Incorporating Inverse Square Law In Gravity

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Discussion Overview

The discussion revolves around the application of the inverse square law in the context of gravitational force calculations. Participants explore the relationship between distance and gravitational force, particularly how changes in distance affect the force experienced by an object due to gravity. The conversation includes mathematical reasoning and conceptual clarifications regarding the inverse square law.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses their understanding of the inverse square law, suggesting that if a number increases, the other number is the square of that increase but in a negative sense.
  • Another participant attempts to clarify the inverse square law by explaining that if the distance between two objects doubles, the gravitational force is reduced to a quarter of its original value, emphasizing the relationship between distance and force.
  • Participants present calculations for gravitational force on a 1kg object, questioning how to adjust the radius in the gravitational force equation when the distance is increased by 10 meters.
  • One participant points out that the calculations presented are correct but highlights a misunderstanding regarding the units used, noting that the weight of a 1kg object is approximately 9.8 N, indicating a significant error in the initial calculations due to unit conversion issues.
  • Another participant reinforces the importance of using correct units and clarifies that the height difference should be considered in meters, not kilometers.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial calculations and interpretations of the inverse square law. There are multiple viewpoints regarding the correct application of units and the implications of the inverse square law in gravitational force calculations.

Contextual Notes

There are limitations regarding the assumptions made about unit conversions, particularly between kilometers and meters, which affect the calculations presented. The discussion also reflects varying levels of understanding among participants, particularly regarding the mathematical representation of the inverse square law.

Leoragon
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First of all, I'm 13 so I might not comprehend the complex vocabulary or symbols others might use. Second, I just joined!

Okay, let's get to it.

I think I know what the inverse square law is: if a number goes up by x, then the other number is the square of x but in the negative side. Right?

I also think that the force of gravity on a 1kg object on the surface of the Earth (from the average radius) is 9,818,373.084 N . Now what if the distance gets increased by 10 metres? In the equation for the force: F=Gm1m2/r2, do you just add 10 to the r? So it would equal 9,787,623.422 N?

1st equation)
F = 9,818,373.084 N
G = 6.673 X 10-11
m1= 1kg
m2= 5.97219 X 1024 (Earth)
r2 = 40,589,641 (63712)

2nd equation)
F = 9,787,623.422 N
G = 6.673 X 10-11
m1 = 1kg
m2 = 5.97219 X 1024 (Earth)
r2 = 40,717,161 (63812)

If so, where does the inverse square law go?

And is this correct?
 
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Leoragon said:
First of all, I'm 13 so I might not comprehend the complex vocabulary or symbols others might use. Second, I just joined!

Okay, let's get to it.

I think I know what the inverse square law is: if a number goes up by x, then the other number is the square of x but in the negative side. Right?

I'm not sure what you mean by "in the negative side", so let me try to explain it for you in words.

Suppose I said the magnitude of some force between two objects increases linearly with separation. That means that if I double the distance between the two objects, the force between the two objects doubles. That is, the factor by which the force increased is the same as the factor by which the separation increased.

Now suppose I said that the magnitude of some force between two objects decreases inversely with the separation. This means that if I double the distance between the two objects, the force between the two objects is cut in half. The factor by which the force changed is 1 over the factor by which the separation changed.

Now suppose I said that the magnitude of some force between two objects increases as the square of the separation. Then if I double the separation between the objects, the force between them is quadrupoled (four times): the factor by which the force increases is the square of the factor by which the distance increased.

If we put these three observations together, you can conclude that if the force between two objects follows and inverse square relationship, then if we double the separation between two objects, the force between them is reduced by 1/4: the factor by which the force is decreased is 1 over the factor by which the separation increased.

Does this make it clearer?

I also think that the force of gravity on a 1kg object on the surface of the Earth (from the average radius) is 9,818,373.084 N . Now what if the distance gets increased by 10 metres? In the equation for the force: F=Gm1m2/r2, do you just add 10 to the r? So it would equal 9,787,623.422 N?

1st equation)
F = 9,818,373.084 N
G = 6.673 X 10-11
m1= 1kg
m2= 5.97219 X 1024 (Earth)
r2 = 40,589,641 (63712)

2nd equation)
F = 9,787,623.422 N
G = 6.673 X 10-11
m1 = 1kg
m2 = 5.97219 X 1024 (Earth)
r2 = 40,717,161 (63812)

If so, where does the inverse square law go?

And is this correct?

The calculations look correct. The inverse square law is the 1/r2 part of the equation for the force: F=Gm1m2/r2.

If the force of gravity were inversely proportional to the separation, then the equation would involve 1/r instead of 1/r2.
 
Leoragon said:
First of all, I'm 13 so I might not comprehend the complex vocabulary or symbols others might use. Second, I just joined!

Okay, let's get to it.

I think I know what the inverse square law is: if a number goes up by x, then the other number is the square of x but in the negative side. Right?

I also think that the force of gravity on a 1kg object on the surface of the Earth (from the average radius) is 9,818,373.084 N . Now what if the distance gets increased by 10 metres? In the equation for the force: F=Gm1m2/r2, do you just add 10 to the r? So it would equal 9,787,623.422 N?

1st equation)
F = 9,818,373.084 N
G = 6.673 X 10-11
m1= 1kg
m2= 5.97219 X 1024 (Earth)
r2 = 40,589,641 (63712)

2nd equation)
F = 9,787,623.422 N
G = 6.673 X 10-11
m1 = 1kg
m2 = 5.97219 X 1024 (Earth)
r2 = 40,717,161 (63812)

If so, where does the inverse square law go?

And is this correct?

It is a good idea to use (and keep track of) units.
The weight of a 1 kg object on the surface of the Earth is around 9.8 N.
You are off by 6 orders of magnitude. The reason is using the radius in km whereas G is in standard SI units (N*m^2/kg^2).

For the meaning of the inverse square law, Mute has explained it already.
 
The same km<->m-issue occurs at the height difference: It is 10m (6371.000 -> 6371.010), not 10km (6371->6381).
 

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