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Minecraft is a game of mining where one digs and finds ore (to put it simply). the world is composed of "blocks" 3 dimensional cubes that have different "pictures" on them, that make them appear like grass or rock or Steele or diamonds...ect. In minecraft there is also TNT which is a block just like all others except it explodes. That does not relate to the question however which is, How do i find the gravitational constant? I mostly need the formula, which my 10th grade physics class has not taught me yet so i can find the neccesary factors to solve it. Aside from just giving me the formula what ways do you think I could find G? Also, just for more info, most block dont fall but sand and gravel do so they could be key to this problem perhaps.

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closed as unclear what you're asking by Chris White, Brandon Enright, Dilaton, Manishearth Feb 20 '14 at 9:06

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

See m.youtube.com/watch?v=aE9_YAXao3I –  Brandon Enright Feb 20 '14 at 8:12
You can't because gravity is constant. But if you dig below the bedrock, you fall which means that there is another source of gravity besides the main map. –  jinawee Feb 20 '14 at 9:23

1 Answer 1

up vote 1 down vote accepted

Just use the free-fall equation. The time spent in falling from a height $ h $ verifies this:

$ h - \frac{g}{2} t^2 = 0 $

So you get:

$ g = \frac{2h}{t^2} $

Note that you have to determine the height $ h $. How?, maybe you can estimate it thinking that the game's character is about 1.80 m.

The $ g $ you are obtaining here is the acceleration of the gravity. In order to get the "universal" gravity constant for the game's universe, $ G $, you'll need to know, or at least estimate, the mass of the planet and the radius of the planet... or the ratio $\frac{M}{R^2}$

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