Calculate Magnitude & Direction of Gravitational Field at 4th Corner

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SUMMARY

The discussion focuses on calculating the gravitational field's magnitude and direction at the fourth corner of a square formed by three equal masses located at the other corners. The gravitational field equations used are g = Gm/r², with specific calculations for the gravitational contributions from each mass. The user initially miscalculated the x and y components of the gravitational field, particularly the sine and cosine values for 45 degrees. The correct approach involves using sin(45) = cos(45) = 1/√2 to accurately determine the resultant gravitational field.

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  • Understanding of gravitational field equations, specifically g = Gm/r²
  • Familiarity with vector components in physics
  • Knowledge of trigonometric functions, particularly sine and cosine
  • Basic concepts of gravitational interactions between masses
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Students studying physics, particularly those focusing on gravitational fields and vector analysis, as well as educators looking for problem-solving strategies in gravitational interactions.

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1. Three objects of equal mass are located at three corners of a square of edge length d. Find the magnitude and direction of the gravitational field at the fourth corner due to these objects. (Use the following as necessary: G, d, and m.)

2. Homework Equations :

g = Gm/r2

3. My attempt at the solution: I labeled the masses as 1, 2, and 3, where 1 and 3 are adjacent to the fourth corner and 2 is in the opposite corner from the fourth.

g1 = g3 = Gm/d2
g2 = Gm/(d[tex]\sqrt{}2[/tex]2)

gy = g1 + g2y

= Gm/d2 + g2 sin 45

= Gm/d2 + Gm/(d[tex]\sqrt{}2[/tex])2

gx = g3 + g2x

= Gm/d2 + g2 cos 45

= Gm/d2 + Gm(d[tex]\sqrt{}2[/tex])2

g2 = gx2 + gy2

g = [tex]\sqrt{}(2Gm/d<sup>2 + Gm/d^2 cos 45</sup>[/SUP])[/tex]
 
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your calculations for the x and y components of g are wrong. use the correct values of
sin and cos

[tex]\sin(45)=\cos(45)=\frac{1}{\sqrt{2}}[/tex]

reworkout that
 

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