# Difference in gravity on both sides of the moon How much (if any) difference in gravity would you feel if you are on the surface of the moon and:

1. standing right 'under' earth (exaclty between the moon and the earth) and
2. on the opposite side of the moon (So both, the moon and earth are exactly under you)

## Very, very little

It's important to keep in mind that the Moon is constantly accelerating toward the Earth. The component of gravitation that one feels on the surface of the Moon is the difference between the gravitational acceleration toward the Earth at the point of interest and the Moon's gravitational acceleration toward thbe Earth as a whole.

At the sub-Earth point (the point on the Moon closest to the Earth), the gravitational acceleration toward the Earth is slight greater than is the gravitational acceleration of the Moon as a whole toward the Earth. The Earth very slightly reduces the gravitational acceleration toward the center of the Earth at the Moon's sub-Earth point.

At the antipode to the sub-Earth point (the point on the Moon furthest from the Earth; I'll call this the anti-sub-Earth point), the gravitational acceleration toward the Earth is slight less than is the gravitational acceleration of the Moon as a whole toward the Earth. The Earth once again very slightly reduces the gravitational acceleration toward the center of the Earth at the Moon's anti-sub-Earth point. The effect is nearly (but not exactly) identical to that at the sub-Earth point.

There are places on the Moon where the gravitational acceleration toward the Earth does slight increase the observed gravitational acceleration, which is the set of points where the Earth is more or less on the horizon. This effect here is about half that (in magnitude) of the effects at the sub-Earth and anti-sub-Earth points.

## A qualitative discussion

Ignoring that the Moon is rather lumpy in a gravitational sense, the gravitational acceleration on the surface of the Moon due to the Moon itself is $$vec a_M = -frac{mu_M}{{r_M}^2}hat r_M$$ where $$mu_M=G M_M$$ is the Moon's gravitational parameter, $$r_M$$ is the Moon's radius, $$hat r_M$$ is the vector from the center of the Moon to some point on the surface of the Moon, and $$oldsymbol a_M$$ is the gravitational acceleration due to the Moon at that point.

At that point, the gravitational acceleration toward the Earth is $$vec a_E = frac{mu_E}{||R_E hat r_E - r_M hat r_M||^3}(R_E hat r_E - r_M hat r_M)$$ where $$mu_E=G M_E$$ is the Earth's gravitational parameter, $$R_E$$ is the distance between the centers of mass of the Moon and the Earth, $$hat r_E$$ is the vector from the center of the Moon to the center of the Earth, and $$oldsymbol a_E$$ is the Newtonian gravitational acceleration toward the Earth at that point.

The Moon itself accelerates Earthward due to Earth gravity by $$vec a_{M,E} = frac{mu_E}{||R_E hat r_E||^2}hat r_E$$ The specific weight (force measured by a spring scale divided by mass) at the point of interest is egin{align} vec g &= vec a_M + vec a_E - vec a_{M,E} &= -frac {mu_M}{{r_M}^2}left(hat r_M-frac{M_E}{M_M}left(frac{r_M}{R_E} ight)^2left(frac{hat r_E - frac{r_M}{R_E}hat r_m}{||hat r_E - frac{r_M}{R_E}hat r_m||^3}-hat r_E ight) ight) end{align} ag{1} In the case where $$hat r_E = hat r_M$$ (the point on the Moon closest to the Earth) equation (1) simplifies to egin{align} vec g &= -frac {mu_M}{{r_M}^2}left(1-frac{M_E}{M_M}left(frac{r_M}{R_E} ight)^2left(frac{1}{||1 - frac{r_M}{R_E}||^2}-1 ight) ight)hat r_m &approx -frac {mu_M}{{r_M}^2}left(1-2frac{M_E}{M_M}left(frac{r_M}{R_E} ight)^3left(1+frac32frac{r_m}{R_E}+cdots ight) ight)hat r_m end{align} In the case where $$hat r_E = -hat r_M$$ (the point on the Moon furthest from the Earth) equation (1) simplifies to egin{align} vec g &= -frac {mu_M}{{r_M}^2}left(1-frac{M_E}{M_M}left(frac{r_M}{R_E} ight)^2left(1-frac{1}{||1 + frac{r_M}{R_E}||^2} ight) ight)hat r_m &approx -frac {mu_M}{{r_M}^2}left(1-2frac{M_E}{M_M}left(frac{r_M}{R_E} ight)^3left(1-frac32frac{r_m}{R_E}+cdots ight) ight)hat r_m end{align} The dominant perturbing factor, $$2frac{M_E}{M_M}left(frac{r_M}{R_E} ight)^3$$ is the same for both points and is rather small, about $$1.5 imes10^{-5}$$. People at these two points would weigh a tiny bit smaller than they would if the Earth was not present. The difference between the two points is tinier still, a factor of about $$2 imes10^{-7}$$.

Very little

Earth's surface gravity is $$1g$$, felt at a radius of about $$4,000$$ miles from the planet's center. The distance to the moon is roughly $$240,000$$ miles, which is ~$$60$$ times the earth's radius. Because gravity decreases with the square of distance, the earth's pull on the moon is only $$frac{1}{3600}g$$.

The moon's surface gravity is ~$$frac{1}{6}g$$, and the earth's gravity will make you lighter/heavier by ~$$frac{1}{3600}g$$ depending on if you're standing on the near, where the Earth is pulling away, so the gravity is $$(frac{1}{6}-frac{1}{3600})g$$ or far side of the moon, where it is $$(frac{1}{6}+frac{1}{3600})g$$. Your apparent weight will only vary by a few tenths of a percent between the near and far side of the moon.

I've ignored the radius of the moon itself here (~$$1000$$ miles), as it is relatively small compared to the earth-moon distance. But there is an even smaller tug from earth on the far side than the near side, further reducing the change in apparent weight.

EDIT: This answer is wrong, but will remain here as testament to incorrect reasoning. @David Hammen has it correct - the earth's pull actually decreases apparent weight on both the near and far side of the moon, much like how the moon creates a tidal bulge on both sides of the earth. The near side of the moon gets pulled more than the moon as a whole, while the moon as a whole gets pulled more than the far side. The overall scale of the effects is correct though, with the earth only exerting a few ten-thousandths of a g at that distance.

To a good first approximation, there is no difference. There's an easy way to see this with a thought experiment: Imagine the Moon was a sphere of liquid. Its geoid (selenoid?) is the surface where the water has no tendency to flow elsewhere due to gravity. If gravity was higher in one place, water would flow that direction raising the surface elevation and decreasing the pull of gravity until it was the same everywhere again.

Dirt and rock flows also, though much more slowly. The Moon's surface adjusts through isostasy so that it follows the geoid surface over large areas. (Over smaller areas, the strength of the rock is enough to hold things in place for a very long time, hence craters and mountains.) Large areas of highland is mostly due to the rock underneath being less dense and hence bouyant

## Difference in gravity on both sides of the moon - Astronomy

Why are tides higher not just during a New Moon, but also during a Full Moon?

I understand this has to do with the alignment of the Sun, the Earth and the Moon, but I would expect that the gravitational effects of the Moon would be weaker during a Full Moon as the Sun is "pulling" from the opposite direction?

That's a great question! Tides are caused by tidal forces, and the answer to your question lies in the definition of a tidal force. A tidal force is related to gravity, but it isn't the same thing. It's really the difference between the the strength of gravity at two locations.

The gravitational attraction between two objects (say the Earth and the Moon) decreases with distance. This means that the Moon's gravity pulls most strongly on the side of the Earth closest to the Moon and least strongly on the side of the Earth farthest from the Moon. Tidal forces on the side of Earth closest to the Moon pull material (mostly water) toward the Moon. Tidal forces on the other side of Earth actually pull material away from the Moon. The resulting deformation of Earth looks the same when the moon is at opposite sides of its orbit, like full moon and new moon or first quarter and third quarter, as shown in the diagram on this page. That's why tides around the equator are higher during both a new moon and a full moon (spring tide).

The Sun also affects the Earth's tides. However, tidal forces due to the Sun are about half as strong as those due to the Moon. This seems strange, because the Sun's gravity at Earth is much stronger than the Moon's. But remember that tides concern the difference between gravity's pull at opposite sides of the Earth. The radius of the Earth is a very small fraction of the distance between the Sun and the Earth, about 0.005%. As a result, the difference between the Sun's gravitational pull on either end of the Earth is small. In contrast, the radius of the Earth is about 1.7% of the distance between the Earth and the Moon. So even though the Moon's gravity isn't as strong as the Sun's, lunar tidal forces are stronger than solar tidal forces, so lunar tides are stronger than solar tides.

#### Michelle Vick

Michelle is a second year astronomy graduate student at Cornell. She works with Professor Dong Lai to study tidal interactions between white dwarfs and black holes.

First of all, tides are not as simple as the "two-bulge" simplification. In reality, the diagram shown is misleading. The two bulges appear assuming an ocean of constant depth covers the entire surface of Earth. Clearly that is not the case and in the diagram you can see the continents. Considering the different sizes of the basins and the distinct frictional characteristics in each location, the resulting tidal effect is much more complex. The difference in phase and amplitude is shown here and it clearly shows that the the tide varies for the same longitude. That wouldn't be the case in the simple explanation above. Source Wikipedia.

Looking at this tidal animation from TPXO is also illustrative.

The simple "two-buldge" explanation would result in a pure two peak daily tide. That is certainly not the case in places like the Gulf of Mexico.

As mentioned in Camilo Rada's answer, the bulges are a consequence of the tidal force. This apparent force result from the difference in strength in the gravitational field. The result is that Earth's body is stretched toward and away from the center of mass of the Earth-Moon system. The water thus adjust to this difference in geopotential giving rise to the tides.

The explanation of the two-bulge tide comes from the fact that the Moon and Earth form a two-body system that rotates about an axis located within Earth.

The bulge of water on the side of Earth that faces the Moon is easily explained. It is due to the gravitational attraction between the Moon and Earth, including the water on Earth. This attraction pulls water toward the Moon and creates a “bulge” on the surface of Earth. The bulge on the other side of Earth is due to inertia. Inertia is the tendency of an object at rest to stay at rest and the tendency of a body in motion to continue its motion in a straight line.

There is an inertial tendency resulting from the rotation of the Earth-Moon system for objects (water among them) to move away from both sides of Earth—the side facing toward the Moon and the side facing away. The model demonstrates that the effect of things moving away from Earth is much greater on the side facing away from the Moon.

Many textbooks and other sources use the concept of “centrifugal force”— which is actually a preconception—to explain the effects of inertia. According to this preconception, there is a force that acts on all objects that are in circular motion, and this force pushes or pulls the object out from the circle. There is no such force. The preconception arises from our own experience with circular motion.

The gravitational forces of Earth, Sun and Moon cause a bulge of water on the nearest side and an equal bulge on the other side. Thus, in this simple scenario, the tide is composed of two bulges of water (four, in fact), traveling around the world as the world spins. When Moon and Sun aligned, their respective bulges add together to form "spring tides" every two weeks. When the Moon and Sun are at right angles, we encounter "neap tides", as the bulge of the sun adds to the low lunar tide, resulting in higher low tides but lower high tides.

The limitations of this model are:

• It cannot explain that there are places without tides, with one daily high, and most with two tidal highs each day.
• Tidal height is not maximal at the Equator (and minimal at the poles) as the simplification suggests.
• High tide is not associated with the position of the Moon. It occurs at different times of the lunar cycle depending on the location.
• If continents are included, the tidal wave would reflect off the continental shelf as it reaches a continent. A tidal wave of almost equal magnitude will be propagating in the opposite direction, which is not observed.
• The tidal waves required for this model would have to travel at much faster speeds that are possible in reality.

In reality, the tides instead of running east to west as Earth rotates, tidal waves propagate around in circles around islands, and certain points in the sea, called tidal nodes or amphidromic points. These nodes can be seen in the first figure from this answer.

Thus, the tidal patterns in the ocean are a set of rotating standing waves. These waves have periods that represent the natural resonance periods of the ocean basins. These waves can be considered modes of "vibration" and can be decomposed using a Fourier decomposition. That is the source of the different tidal constituents that are used currently for tidal prediction.

## Astronomy Picture of the Day

Discover the cosmos! Each day a different image or photograph of our fascinating universe is featured, along with a brief explanation written by a professional astronomer.

2011 November 1
Hammer Versus Feather on the Moon
Image Credit: Apollo 15 Crew, NASA

Explanation: If you drop a hammer and a feather together, which reaches the ground first? On the Earth, it's the hammer, but is the reason only because of air resistance? Scientists even before Galileo have pondered and tested this simple experiment and felt that without air resistance, all objects would fall the same way. Galileo tested this principle himself and noted that two heavy balls of different masses reached the ground simultaneously, although many historians are skeptical that he did this experiment from Italy's Leaning Tower of Pisa as folklore suggests. A good place free of air resistance to test this equivalence principle is Earth's Moon, and so in 1971, Apollo 15 astronaut David Scott dropped both a hammer and a feather together toward the surface of the Moon. Sure enough, just as scientists including Galileo and Einstein would have predicted, they reached the lunar surface at the same time. The demonstrated equivalence principle states that the acceleration an object feels due to gravity does not depend on its mass, density, composition, color, shape, or anything else. The equivalence principle is so important to modern physics that its depth and reach are still being debated and tested even today.

## Could a Long-Ago Collision Explain Our Two-Faced Moon?

By: Javier Barbuzano May 28, 2019 1

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New research shows that a planetoid impact could have made our Moon asymmetric. Artist's depiction of a collision between two planetary bodies.
NASA / JPL-Caltech

Scientists have been scratching their heads about the striking differences between our Moon’s near and far sides for 60 years, since the Soviet satellite Luna 3 orbited the Moon in 1959. When it sent back the first pictures of the Moon’s long-hidden farside, scientists saw it looked nothing like the “Man in the Moon” we’re used to seeing in the sky. Now, a study purports to explain the dichotomy.

Because the Moon orbits Earth so closely, it’s tidally locked, keeping one hemisphere constantly turned toward us and the other turned away. The nearside shows two distinct types of terrain: the lighter-toned highlands and the dark maria (Latin for “seas”). The highlands formed when the Moon cooled down shortly after its formation 4½ billionyears ago the maria formed much later, about 3.1 to 3.9 billion years ago, when volcanic eruptions filled preexisting impact basins. The topographic (A), crustal thickness (B), and thorium distribution of the Moon show a dramatic difference between the nearside and farside. The star on the nearside represents the center of the proposed impact basin. The black dashed lines represent the boundary of Imbrium (Im), Orientale (Or), and Apollo (Ap) basin, respectively.

JGR: Planets / Zhu et al. 2019 / AGU

In contrast, the heavily cratered lunar farside has almost no maria. While impact basins scar both sides of the Moon equally, lava filled only those on the nearside. NASA’s Gravity Recovery and Interior Laboratory (GRAIL) orbiter has found that an extra layer of material lies on top of the primordial crust.

Scientists have proposed many theories over the years to explain these differences. Some have pointed to the influence of the Earth’s gravity, which might have thinned the crust and increased volcanic activity on the nearside. Others have suggested some sort of cataclysm made the sides so different, such a collision with a sub-Moon or an incoming planetoid.

Meng-Hua Zhu (Macau University of Science and Technology) and colleagues reported May 20th in the Journal of Geophysical Research: Planets that the latter is a real possibility. The team ran a series of computer simulations to show that, under certain conditions, an impact with a dwarf planet slightly smaller than Ceres could produce a two-faced Moon like the one we have today.

For this scenario to work, the impact would have happened after the Moon had formed a solid crust but before that crust became too stiff otherwise, the surface would have preserved scars from such a massive hit.

In the simulations, the team found that a body 780 kilometers (480 miles) in diameter — a bit smaller than Ceres — would need to hit the Moon’s nearside at 22,500 kilometers per hour (14,000 mph) to reproduce the asymmetry in the crust. Such an impact would have been relatively slow, a quarter of the speed with which meteorites typically hit Earth.

The simulations showed that the impact would have ejected vast amounts of material, which would have landed mostly on the Moon’s farside, burying its original crust in 5–10 kilometers of debris, equivalent to the extra layer detected in GRAIL observations. An impactor 780 km in diameter (with a 200-km-diameter iron core) hits the Moon going 22,500 km per hour (14,000 mph). In each panel, the left halves represent the materials used in the model: gabbroic anorthosite (pale green), dunite (blue), and iron (orange) represent the lunar crust, mantle, and core, respectively. The gabbroic anorthosite (pale yellow) also represents the impactor material. The right halves represent the temperature variation during the impact process. The arrows in (C) and (D) represent the local materials that were moved and formed the new crust together with deposits of material that was blasted from the impact.
JGR: Planets / Zhu et al. 2019 / AGU

However, giant-impact models such as this one not only have to get the mechanics right, they also need to match everything else we know about the Moon.

“It’s very difficult to get a model that claims that it can match all the constraints,” says William Bottke (Southwest Research Institute), who wasn’t involved in the study. There are a lot of constraints: We know the thickness of the Moon’s crust, thanks to GRAIL observations, but we also know chemical and isotopic composition of the surface rocks. The Moon and Earth have strikingly similar isotopic compositions, suggesting that their building blocks already had the same isotopic composition or that some unknown mechanism stirred everything together right after the Moon formed. If a giant impact, such as the one proposed by Zhu and colleagues, added a substantial amount of material to the Moon after it formed, it’s very unlikely that the Moon would still have an isotopic composition identical to Earth’s. “If they can indeed fulfill all these constraints, then I think it’s looking pretty interesting,” Botkke says.

David Stevenson (Caltech), who was also not involved in the study, calls the scenario “unlikely,” — not only because of the isotopic similarity between the Moon and Earth, but also because of the “just-right” timing of the impact. He also finds it too convenient that the impact would produce two distinct hemispheres.

Ultimately, samples from the far side of the Moon could change our understanding of the differences between the Moon’s nearside and farside. At this very moment, the Chinese spacecraft Chang'e 4 is collecting data from the lunar farside. After landing on the Moon on January 3rd, it deployed a small rover called Yutu-2, which has detected signs of excavated mantle material on the farside’s surface. Chang’e 4 isn’t a sample return mission, but the upcoming Chang’e 5 will be, and it’s expected to launch later in 2019.

## Gravity Maps Reveal Why the Moon's Far Side Is Covered with Craters

When the Soviet probe Luna 3 sent back the first shots of the dark side of the Moon, they showed that it was noticeably more pockmarked by craters than the near side. The nearside crust, by contrast, had more large, shallow basins. More than 50 years after those images first baffled researchers, a study published today in Science explains the observations.

Some theories suggest that the large basins on the near side were caused by impacts from asteroids bigger than those that caused the craters on the far side. But the latest study suggests that the observed basins do not accurately reflect the size of the initial impact, because as asteroids battered the lunar surface in the early history of the Solar System, the Moon's warmer and softer nearside crust melted like butter, producing giant lava flows that filled the impact craters and transformed them into basins.

To improve on previous estimates of the size and distribution of basins, the team behind the study used data from NASA&rsquos Gravity Recovery and Interior Laboratory mission (GRAIL), two satellites that since 2011 have been orbiting the Moon and mapping subtle variations in the strength of its gravitational field. Basins are characterized by thinner crust, says first author Katarina Miljkovi, a planetary scientist at the Paris Institute of Earth Physics. The team used GRAIL's gravity maps to find such thin crust and measure the true size of the basins.

&ldquoWe didn&rsquot have to look at topography nearly at all, just at the crust thickness,&rdquo says Miljkovi. The researchers found that although both sides of the Moon had the same total number of impact craters, the near side had eight basins larger than 320 kilometres in diameter, whereas the far side had only one.

Hot hit
The asteroid bombardment should have battered both sides equally, Miljkovi points out. The asymmetry could have arisen from comparatively small objects punching above their weight on the near side, producing basins more easily than on the far side.

Simulations showed that if the largest dark area on the near side &mdash the plain of volcanic rock known as Oceanus Procellarum &mdash was hundreds of degrees hotter than crust on the far side, impacts there would produce basins up to twice as large as impacts from similar-sized bodies on the far side (see video below).

Harder and softer landing: These simulations show that the crash of an asteroid onto the surface on the young Moon would have created a larger crater in a region where the crust's temperature is heightened by radioactivity (right) than it would in a region at normal temperature (left).

And indeed, around 4 billion years ago, or 500 million years after the Moon formed, the near side could have been warmer than the far side. Researchers looking at the near side have detected the presence of radioactive isotopes their decay would have heated up the rock, explains study co-author Maria Zuber, a planetary scientist at the Massachusetts Institute of Technology in Cambridge and principal investigator of GRAIL.

The findings fit well with the observations, but &ldquothere is no consensus&rdquo as to what caused the startling asymmetry in isotope content between the near side and the far side, says Jeffrey Taylor, a lunar scientist at the University of Hawaii in Honolulu. One leading theory posits that material rich in radioactive elements rose in a gigantic volcanic plume and formed a magma basin another that it came from a collision with a sister moon around 1,000 kilometres in diameter.

William Bottke, a lunar scientist at the Southwest Research Institute in Boulder, Colorado, says that the work could lead researchers to revise just how dramatic asteroid bombardments were in the early Solar System. "This can be used to more accurately derive what the small-body populations were like four billion years ago.&rdquo

Davide Castelvecchi is a senior reporter at Nature in London covering physics, astronomy, mathematics and computer science.

## There is only one earth rotation, so why are their two low and high tides each day?

The reason for this is that there are two astronomical bodies that affect the tides. Because the sun and the moon both have their own gravitational gradient that alters depending on the position of each of them.

There are two high and low tide cycles in one day because the two will overlap each other as the moon orbits the earth throughout the day. And as the sun orbits the sun at the same time.

## When the moon is directly overhead, does it's gravity make us weigh less by a measurable amount? Obviously the moon's gravity effects things on earth (tides), but do objects on the earth's surface actually weigh less when the moon is directly overhead? Are there any applications that require weight measurements so precise that this needs to be accounted for? *Final Edit: I'll leave this calculation here for fun, but it has errors. The final answer is below in my response to AtTheLeftThere's question. Today, I re-learned how tides work.

Interesting question. We can work this out, though. Let's assume an object at the equator. To get maximal effect, let's look at the Moon at perigee when it's directly overhead and when it's on the other side of the Earth.

At the equator, acceleration due to gravity is 9.7805 m/s² (Wikipedia).

At perigee, the Moon is 359,861 km from the center of the Earth (source). But the radius of the earth at the equator is 6,378 km (equatorial diameter taken from above Wikipedia link).

This means the Moon at perigee overhead is 359,861 km - 6,378 km = 353,483 km. On the other side at perigee, the Moon is 359,861 km + 6,378 km = 366,239 km

Now use Wolfram Alpha to see the acceleration on one side plus the acceleration on the other (it's a difference of accelerations, but they go in opposite directions, so add the magnitudes together): WolframAlpha. This yields 7.58*10 -5 m/s².

As a fraction of the acceleration due to gravity at the equator, that's 7.75 * 10 -6 , or 0.000775%.

That's a larger effect than I would have guessed, and I imagine it might matter for some applications. It's also wholly possible I messed up the calculation -- it's been a long time since I thought I was going to be a physics major.

*Edit: See benjaffe's comment and Astrokiwi's comment. The acceleration of the Earth itself would largely cancel out this effect.

Here's an updated calculation subtracting out the acceleration of the Earth from each direction: WolframAlpha = 7.14 * 10 -8 m/s². This is over 1,000 times smaller.

But it's still larger than the pull of one mosquito 10 cm away (oats2go's comment, WolframAlpha calculation), so I wonder if I'm still missing something.

This wouldn't actually be the difference in the weight you measure however, because the moon is accelerating the Earth beneath you as well. What you really need to measure is the difference between the moon's gravity on the floor beneath you, and the moon's gravity on you - i.e. what is the tidal force on the scale of a person. That will be much much smaller.

How do the tides get so high if the moon's pull is negligible on us? Youɽ think it would be more pronounced.

I could read all that, or you could just give me a yes or no answer and save us all some time.

(I got to go to sleep in 10 minutes, so I am lazy and really just want the answer!)

Yes. I'm not sure of a source, but I do remember reading this somewhere. I'll try and find it.

I was actually wrong. According to Scientific American:

the late astronomer George Abell of the University of California, Los Angeles, noted, a mosquito sitting on our arm exerts a more powerful gravitational pull on us than the moon does. Yet to the best of our knowledge, there have been no reports of a “mosquito lunacy effect.” Second, the moon’s gravitational force affects only open bodies of water, such as oceans and lakes, but not contained sources of water, such as the human brain.

As worded, that quote is wrong. The moon's gravitatonal force affects everything in the universe (according to Newton, and pleading ignorance about relativity's influence) -- in proporton to the object's mass and inverse proportion to the square of its distance.

Since we're at the same distance from the Moon as the oceans are, we are affected to the same extent as the oceans. That also goes for water anyplace on Earth, contained or not.

Perhaps the author meant that contained water will not show tidal fluctuations, which is correct (See footnote).

BTW, I was puzzled for a long time why the tide should be high, for instance, on both sides of the Earth. How could gravity push away the water on the far side of the Earth?

But the explanation turned out to be pretty reasonable, and has to do with the difference in the Moon's gravitational force due to the diameter of the Earth.

It might help to define some terms. Imagine a large circle (the Earth) with a smaller circle (the Moon) some distance away. Call the spot on the Earth where the Moon is directly overhead, the Zenith point. Call the point on the Earth directly opposite the Nadir point. And call the points on the Earth halfway between Zenith and Nadir, the Middle points. (Takes a deep breath).

The Moon's gravity exerts a force at the Middle points, call it the Middle force. At the Zenith point, the gravitational force will be greater, but at the Nadir point the Moon's gravitational force will be less because the Nadir is farther away from the Moon.

So at the Nadir, the ocean feels less of a pull toward the Moon than it does anywhere else: so it can rise higher than at the Middle points. Hope that's clear, it was harder to explain than to visualize.

(Footnote:) Except for me: I'm convinced that my brain has its high and low tides, and currently seems to be pretty low.

## How much does Earth's gravity affect the gravity that we feel on the Lunar Surface?

Earth R: 6,371 km
Moon R: 1,737.1 km
d1: 384,400 km (center to center)
d2: 382,663 km (E core to near side)
d3: 386,137 km (E core to far side)
Assuming Earths gravity is 1g, M and E are perfect spheres, orbits are circular, distribution of mass is perfectly uniform, no other forces complicating our system like for example the pull of the sun.
---

A man at the center of the near side of the moon, is standing 60.06 times further away from the Earths CORE compared to a person standing at the surface of Earth. Applying inverse square law we can find that: Earths gravityis 3,488.4 times weaker up there.
Moons surface gravity averaging

0.166g, the Earths pull is -0.0002772g at that point and has a completely oposite orientation (1.66% in comparison),

Simmilar calculations for the far side of the moon (at the point where Earths and Moons gravitational vectors align): 386,137/3671=60.60|sqr(60.60)=3,673.39|That is: +0.000272g or 1.638% in comparison to Moon Average surface gravity

To put that in perspective a cargo box weighting 1000kg will have an average weigth 166Kg on the Lunar surface,
But at the center of NEAR side it will weight (

1.66% less) 163.24Kg, and (+1.638%)168.72Kg @ the center of opposite side,

Thats a whole 5.48 Kg difference for that container, or 548 grams difference for a man that originally weights 100Kg on Earth

(3.35% icrease in weigth) just because you changed location on the Moon, isnt that crazy considering we just took in account the change due to Earths gravity gradient?

## The end of a mission

GRAIL's mission ended as the two spacecraft ran out of fuel. The team decided to send the pair out in a blaze of glory, deliberately crashing them into the moon's surface with the hope of revealing more about the lunar composition upon impact.

But before the collision, GRAIL completed one last experiment. Fifty minutes prior to the Dec. 17, 2012, impact, the two spacecraft fired their engines until their propellant was depleted, a maneuver designed to precisely determine the amount of fuel remaining in the tanks. This final act helped NASA engineers validate computer models and improve predictions of fuel needs for future missions.

Because the moon is about 236,000 miles (380 km) away from Earth, it would have been a challenge to observe GRAIL's impact plumes from our planet. So, three weeks before the scheduled impact, GRAIL researchers contacted NASA's Lunar Reconnaissance Orbiter (LRO) team, which was also studying the moon.

The LRO team scrambled to get its orbiter in place to witness GRAIL's fiery demise. When Ebb and Flow crashed into the surface, LRO was only about 100 miles (160 km) from the lunar surface. Because the crash site was in the shadows, the team had to wait for the plumes to rise high enough to be visible in the sunlight. LRO's instruments revealed mercury and hydrogen in the plume.

Ebb's and Flow's resting places are both relatively small, only about 13 to 20 feet (4 to 6 meters) in diameter. The material blown up by the impacts created unusual dark patterns on the moon's surface.

"Fresh impact craters on the moon are typically bright, but these may be dark due to spacecraft material being mixed with the ejecta," LRO team member Mark Robinson, of Arizona State University, said in a statement.

Both impact sites lie on the southern stope of an unnamed mountain and are about 7,218 feet (2,200 m) apart. Ebb slammed into the ground about 30 seconds before Flow, which created its crater to the west and north of its twin.

Not long after the pair's demise, their craters were named in honor of former astronaut and GRAIL team leader, Sally Ride, who passed away in July of 2012 while the mission was in progress.

"Sally was all about getting the job done, whether it be in exploring space, inspiring the next generation or helping make the GRAIL mission the resounding success it is today," Zuber said in a statement. "As we complete our lunar mission, we are proud we can honor Sally Ride's contributions by naming this corner of the moon after her."

Over its 15-month mission, GRAIL improved our understanding of the moon, taking some of the steps necessary for eventually bringing humans back to the lunar surface.