When is the force of gravity 0?

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Homework Help Overview

The discussion revolves around determining when the force of gravity is zero in the context of gravitational interactions between the Earth and the Moon. Participants explore the implications of gravitational equations and the conditions under which gravitational forces may balance.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the concept of gravitational equilibrium and whether to set gravitational forces or accelerations to zero. There are attempts to derive relationships between gravitational forces from the Earth and the Moon.

Discussion Status

The discussion is active, with various interpretations being explored. Some participants suggest methods for setting up equations based on gravitational forces, while others question the assumptions regarding distances and the nature of the forces involved.

Contextual Notes

There are mentions of specific gravitational values and distances, as well as concerns about the validity of certain assumptions, such as the relationship between the gravitational pull of the Earth and the Moon at a given distance.

Alameen Damer
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Homework Statement


upload_2015-9-6_0-24-11.png


Homework Equations


Fg=Gm1m2/r^2
g=Gm/r^2

The Attempt at a Solution


I am having trouble finding a way to start. I can't set the g to 0 because that would not let me solve for radius.
 
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The net Gravity is zero when the two forces are in equilibrium. Set gravity from Earth equal to gravity from moon and solve for r.
 
Do i make it so the forces of gravity equal to 0 or the accelerations of gravity equal to 0
 
Alameen Damer said:
Do i make it so the forces of gravity equal to 0 or the accelerations of gravity equal to 0
It's late, and I'm not overthinking this but I do believe that the answer is: yes!

F=ma, but you will find that m will drop out of the equation (since it does not matter how massive the rocket is).
 
Ok so the forces of gravity must cancel out:

FgE=FgM

9.8m=gMm

gM must equal 9.8

9.8=Gm1/r^2
r=root(Gm1/9.8)
r=709685

Shouldn't that be the answer? As at that distance the g of the moon equals 9.8?
 
Alameen Damer said:
Ok so the forces of gravity must cancel out:

FgE=FgM

9.8m=gMm

gM must equal 9.8

9.8=Gm1/r^2
r=root(Gm1/9.8)
r=709685

Shouldn't that be the answer? As at that distance the g of the moon equals 9.8?
But what will the acceleration due to Earth's gravity be there?
(Besides, that's less than the moon's radius.)
 
If I were going to solve this problem then I'd start by drawing a diagram with the Earth and the Moon separated by the orbital radius of the moon that is given in the question. Next, I'd mark a point somewhere between the Earth and the Moon were the pull towards the Earth is the same as the pull towards the moon. We don't know where this point is yet, but we know that it is some distance r1 from the Earth and some other distance r2 from the Moon.

Then I'd pretend that I was putting a small mass, m, at that point. We already have a formula to calculate the force on the small mass due to the Earth. The same formula works for the force pulling towards the Moon. All we have to do is write down an equation (don't put in any numbers yet!) setting these two forces equal to each other. You might notice that some terms can be canceled out.

We can also write down another equation. What would you get if you added together r1 and r2?

Since we have two equations you might want to think about how we can now solve for the two unknown quantities r1 and r2. If you can remember solving simultaneous equations in a maths class then this will be useful.
 
If you let R be the distance between the centers of the Earth and moon, then the only unknown for distance is r, the distance from Earth's center. The distance from the moon's center is necessarily R-r.
 

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