Distance - Finding Equilibrium Point for Net Force

AI Thread Summary
To determine the distance from Earth where an astronaut feels no net force due to the gravitational pulls of both the Earth and the Moon, a force analysis is necessary. The astronaut will experience no net force when the gravitational forces from the Earth and Moon are equal. The equations for gravitational force can be rearranged to express the relationship between the distances from the astronaut to both celestial bodies. By substituting the distance from the Moon to the astronaut as the total distance minus the distance from the Earth, the problem can be simplified. The final calculation yields a distance of approximately 3.46 x 10^8 meters from Earth for the astronaut to experience equilibrium.
ecthelion4
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Homework Statement



At what distance from the Earth should an astronaut be placed so that he will feel no net force when the Earth and the Moon and he are aligned?

Homework Equations



Fnet=ma
F=(Gm1m2)/d^2

The Attempt at a Solution



I assume I'm going to do a force annalysis, I'm just not sure how to. Also I thought that maybe if the pull from the moon to the astronaut and the pull from the Earth to the astronaut were equal, he should feel no force.
 
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Thats exactly right. Youve got the equations, youve figured it out. Just do the math.
 
I got a problem though. I have no way of knowing the distance from the moon to the astronaut. If the forces are equal, and I need to know the distance from the astronaut to the earth, then the resulting equation cleared for d1 (d1 being the distance from the Earth to the astronaut), ends up as this:

d1^2=(m2*d2^2)/m3

m2 being the mass of the earth, m3 being the mass of the moon and d2 being the distance from the astronaut to the moon. I got that equation clearing this in terms of d1 :

(G*m1*m2)/d1^2=(G*m1*m3)/d2^2

G is the gravitational constant and m1 is the mass of the astronaut.
 
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ecthelion4 said:
I have no way of knowing the distance from the moon to the astronaut.
Hint: What's d1 + d2 equal?
 
Chaos, he has the right equations however there's some things that aren't apparent.

Draw a picture first of all of this happening and some things will become apparent.
 
I did it :biggrin: All I needed was what Doc Al said. d2 would be the distance from the Earth to the moon minus d1, and since I had only one variable left, the rest was algebra. Thanks!
 
I'd solve it like this instead ->

( G*Mm*m ) / Rm^2 = (G*Me*m) / Re^2

Rm = Re * sqrt( Mm / Me );

Alpha = Rm + Re = 3.84x10^8 m.

Alpha = Re * sqrt(Mm / Me) + Re; Algebraically pull out the Re.

Alpha = Re * [ sqrt(Mm / Me) + 1 ]

Re = Alpha / [ sqrt(Mm / Me) + 1 ]

Mm = 7.35x10^22 Kg; Me = 5.98x10^24 Kg
Re = 3.84x10^8 / [ sqrt(7.35x10^22 / 5.98x10^24) + 1 ]

Re = 3.46x10^8 m
 
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