Gravitational Attraction Question

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Homework Help Overview

The discussion revolves around gravitational attraction between the Earth and the Moon, specifically focusing on finding the point where the gravitational forces from both bodies cancel each other out. The problem involves calculations related to gravitational force equations and the distances involved in the Earth-Moon system.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the gravitational forces acting on an arbitrary mass located between the Earth and the Moon. They discuss the need to express these forces symbolically before substituting numerical values.

Discussion Status

The conversation has progressed through various attempts to set up the equations for gravitational forces. Participants are encouraged to keep their expressions in symbolic form and have begun to derive relationships between the distances and masses involved. There is ongoing clarification regarding the proper setup of the equations.

Contextual Notes

Participants are reminded to avoid numerical answers until the symbolic equations are fully established. There is an emphasis on correctly identifying variables and maintaining clarity in the equations as they relate to the gravitational forces from both the Earth and the Moon.

  • #31
the { } is square root
 
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  • #32
wilson_chem90 said:
the { } is square root
Ah, okay. Then yes, this is correct:
wilson_chem90 said:
alright so the new equation should be X/{Me} = D-X/{Mm}
Next is to solve that equation for X.
 
  • #33
i got to X = D-X{Me} / {Mm}
 
  • #34
wilson_chem90 said:
i got to X = D-X{Me} / {Mm}
OK, write that as:
X = (D-X) {Me/Mm} (using {} for square root)

Now solve for X. (Get all the X terms on the same side.)
 
  • #35
so would it just be X+X = D {Me/Mm}
 
  • #36
wilson_chem90 said:
so would it just be X+X = D {Me/Mm}
No. Realize that:

X = (D-X) {Me/Mm} →

X = D{Me/Mm} - X{Me/Mm}
 
  • #37
oh okay, yeah i remember that now, is that the final equation though?
 
  • #38
What do you mean by "final"? You still have to solve for X.
 
  • #39
well i have to isolate the other x right, but the only think i can think of is it would be X^2 = D{Me/Mm} - {Me/Mm}
 
  • #40
wilson_chem90 said:
well i have to isolate the other x right, but the only think i can think of is it would be X^2 = D{Me/Mm} - {Me/Mm}
No. You must isolate X, but you don't end up with X^2.

How would you solve this one?

x = a -bx (a and b are just constants)

It's the same basic equation.
 
  • #41
would it be x = a/b?
 
  • #42
wilson_chem90 said:
would it be x = a/b?
No. Do it step by step:

x = a - bx

group the x terms on the left:
x + bx = a

factor out the x:
x (1 + b) = a

isolate the x by dividing both sides by (1 + b):
x = a/(1 + b)

done!
 
  • #43
YES! finally, thank you so much. i appreciate your time and patience with me
 

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