# Calculating Gravitational Balance Point between Earth & Moon

• slybuster
In summary, the conversation discusses the calculation of the point where the force of gravitational attraction between the Earth and the Moon cancels out. The equation used is FeG=FmG, and it yields two solutions: one between the Earth and the Moon and the other beyond the Moon. The first solution is the Lagrange point, where the forces cancel out, while the second solution corresponds to the point beyond the Moon where the forces are in the same direction. The conversation also mentions the use of a quadratic equation in solving for the distances and the difficulty in isolating the variable. Some clarification on the concept of Lagrange points is also provided.
slybuster

## Homework Statement

The mass of the Moon is 7.35*10^22kg. At some point between the Earth and the Moon, the force of gravitational attraction cancels out. Calculate where this will occur, relative to Earth.

Me = 5.98*10^24 kg
Mm = 7.35*10^22 kg
D = 3.84*10^8 m
g = 6.67*10^-11 Nm^2/kg^2
**Mx(object) = 1 kg

## Homework Equations

r(moon) + r(earth) = D
r(earth) = D-r(moon)

FeG = FmG
G(MeMx)/r^2 = G(MmMx)/(D-r)^2 *I had some trouble deciding where to put the (D-r)

## The Attempt at a Solution

r^2(Me-Mm) - 2DMe*r +MeD^2 = 0

Plugging in the numbers, I get:
r = 4.319*10^8 m or r = 3.46*10^8 m (this appears to be the correct answer).

I researched this question and found the equation:

r = (D[Me-(MeMm)^1/2]) / (Me-Mm)

I was wondering how to isolate r and simplify the equation in order to get it to look like the one above. I am having trouble understanding where the 'Mm' term in the '(MeMm)^1/2' came from; also, does the '^1/2' come from taking the square root or does it have something to do with the '2' in the initial equation being taken out? Aside, what is a suitable explanation regarding the existence of the fallacious answer in the initial unsimplified quadratic?

Thank you very much, I really appreciate the help.

Shown symbolically only,

D is distance between centers of Earth and moon;
re is NOT radius but distance from 1 kg mass to center of earth;
rm is NOT radius but distance from 1 kg mass to center of moon;
D=re+rm
Mx=1, the 1 kg mass between Earth and moon;

Forces between the mass and either large body are equal will be equal at some re and rm

$$\frac{{GM_e M_x }}{{r_e^2 }} = \frac{{GM_m M_x }}{{r_m^2 }},\quad \quad D = r_e + r_m {\rm }$$
Solve for either re or rm .

Your quadratic equation gives two distances from the Earth where the forces are equal: one distance shorter, the other distance longer that the distance between Moon and Earth. There are tow points really where both the Earth and the Moon attracts a body with equal forces. But you were asked about the point in between, so choose the smaller root of your quadratic equation.

ehild

While researching, I came across the concept of "Lagrange Points." While the answer, which I labeled correct, of the quadratic conforms to one of these (i.e. the point between the Earth and the Moon) the other answer seems to correspond to the point beyond the moon. Is this correct? Simple if true I guess, I should have seen it.

Also, I have been out of the math game for quite some time...just coming back to it after several years. The main problem I'm having is changing the quadratic into the isolated variable equation I presented. The book I saw it in does not do this. I've tried taking derivatives as well as various algebraic manipulations (more than a few of which I'm sure are incorrect) but am unable to figure out how it was done. Could someone perhaps point me toward a specific online source (or even a printed one I can order on amazon) that will detail such operations or point me in the right direction (with specific attention to those areas I mention...i.e. the (MeMm)^1/2 section).

Thank you so much for your explanation ehild, it was just enough to point me in the right direction while not spoiling the benefit of thinking out the problem for myself. Thank you.

No, Lagrange points are different. You found the two places, on a straight line through the two centres, where the magnitudes of the forces are equal. At one of those points the forces cancel, but at the other they add.
At a Lagrange point the forces do not cancel. Instead, the (vector) sum of the forces is the same as would be experienced if the small satellite were to replace the smaller of the two other bodies. That is, it is a point where the satellite can orbit the larger of the other two with the same period as the smaller. Given any two bodies in orbit around each other, there are five Lagrange points: three on the line through their centres (one between them, one each beyond them), one 60 degrees ahead in orbit and one 60 degrees behind. At the Lagrange point between Earth and moon, the net gravity is towards the Earth; the gravity cancellation point would be closer to the moon.

Returning to your question, it seems to be just a matter of solving quadratic equations. I suggest you Google that.

slybuster said:
While researching, I came across the concept of "Lagrange Points." While the answer, which I labeled correct, of the quadratic conforms to one of these (i.e. the point between the Earth and the Moon) the other answer seems to correspond to the point beyond the moon. Is this correct? Simple if true I guess, I should have seen it.
Your equation "FeG = FmG" means forces equal in magnitude, and says nothing about the direction. And there are two points along the line connecting Earth and Moon, where the forces are of the same magnitude. Between Earth and Moon, these forces are opposite, so they cancel. A the point beyond the Moon, they point in the same direction, toward the Earth.
slybuster said:
Also, I have been out of the math game for quite some time...just coming back to it after several years. The main problem I'm having is changing the quadratic into the isolated variable equation I presented. The book I saw it in does not do this. I've tried taking derivatives as well as various algebraic manipulations (more than a few of which I'm sure are incorrect) but am unable to figure out how it was done. Could someone perhaps point me toward a specific online source (or even a printed one I can order on amazon) that will detail such operations or point me in the right direction (with specific attention to those areas I mention...i.e. the (MeMm)^1/2 section).

The URL below explains the quadratic formula: The solution of a quadratic equation. You can use the formula also with symbols.

The general form of the quadratic equation is ax2+bx+c=0
Your equation is : r^2(Me-Mm) - 2DMe*r +MeD^2 = 0
a=Me-Mm b=-2DMe c=MeD2.

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

is applied to your equation, giving

$$r=\frac{2DM_e\pm\sqrt{(2DM_e)^2-4(M_e-M_m)M_eD^2}}{2(M_e-M_m)}$$

Expand the parentheses and simplify the expression under the square root:

$$\sqrt{(2DM_e)^2-4(M_e-M_m)M_eD^2}=\sqrt{4DM_e^2-4DM_e^2+4M_mM_eD^2}=\sqrt{4D^2M_eM_m}$$
You can factorize the square root:
$$\sqrt{4D^2M_eM_m}=\sqrt{4D^2}\sqrt{M_eM_m}=2D\sqrt{M_eM_m}$$

I think you can proceed from here.

ehild

Ha...it never donned on me to put it into a quadratic and then simplify it. Thank's a lot...I look forward to playing with it and manipulating it around.

You're truly great--I really appreciate it. Thanks again ehild... ****stars

You are welcome. And thank you for the compliment.

ehild

## What is the gravitational balance point between Earth and Moon?

The gravitational balance point between Earth and Moon, also known as the barycenter, is the point at which the gravitational forces of both objects are equal. It is the point around which they both orbit.

## How is the gravitational balance point calculated?

The gravitational balance point is calculated using the masses and distances of Earth and Moon from each other. This can be done using Newton's law of universal gravitation or the formula for finding the center of mass.

## Where is the gravitational balance point located?

The gravitational balance point is located between Earth and Moon, but closer to Earth due to its larger mass. It is approximately 4,671 km from the center of Earth and 384,400 km from the center of Moon.

## Does the gravitational balance point ever change?

Yes, the gravitational balance point between Earth and Moon is constantly changing as their positions and distances from each other change. It also changes as the Moon's orbit around Earth is affected by other celestial bodies.

## Why is the gravitational balance point important to study?

The gravitational balance point is important to study because it helps us understand the dynamics of the Earth-Moon system. It also has practical applications, such as in spacecraft navigation and mission planning.

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