The Moon is orbiting the Earth in a circular motion. To keep any object in a circular motion we need energy. Hence, does the Earth lose energy by pulling the Moon? In that case, does the Earth's gravity reduce constantly? If it does, then as gravity is proportional to mass, does the Earth lose mass as well?
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$\begingroup$ See: astronomy.stackexchange.com/a/32553/35206 $\endgroup$– Nilay GhoshCommented Mar 8, 2023 at 6:40
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5$\begingroup$ related in Earth Science SE: Is Earth getting heavier or lighter? and links below it to questions in other sites. $\endgroup$– uhohCommented Mar 8, 2023 at 15:24
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2$\begingroup$ The actual, dominating mass-loss from Earth is via loosing atmosphere, around 500-1000g/s both in H and O atoms (see review numbers given in ui.adsabs.harvard.edu/abs/2018A%26A...614L...3G/abstract ) $\endgroup$– AtmosphericPrisonEscapeCommented Mar 9, 2023 at 16:43
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2$\begingroup$ Prajawl. The very simple answer to the question the way you're thinking about it is just NO. Say you are in deep space, and there is absolutely no friction. Say an object (spaceship, whatever) is "moving" at 100 mph. It will just keep moving at 100 mph forever It's that simple. That's all your asking. Similarly, if an object is spinning, it will just keep spinning forever. (The "earth-moon" is simply an object spinning.) Objects keep moving, spinning, forever if there is no friction. $\endgroup$– FattieCommented Mar 10, 2023 at 12:58
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2$\begingroup$ Gravity and the other forces that keep things together don't "use" energy. It's like a rod connecting them. $\endgroup$– FattieCommented Mar 10, 2023 at 12:59
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4 Answers
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You are wrong that "to keep any object in circular motion requires energy"
In a circular orbit, the force of gravity is always perpendicular to the motion of the moon, so no work is done by gravity, and no energy is required.
This is analogous to the string in a pendulum, which also does no work.
In an elliptical orbit, energy is constantly converted between potential and kinetic energy, but the total energy is constant, and over the full orbit, no work is done.
So, no, the gravity doesn't run out, and Earth doesn't lose mass.
In a three-body interaction (for example with a comet, the Earth and the sun) the perturbing effect of the sun can cause the Earth to transfer a bit of its kinetic energy to the comet. In this case the Earth can lose (or gain) some energy. The amount of energy lost is minuscule in comparison to the total amount of energy stored in the motion of the Earth. The mass (by E=mc²) is very very small. This process of a three body interaction causing the Earth to do work on a comet is called a "gravitational slingshot". Again the amount of energy lost in these interaction is very very small, and there is not a snowball's chance that the Earth will ever run out of energy.
(Earth does lose a little mass in other ways: a little bit of air is always escaping into space, but there is plenty left, and we won't ever run out.)
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14$\begingroup$ Earth also constantly gains mass from meteor impacts -- I believe at a higher rate than atmosphere loss, but I'm not sure. $\endgroup$ Commented Mar 8, 2023 at 15:57
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4$\begingroup$ @PrajwalDM Both the comet and Earth will exert a gravitational force on each other, and both of their orbits will be changed. (Earth's only fractionally so, because it's much larger; a comet's orbit can be dramatically changed by an encounter with a planet because of its lower mass.) Earth's orbit might either gain or lose energy depending on the exact movement involved, but that energy doesn't come from Earth, it comes from (or goes to) the comet's orbit. $\endgroup$– CadenceCommented Mar 8, 2023 at 18:56
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5$\begingroup$ I'd take issue with conflating energy and mass. Unless you're talking about relativistic mass (which has largely fallen out of favor as a useful concept), the Earth's mass doesn't change when objects perturb its orbit or when it radiates gravitational waves. $\endgroup$– StenCommented Mar 8, 2023 at 22:55
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2$\begingroup$ From physics.stackexchange.com/a/412990/123208 the Earth-Sun system radiates ~196 watts as gravitational waves, which is equivalent to ~2.18 picograms per second (just under 69 micrograms per Julian year). But that doesn't affect the rest mass of the Earth or Sun, it just decreases their gravitational potential energy. I suppose you can say it reduces the rest mass of the Sun-Earth system, which is less than the sum of the rest masses of the Sun & Earth, the deficit being equal to their gravitational binding energy. $\endgroup$– PM 2RingCommented Mar 9, 2023 at 2:50
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2$\begingroup$ And of course that loss of mass through diffusion of the atmosphere and the little bit we shoot up in the form of rockets is at least in part compensated for by asteroids and other objects being captured by the earth's gravitational field and being pulled in. $\endgroup$– jwentingCommented Mar 9, 2023 at 7:34
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Earth does not lose mass to the moon's orbit. However, there are a few ways it can lose energy to gravitational effects. One way it can lose energy is to gravitational slingshotting, from satellites, asteroids or comets. In the case of a gravitational slingshot, Earth doesn't lose mass, it simply slows down. But since Earth is so massive in comparison that slowdown is extremely slight.
The Earth also loses some rotational energy to tidal forces from the moon. In this case, the Earth simply turns very slightly more slowly. Still, no mass is lost.
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$\begingroup$ How could we still be losing rotational energy to the moon when the earth and moon are already tidally locked? $\endgroup$– CruncherCommented Mar 9, 2023 at 15:38
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6$\begingroup$ @Cruncher The moon is tidally locked to the earth, but the earth is not locked to the moon. That's why we have tides that move around the earth. The land also has tides, although much, much less noticeable. These tides ultimately convert Earth's kinetic energy to heat. Those earth-tides are also hoisting the moon into a higher orbit, so the earth does work on the moon, drawing from its kinetic energy. $\endgroup$– ericksonCommented Mar 9, 2023 at 15:45
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No energy is needed for the moon to keep going straight ahead in space that is curved by earth's gravity. Which makes "straight ahead" being orbital around the earth.
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4$\begingroup$ Yes, gravity curves spacetime, but the spatial curvature of the Moon's orbit is much greater than the spacetime curvature of its worldline. Please see physics.stackexchange.com/q/586835/123208 $\endgroup$– PM 2RingCommented Mar 9, 2023 at 14:15
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2$\begingroup$ This answer seems correct according to the theory of general relativity. Gravity isn't a force. An object in motion stays in motion, and because spacetime is curved around the earth, the moon will continue to 'keep going straight' in curved spacetime around the earth. When something keeps going straight there's no energy lost because it doesn't accelerate in any direction. The moon isn't changing directions and no force is acting upon it. Where there is no force, no energy is used. $\endgroup$ Commented Mar 9, 2023 at 21:07
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1$\begingroup$ @Nick Yes, the Moon's worldline is a timelike geodesic, but Kjetil's answer implies that the Moon's trajectory in space is straight, and that's not correct. $\endgroup$– PM 2RingCommented Mar 11, 2023 at 1:00
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Apart from the mainly correct replies so far, and just considering the mass of the planet in terms of mass lost to space, or gained from outside, we keep losing some mass from our atmosphere, both from molecules blasted away by solar wind etc, and from the fact that molecules much lighter than N2 (eg water) have typical speeds exceeding Earth's escape velocity.
However, we also are constantly gaining mass from space dust and meteoroids that collide with the planet, thereby becoming meteorites; and I suspect (but have no data) that our input from solar wind plus cosmic "rays" (largely matter particles, such as protons and electrons in terms of mass) exceed our constant leakage into space.
If I am right, then you could say that we still are in a state of accretion as we were early in the history of the solar system, but by now in a very late stage.
In some more billions of years (I have seen various figures) we should find ourselves inside the outskirts of the sun's atmosphere, and accordingly piling in lots more mass, mainly light elements of course. I suspect that when the sun shrinks back to red dwarf status after a few more millions of years, our planet still will be in orbit, and significantly more massive than before, and I would love to see the outcome, but in spite of all my kicking and screaming and writing to congressmen and lawyers, it seems to be very difficult to get ringside tickets.
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$\begingroup$ Do you have any numbers to assert your notion that we are gaining net mass? Also the sun will never become a red dwarf. And while Earth, by the time the sun expands to 100-200 solar radii, might accrete some of the solar wind, it will have had most of its atmosphere blasted away by the enormous luminosity of the star. $\endgroup$ Commented Mar 15, 2023 at 10:15
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$\begingroup$ @AtmosphericPrisonEscape Earth's mass influx due to meteorites is around 50000t/yr: lpi.usra.edu/books/MESSII/9021.pdf (end of §3.4) $\endgroup$ Commented Mar 15, 2023 at 10:22
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$\begingroup$ Other authors arrive at a factor of 10 less (a few 1000t/yr): agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2019JE006241 $\endgroup$ Commented Mar 15, 2023 at 10:31
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$\begingroup$ So if we take your upper number, that translates to ~1500 g/s, compared to an estimated 500 g/s and 800 g/s atmospheric loss from H and O escape (Gunell+2018). Seems like it could be a close race.. $\endgroup$ Commented Mar 15, 2023 at 10:34
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1$\begingroup$ @Jon not sure I'd agree to call todays influx from meteorites and zodiacal dust as 'accretion'. Accretion is usually understood as a phase of significant mass gain. We are talking here of a fraction of around 10^-18 .. 10^-17 earth masses per year while the actual planetformation takes place inless than ~10 million years, so accretion rates of 10^-7 earth masses per year. Thus whatever hasn't grown till then, will most likely not grow much more. Today we simply have a mass exchange with the interplanetary medium. $\endgroup$ Commented Mar 15, 2023 at 12:36