AL.PHYSICS Gravition

From the given, we have, at the moon, its distance is approximately 60 times the radius of the earth.
Therefore, from the relation
g = GM/r2 where
G = Gravitational constant
M = Earth mass
r = Earth radius
The earth's gravitational field strength at the moon is:
g' = GM/(60r)2
= g/3600
= 0.00278 m/s2

2008-10-25 20:17:40 補充：
A mistake from physics8801:
We can regard that the moon is a point, then counting from the earth centre to the moon, the distance is 60 times of the earth's radius.
And at the earth surface, its distance from the earth centre is the earth radius.
Therefore the distance ratio is 60:1

The answer from physics8801 is wrong...
Although counting from the earth surface, the ratio of the earth-moon distance to that of the radius of the earth is 59 : 1, whenever we are regarding g, we start from the the centre of the earth as the earth is regarded as the centre of the system.

2008-10-25 20:26:51 補充：
Hence, we should start from the centre of the earth, not from the surface of the earth. Hence, the answer from 飛天魏國大將軍張遼 is correct.

Let R be the radius of the earth.

Since g = GM/R^2
where M is the mass of the earth
G is the Universal Gravitational constant

At the distacne of the moon, the acceleration due to gravity becomes

g' = GM/(30x2R-R)^2 = GM/(59R)^2 = g/59^2 = 10/59^2 m/s2

2008-10-24 23:00:54 補充：
Bw aware that g is the acceleration due to gravity at the SURFACE of the earth, NOT that at the centre of the earth. The moon is at a distance of (60R-R) = 59R from the earth surface.