where is the gravity neither - nor + between the earth and the moon?

Answer 1

I think you might be looking for the Earth-Moon Lagrange point, L1, which is about 323,110 km from the Earth, or 84% of the way to the Moon.

Answer 2

Since gravity is affected by mass over separation squared, there must be a place between the Earth and the moon where the forces of gravity from both are the same and pointing in opposite directions. It would be somewhere between an imaginary line drawn from the center of the earth to the center of the moon.

Answer 3

It's actually negative - the moon is drifting away a few inches each year.

It has to do with gravity wells - the earth makes a big dent in spacetime. This dent is called a gravity well. The moon is rotating around the earth at just the right speed (give or take a little bit) that it doesn't fall INTO the gravity well, and it doesn't fly OUT of the gravity well either. It just spins around and around, happy as a clam.

A very big clam.

In space.

Made of rock.

Answer 4

The centre of gravity between the two is actually inside the Earth some 1700km below the surface. (the barycentre).

The Earth is 81.3 times more massive than the Moon, that's why.

Answer 5

The java planet calculator (: http://www.a2dvoices.com/realitycheck/Coriolis/ lists) explains and shows the distances for L1-L5 earth-moon lagrangian locations.

Note: You need to select the "Earth Orbit" tab on the java tool and then select the "L1/L2 Earth-Moon Lagrangian" or "L3/L4/L5 Earth-Moon Lagrangian".

You will also find the earth-sun Lagrangian locations.

Answer 6

there is, but when a shuttle or what not is in space, it is in orbit, basically a constant state of falling around the planet. so the astronauts are falling with the craft, creating zero gravity.

Answer 7

The force of gravity is given by

F = -GMm/r^2 where G = Newton's constant, M = mass of central body, m = mass of "test" body, r = distance between centers of masses.

Now the gravitational force of the moon must be equal and opposite that of earth's for theret o be no net gravitational force:

Mem/(r^2)em = Mlm(r^2)lm where Me = mass of earth, Ml = mass of moon, (r^2)em = distance squared between earth and mass, and (r^2)lm = distance squared between moon and mass.

Let R = distance between center of earth and moon. Then,

rlm = R - rem = R -r where I can drop the "em" now.

SO Me/r^2 = Ml/(R-r)^2

r^2 = Me/Ml (R-r)^2 = (Me/Ml) (R^2 -2Rr + r^2)

0 = (Me/Ml - 1)r^2 - 2(Me/Ml) R r + (Me/Ml)R^2

Now you need to use the quadratic formula to find r.

r = (Me/Ml) R +/- sqrt{ (Me/Ml)^2 R^2 - R^2(Me/Ml-1)(Me/Ml)}

= (Me/Ml) R +/- sqrt{ R^2(Me/Ml)} = R (Me/Ml) {1 +/- 1/sqrt(Me/Ml)}

You get two solutions: One lies between earth and the moon, the other lies to the outside of the two bodies. I'll let you plug in the constants and do the arithmetic.