Finding where the gravitational field strength is 0

AI Thread Summary
The discussion focuses on finding the point where gravitational field strength is zero using the equation g = (Gm)/r². A user attempts to set up the equation but encounters a quadratic in the denominator, complicating their solution. Suggestions include isolating the squared terms and considering the fraction of the distance from the center of mass. Clarification is provided on the use of plus or minus in the equation and the importance of correctly solving for the variable a. The conversation emphasizes simplifying the problem by expressing distances as fractions of the total distance r.
HornetHades
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1. Homework Statement

NDdr7pR.png

Homework Equations


g = (Gm)/r2

The Attempt at a Solution


I tried doing (Gm1)/a2 = (Gm2)/(r-a)2 where a is equal to the distance from the position we're trying to find to the center of the most massive object but end up getting a quadratic in the denominator on the right side. The answer in the back of the textbook is this:
rfsYdeU.png


Thank you!
 
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HornetHades said:
I tried doing (Gm1)/a2 = (Gm2)/(r-a)2
Would taking the square root of both sides help? First, you might want to get all of the "squared stuff" on one side of the equation and everything else on the other side.
 
TSny said:
Would taking the square root of both sides help? First, you might want to get all of the "squared stuff" on one side of the equation and everything else on the other side.
Thanks for your (very) prompt reply! Yeah that helps but now i have (r-a)/a±√(m1/m2). It's very close to the answer but I have no idea what to do with the a :/
 
HornetHades said:
Thanks for your (very) prompt reply! Yeah that helps but now i have (r-a)/a±√(m1/m2). It's very close to the answer but I have no idea what to do with the a :/
Did you leave out an equals sign somewhere in (r-a)/a±√(m1/m2) ?

Decide whether the + or the - should be used here.

Then, solve for ##a##.
 
Since you're solving for a fraction of the distance r, why not let a be that fraction so that 0 < a < 1. Then one distance will be ar and the other (1-a)r. Otherwise I think you're going to struggle more than necessary to get the distance differences into fractional form.
 
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