earth moon relationship?

Question

Someone said the pull of gravity between the earth and the moon balance out at 90% of the distance to the moon. I know the force of gravity is about 1/6 that of the earth, why isn't the balance point at about 83.5% of the distance?

Answer 1

I just answered essentially this question elsewhere. I'll repeat that answer, below.

But first, what's wrong with your own answer? What you are quoting are the SURFACE GRAVITIES (only) on the Earth or on the Moon. But both gravitational attractions DECREASE as the INVERSE SQUARE OF THE DISTANCE away from each particular body. Your argument doesn't allow for the effects of such decreases, which will be different at some arbitrary intermediate point, so it simply can't be correct.

The scientific result and reasoning now follow:

In quite close terms, the answer is at a point 90% of the distance to the Moon. So, at that point you're 90% of the full Earth-Moon distance away from the Earth, and 10% of it away from the Moon.

The reason for this is that gravitational attraction towards any one body is proportional to M / D^2. (We can ignore G, the Gravitational Constant, which is common to both attractions in this problem.) So, to find the "balance point" where the two attractions of the Earth and the Moon precisely cancel one another out, you need to solve:

M_e / (D_e)^2 = M_m / (D^m)^2 or (D_e / D_m)^2 = M_e / M_m,

where suffixes 'e' and 'm' stand for Earth and Moon respectively.

Now, extremely conveniently, the Earth is very closely 81 times as massive as the Moon, and 81 is a perfect square! So, clearly,

D_e / D_m is very closely 9.

How convenient in a decimal system!: the distance from the Earth to the balance point, D_e, is therefore ~ 90% of the full Earth-Moon distance, and the distance to the Moon, D_m is only ~ 10% of that full distance.

(More accurately, Earth is ~ 81.3 times more massive than the Moon. Sqrt(81.3) is ~ 9.02. So the fractions are ~ 9.02/10.02 and ~1.00/10.02; to 3 sig. figs., the answers found above are almost unaltered: 90.0% and 9.98%. [They don't quite add up to 100% because of the next, but "insignificant" figure in the first one making it more like 90.02%.] )

Live long and prosper.

POSTSCRIPT: Please note that scythian1950's suggested links refer to a DIFFERENT POINT, the "Lagrangian point L1."

There's some confusion about this. L1 is not simply where the Earth's and the Moon's two individual pulls of gravity "balance out" (which is the question that you asked), but rather where the combined earth and moon gravities PLUS a "centrifugal force term" due to the combined Earth-Moon orbital rotation rate all has a stationary value. Because of the "centrifugal force term," this L1 point ISN'T at the same place as the (mere) "equal but oppositely directed gravities" point.

"Lagrangian points" (there are 5 for each pair of main attracting bodies) are more important in practice than the mere "gravitational balancing points" because small motions in their vicinity can be stable. That's why L4 and L5 (making equilateral triangles with the positions of the two main bodies, one preceding and the other trailing in orbit) are sometimes spoken of as possible places to set up stably orbiting space stations or colonies. Indeed, Jupiter is "accompanied" in its orbit around the Sun --- if at the very distant L4 and L5 points that it possesses mutually with the Sun --- by families of objects called the "Trojan asteroids" making such stably librating captured orbits.)

Answer 2

Maybe you're better than you think? Check the Space.com article, which gives a figure of about 84% of the distance. Also another link to some calcs which give the same figure. Give yourself points for this one.

Addendum: Dr Spock has given a good and lengthy response to this question, but it does raise the question, "How did the Asker arrive at the 83.5% figure, without understanding it's for Lagrange point L1?" Now, I'l have to go back and figure out how that came about.

Answer 3

Gravity is a property of mass. The Moon's mass is 1/81 that of the Earth. Expressed as a decimal, this is 0.0123456, or 1.23456% of Earth's mass. The reference to 1/6 of Earth's gravity has to do with relative surface gravity escape velocity, which is 1.2 m/s2 on the Moon and 7 m/s2 on Earth.
(and Jerry, you are quoting center of MASS not center of gravity, of the earth-moon system)

Answer 4

Center of mass in a 2-body system is given by:

m1*d1= m2*d2

where m1 and m2 are the masses of the bodies, and d1 and d2 are the distances of thje bodies from the center of mass.

Of course, d1 + d2 = R, the distance between the centers of mass of each body.

So, the mass of the moon is 1/81 that of the earth. That puts the center of gravity between the bodies *inside* the sphere of the earth!.

Answer 5

aside from the tides that help marine existence, and so on., the moon additionally makes existence on earth available via giving flora and animals friendly climate. The presence of the Earth`s satellite tv for pc stabilizes our planet`s wobble. This has helped the planet to have a much greater stable climate over billions of years, that would have affected the form and enhance of existence on earth.

Answer 6

I honestly don't know.
The answer may be contained in your "about". Or:
Also, gravity is not linear. Depends on the square of the distance.

Answer 7

Beats me.

Answer 8

Good question.

Answer 9

perhaps this 'someone' was wrong?

Answer 10

wo0ow--eyYy!!