Calculate Where Earth & Moon's Gravitational Attraction Is Canceled

AI Thread Summary
To determine the point between Earth and the Moon where their gravitational attractions cancel, the masses of both celestial bodies and the distance between them are essential. The gravitational force equations for both Earth and the Moon can be set equal to each other, assuming a test mass of 1 kg. The user attempts to derive a quadratic equation to solve for the distance from Earth, but is unsure about the algebraic manipulation required to isolate the variable. They express concern about the complexity of the equation and seek clarification on the next steps in solving for the distance. The discussion highlights the need for careful algebraic handling to find the correct solution.
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Homework Statement


The mass of the Moon is 7.36 x 10 ^ 22 kg. At some point between Earth and the Moon, the force of Earth's gravitational attraction on an object is canceled by the Moon's force of gravitational attraction. If the distance between Earth and the Moon (center to center) is 3.84 x 10 ^ 5 km, calculate where this will occur, relative to Earth.


Homework Equations



Fg = Gm1m2 / r ^ 2



The Attempt at a Solution



Given:
Mass of Earth = 5.98 x 10 ^ 24 kg
Mass of Moon = 7.36 x 10 ^ 22 kg
Radius between Earth and Moon = 3.84 x 10 ^ 5 km = 3.84 x 10 ^ 8 m

So at this point, Fgmoon = Fgearth, but I need another object to calculate the point.
So assuming the object is 1kg. Fg between Moon and object = Fg between Earth and object.
I need to find the r between Earth and object. This distance is the entire distance between Earth and Moon minus distance between object and Moon.

Gm(earth)m(test) / (distance between moon and sun - r) ^ 2 = Gm(test)m(moon) / r ^ 2

I don't know either r.
I'm not sure how to find r with this equation.
 
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The two r's are identical so you have only one unknown.
The form of your equation is like a/b = c/d.
Your first step should be to cross multiply to ad = bc, which eliminates all fractions.
Cancel what you can. Express the quadratic equation in standard form
ar^2 + br + c = 0
Use the quadratic solution formula to find r.
 


Ok, so I have:
(G)[m(earth)][m(test)](r ^ 2) = (G)[m(test)][m(moon)][(distance between moon and sun - r) ^ 2]
(9.8N/kg)(5.98 x 10 ^ 24 kg)(1kg)(r^2) = (9.8N/kg)(1kg)(7.36 x 10 ^ 22 kg)(3.84 x 10 ^ 8 m)

Wait, so the (G) and the [m(test)] can cancel out yes?
So then,

[m(earth)](r ^ 2) = [m(moon)][(distance between moon and sun - r) ^ 2]
(5.98 x 10^24kg)(r^2) = (7.36 x 10^22kg)(3.84x10^8m - r)^2
(5.98 x 10^24kg)(r^2) = (7.36 x 10^22kg)(14.75x10^16m^2 - 7.68x10^8mr - r^2)
Is this correct?
Can I now just take (14.75x10^16m^2 - 7.68x10^8mr - r^2) to find r or do I need to solve the rest of the equation? Sorry my algebra is very bad.
 
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