Finding where the gravitational field strength is 0

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Homework Help Overview

The discussion revolves around finding the point where the gravitational field strength is zero, involving gravitational equations and distances from two masses.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss setting up the gravitational equations and manipulating them to find a solution. There are attempts to simplify the equation by isolating squared terms and considering the implications of taking square roots. Questions arise about the correct use of signs and the representation of distances as fractions.

Discussion Status

Participants are actively engaging with the problem, offering suggestions for manipulating the equations and exploring different representations of the variables involved. There is a focus on clarifying the setup and ensuring the mathematical expressions are correctly formed, but no consensus on a final approach has been reached.

Contextual Notes

There is mention of textbook answers and the constraints of the problem, including the need to express distances as fractions of a total distance.

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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