Where Does the Moon's Gravitational Pull Overtake Earth's?

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Homework Help Overview

The problem involves determining the distance from the center of the Earth at which the Moon's gravitational pull surpasses that of the Earth. It is situated within the context of gravitational forces and celestial mechanics, specifically relating to the Apollo missions and the Earth-Moon system.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting gravitational forces equal to each other to find the point where the Moon's pull overtakes the Earth's. There are attempts to derive an inequality and manipulate gravitational equations, with some participants questioning the relationships between the distances involved.

Discussion Status

The discussion has seen various approaches to setting up the problem, including the use of gravitational equations and inequalities. One participant indicates they have arrived at a correct answer, suggesting some productive guidance has been exchanged, though the details of the solution are not disclosed.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. The relationship between the distances from the Earth and Moon is a key point of discussion.

Andy111
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Homework Statement



On the way to the moon, the Apollo astronauts reach a point where the Moon's gravitational pull is stronger than that of the Earth's.
Find the distance of this point from the center of the Earth. The masses of the Earth and the Moon are 5.98e24 Kg and 7.36e22 Kg, respectively, and the distance from the Earth to the Moon is 3.84e8 m.

Answer in units of m.

Homework Equations



F=G(m1)(m2)/r(squared)

Where F is force of gravity, G is gravitational constant, m1 is mass of one object, and m2 is mass of second object. And r(squared) is the radius from center to center.

F(C)=mv(squared)/r

where F(c) is centripetal force, m is mass of an object (in Kg), v(squared) is the linear speed of an object (in m/s), and r is the radius from the center of the object being orbited around.

g=Gm1/r(squared)

g is acceleration due to gravity, G is gravitational constant,m1 is mass of object causing gravity, and r(squared) is radius from center to center.

V(squared)=Gm1/R

V(squared) is linear speed, G is gravitational constant, m1 is object causing gravity, and r is radius.

The Attempt at a Solution



I tried using the equation g=Gm1/r(squared) for each the Earth and the moon, and setting them equal to each other, but that didn't really work out.
 
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Setup an inequality

[tex]\frac{G(m_{moon})(m_{apollo})}{r_{moon to apollo}^2} > \frac{G(m_{earth})(m_{apollo})}{r_{earth to apollo}^2}[/tex]

Stuff cancels out and you're left with:

[tex]\frac{m_{moon}}{r_{moon to apollo}^2} > \frac{m_{earth}}{r_{earth to apollo}^2}[/tex]

We know that the distances are related because the Earth and the moon are a set distance apart and the sum of the distance between the astronauts and the moon and Earth must be the same as the distance between the Earth and moon. If you find when the two are equal you when when the moons gravitational pull will be greater
 
Last edited:
If x is the distance from the Earth at which the gravitational pull due the Earth and moon are equal then GMe/x^2 = GMm/(3.84e8 - x)^2. Now solve for x.
 
Thankyou for the help. I got the answer now, and it was correct.
 

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