Please gravitate to my question

[streamline]

If one is walking on the moon, does Earth's presence come into play regarding the pull of the moon's gravity on the walker? That is, does Earth have any effect gravitationally-speaking on objects on its own moon? Are such effects, if any, similar to those that the moon induces on Earth (eg, tides)? What about regarding any/all other celestial bodies further away -- planets, sun..?

Thanks!!!

[Loren Pechtel]

Originally posted by streamline
If one is walking on the moon, does Earth's presence come into play regarding the pull of the moon's gravity on the walker? That is, does Earth have any effect gravitationally-speaking on objects on its own moon? Are such effects, if any, similar to those that the moon induces on Earth (eg, tides)? What about regarding any/all other celestial bodies further away -- planets, sun..?

Thanks!!!

Earth makes *FAR* more tide on the moon than the moon makes on the Earth. That's why the moon only shows one face to us--the tide stopped it.

Tide drops off at the third power of distance so the various planets have basically negligible tidal effects on each other. They do affect each other's orbits to some degree (example: The resonance between Neptune and Pluto.)

[caravelair]

however, while tidal forces certainly effect the moon itself, they would not really effect someone walking on the moon. this is because tidal forces result from a difference in gravitational pull from one side of a body to the other. one side of the moon is about 3.5x10^6 metres farther from the earth than the other, and since gravity drops off proportionally to 1/r^2, the far side feels much less gravitational pull from the earth than the close side does. however, with someone walking on the moon, their head is only going to be at most about 2 meters closer to the earth than their feet. gravity will not drop off very much over such a short distance, so the person will not feel a tidal force from the earth. the person will feel a gravitational pull from the earth, but at the surface of the moon, the earth's gravity is very small in comparison the the gravity of the moon.

specifically, we have g = GM/r^2, for the gravitational acceleration from a body of mass M at a distance r from the center of that body.

G is a constant who's value is about 6.7x10^-11(m^3)/(kg)(s^2)
the earth's mass is about 6x10^24kg.
the moon's mass is about 7x10^22kg.
the moon's radius is about 1700km = 1.7x10^6m
the earth-moon distance is about 385,000km = 3.85x10^8m

so at the surface of the moon, the gravitational acceleration from the moon is:

g1 = G(7x10^22)/(1.7x10^6)^2 (m/s^2)

which equals about 1.6 (m/s^2).

and at the moon's surface, the gravitational acceleration from the earth is:

g2 = G(6x10^24)/(3.85x10^8)^2 (m/s^2)

which equals about 0.0027 (m/s^2)

g1/g2 is about 590, which means if you are walking on the moon, the force of gravity acting on you from the moon will be almost 600 times that of the earth.

[caravelair]

i s'pose i might aswell figure out the various tidal forces involved too...

as mentioned, tidal force is actually the difference between 2 forces.

taking the same equation for acceleration, we have for the close side of the object g1 = GM/r1^2 and for the far side, g2 = GM/r2^2. this gives us a tidal acceleration (is that the correct term?) of:

g1 - g2 = GM(1/r1^2 - 1/r2^2) = GM(r2^2 - r1^2)/((r1^2)(r2^2))

now, (r1^2)(r2^2) is going to be approximately equal to r1^4, so we'll make that approximation, and then we see that the magnitude of tidal acceleration depends heavily on the term (r2^2 - r1^2). let's call this term d.

now, the moon's diameter is about 3.4x10^6m, so r1 = 3.85x10^8m, r2 = 3.884x10^8m, which gives us d1 = 2.6x10^15m^2.

something the size of a person has a diameter closer to 2m. we'll use the same r1 for convenience, and that means our r2 is 3.85000002x10^8m, which gives us d2 = 1.5x10^9m^2.

d1/d2 = 169,000 which tells us that the moon itself feels somewhere around 200,000 times more tidal acceleration than a person on the moon would.

i made a ton of approximations, but it's just to give an idea of magnitude.

[Silent Acorns]

Originally posted by caravelair
something the size of a person has a diameter closer to 2m.
2m in diameter?
Wow! That is one FAT moon man!

Seriously though, we can make a few more approximtations. Starting from:

a = GM(r2^2-r1^2)/r^4

and noting that:

r2^2-r1^2 = (r2+r1)(r2-r1) ~ 2rD

Where D = diameter of the moon/fatman

This is much more accurate for cases where (r2-r1)/(r2+r1) is very small.

then:

a = 2GMD/r^3

Where,
a = tidal acceleration
M = mass of the Earth
r = radius of the Moon's path around the Earth

So, the ratio of the tidal acceleration on the moon itself to the acceleration on a person on the moon is:

ratio = Dmoon/Dman

If Dmoon = 3 400 000 m and Dfatman = 2m then

ratio = 1 700 000

The differece, i think, is due to a calculation error by caravelair at the final step.

[Silent Acorns]

Also, if you want to comapre tides on the earth to tides on the moon then

Aearth = Mmoon * Dearth
Amoon .. Mearth * Dmoon

and assuming that the density of the Earth is roughly the same as that of the moon then:

M a D^3

and therefore:

Aearth/Amoon = (Dmoon/Dearth)^2

According to caravelair, Dmoon = 3.4x10^6 m and I know Dearth = 12.76x10^6 m, then:

Aearth/Amoon = (3.4/12.76)^2 = 0.071

Or, the tidal force on the moon is about 14 times stronger than the tidal force on the earth.