Going clockwise label the masses in a square. M1 top left corner, M2 top right corner, M3 bottom right corner, M4 bottom left corner.
Start with M1 and M2 use Newton's universal gravity formula;
Where G always equals 6.67*10^-11
Back to M1 and M2
Fg=1.04*10^-8 @ 0°
Same exact thing for M1 and M4 but instead of 0° it's 270°
For M1 and M3 it's a diagonal radius so to solve for the Diagonal radius make a right triangle and use the equation a^2+b^2=c^2 where c^2 is what we are finding.
Square root.72 To get the radius of .84
Now solve M1 and M3
Fg=5.31*10^-9 @ 315°
Now you have to add the vectors to get the vector of C^2 by making a right triangle connecting M1,M2,M3 making M2 where the 90° is
From M1 to M2 we already know it's 1.04*10^-8 but she still need to find the cos to find the other part of the line.
Y (aka M2 to M3) equals the same because 45 makes sin and cos the same
After finding that add
Getting; 1.41*10^-8 Remember both lines M1 to M2 and M2 to M3 are equal so they are both 1.41*10^-8
With that said to finalize this problem find C^2 now
(1.41*10^-8)^2 + (1.41*10^-8)^2 =C^2
square root 4.00*10^-16 and you should get 2.0*10^-8
ANSWER: 2.0*10^-8 @ 45°
I had been looking online for the answer to this question for 3 hours and this is the only process that I found was right THANK YOU
You're a blessing Tiffany :D You save me ^_^
This saved me! None of the other processes online were right but this one!
Thanks Tiffany! I wish my teacher could actually teach me this instead of puffing the devil' so lettuce...
Bless! Thank you.