric value, kilometers is the unit.

You're not going to be able to find this gravitational constant in Minecraft, because the world is not a sphere; even if it were, you don't know its mass or radius. $G$ appears in the formula for the gravitational force between two bodies:

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

where $m_1$, $m_2$ are the masses and $R$ is the distance between the objects. What you might be able to find is $g$, the acceleration of gravity at the surface. This constant appears, for example, in the formula to calculate the distance $d$ a body falls in a time $t$:

$$d = \frac12 g t^2$$

This formula ignores air resistance. I don't know whether things in Minecraft have a terminal velocity or not; it might be fun to find out (or you could look at the code).

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What you have listed is not a formula: It's the gravitational constant itself.

Getting an approximate for the gravitational constant should be fairly simple, experimentally. All you need to do is jump.

Here's the equation you need:

$$F = \frac{G M m}{r^2}$$

Where $F$ is the magnitude of the force exerted on each of the two objects in question, $G$ is the gravitational constant, $M$ is the first mass in question (say, the Earth) and m is the second mass in question (say, your character). Oh, and $r$ is the center-of-mass distance between the two.

From here, you should come up with an approximation for the mass per unit area. This is doable as the thickness of a Minecraft world is finite. You will need to do some calculus, as the r between each block and your player is different. Fortunately, this is actually almost the same problem as exactly analogous to the infinite sheet of charge problem, so you can pretty much subvert the calculus by inspecting the similarities between Coulomb's law and the gravitation law.

So at this point, all you will have left to do is figure out $F$ experimentally, by measuring your acceleration when you jump in a Minecraft world.

Here's what you need to measure:

  1. The height of your jump
  2. The time of your jump (i.e that time it took from when your character jumped to when your character hit the ground)

From here, you can use the standard kinematics equations to figure out your acceleration, a. Since

$F = ma$

you can use this (along with your character's approximate mass) to calculate $F$. Now you have one equation with one unknown, and you should be able to calculate $G$.

I hoped this helped!

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Hi DBPriGuy and welcome to the Physics.SE! You can typeset formulae by using $\LaTeX$ via MathJax here. A basic tutorial is here. Otherwise, have fun! –  ACuriousMind Jul 18 at 22:00

This isn't an equation so there isn't anything to "solve". In terms of what the symbols mean "m3 kg-1 s-2" is just giving you the units of the gravitational constant, which in SI are meters cubed per kilogram per second squared. E^-11 is just a short hand for $\times 10^{-11}$ . The whole thing is just a number (indeed a constant!) \begin{equation} G = 6.67384 \times 10^{-11} \mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}\end{equation}

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