Gravitational Constant Problem in Planetary Java Model.

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Discussion Overview

The discussion revolves around the implementation of Newton's law of gravitation in a planetary model, specifically focusing on the gravitational constant (G) and the correct application of the gravitational force formula. Participants explore the formula's components and seek clarification on the units involved.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant proposes using the formula GravityField = G * ((500*20)/(150^2)) to calculate gravitational force but is corrected that this formula represents gravitational force, not the gravitational field.
  • Another participant expresses confusion about the value of G and requests further elaboration on its meaning and units.
  • There is a discussion about the units of measurement for G, with one participant emphasizing the importance of understanding the units (m, kg, s) in the context of the formula.
  • One participant notes that the gravitational force will be small unless dealing with large masses like planets, highlighting the scale of the gravitational constant.
  • There are expressions of disappointment regarding the tone of responses, indicating a mix of frustration and a desire for clearer communication.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the clarity of the explanation regarding the gravitational constant and its units. Some express confusion while others attempt to clarify, indicating a mix of understanding and misunderstanding.

Contextual Notes

There are unresolved questions about the interpretation of the units in the gravitational constant and how they apply to the specific scenario presented. The discussion reflects varying levels of familiarity with physics concepts among participants.

Who May Find This Useful

This discussion may be useful for individuals learning about gravitational forces, the application of Newton's law of gravitation, and the significance of units in physics equations.

eArtist
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Hi there

I am currently trying to implement Newtons law of gravitation into a planetary model.

I have 2 objects with mass of 500 and 20. The distance between them is 150.
From what I gather from wikipedia I could use the following formula:

GravityField = G * ((500*20)/(150^2))
Where G is the gravitational constant.

What is the value of G? I could not figure it out from wiki...
And is the formula correct?

- eArtist
 
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eArtist said:
GravityField = G * ((500*20)/(150^2))
False! That is formula for gravitational force, not field.
Where G is the gravitational constant.
What is the value of G? I could not figure it out from wiki...
Come on! Could it be more obvious?
http://en.wikipedia.org/wiki/Gravitational_constant
 
I am not as into physics as you it seems. Can you please elaborate?
I need to set a value to G.
 
eArtist said:
I am not as into physics as you it seems. Can you please elaborate?
I need to set a value to G.
Just click the link to wiki. You need not to be a physicist to click and read it. It is written there in a VERY BIG characters.
 
I am a bit dissapointed at your hostile attitude. I am sorry if I offended you in any way.

Yes I have studied it for a while, but I don't understand.

6.67384 * 10^-11 m^3 kg^-1 s^-2 = ??
 
You finally found the value you looked for.
What is to be understood in it? You got it expressed using several units of measure, chose whichever fits to your needs or convert it to yet other ones. If you consequently use SI system the first units are easiest to follow.
 
What I don`t understand is:
what value is s?
what value is m?
what value is kg?

6.67384 * 10^-11 m^3 kg^-1 s^-2 = ??
 
They are the units. Your two objects have mass 200 and 50, hopefully that is kilograms, and if the distance of 150 is in metres, then your units are correct, and you can go ahead and use the number 6.67*10^(-11) for G (which will give you a force in Newtons). The units are an essential part of any physically meaningful equation.

Notice that it's a pretty small number, the mutual force between the masses will be pretty small unless you're dealing with planets or something!
 
You really do not know what second (s), meter (m), and kilogram (kg) are?
 
  • #10
Oh i see!
Thank you very much for the clarification :smile:
 
  • #11
xts said:
You really do not know what second (s), meter (m), and kilogram (kg) are?

We all were confused at some point in our lives, and we probably still all would be without patient teachers...
 

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