Distance - Finding Equilibrium Point for Net Force

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SUMMARY

The discussion focuses on calculating the distance from Earth at which an astronaut experiences no net gravitational force due to the Earth and the Moon being aligned. The key equations utilized include Newton's law of universal gravitation, represented as F=(Gm1m2)/d^2, and the net force equation Fnet=ma. The solution involves setting the gravitational forces from the Earth and Moon equal, leading to the derived formula Re = Alpha / [ sqrt(Mm / Me) + 1 ], where Mm is the mass of the Moon (7.35x10^22 Kg) and Me is the mass of the Earth (5.98x10^24 Kg). The final calculated distance Re is approximately 3.46x10^8 meters.

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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.
 
Last edited:
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
 
Last edited:

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