Gravity fail
January 6, 2015 8:10 PM Subscribe
I'm sure you've all seen the quite obvious and silly "gravity will fail" hoax. It got me wondering...if all the planets, stars, asteroids, comets, black holes, etc (basically all known physical matter) were to somehow align on one side of the planet would the cumulative gravity exceed Earth's? I asked my dad (a doctor of nuclear chemistry) and he was stumped. Exclude dark matter for these purposes.
It depends on how far away all the mass in the universe is from the object in question. Gravitational force goes as the inverse of distance squared, so there is some distance at which the gravitational force resulting from all the non-Earth mass in the universe will be less than the gravitational force resulting from Earth.
It happens to be at a distance greater than or equal to that distance for you right now, by the way.
posted by mr_roboto at 8:38 PM on January 6, 2015
It happens to be at a distance greater than or equal to that distance for you right now, by the way.
posted by mr_roboto at 8:38 PM on January 6, 2015
Here is a basic calculation to get a sense of the scale:
The gravitational acceleration due to an object of mass M is given by g = G*M/r^2, where G is the gravitational constant and r is the distance from M. The mass of the observable universe is estimated to 10^53 kg according to Wikipedia and at the surface of the earth, we have g_earth = 9.8m/s^2.
So, if all the matter in the universe was gathered in one point, it would exert a stronger gravitational pull on you than the earth if it was any closer to us than r = sqrt(G*M/g_earth) = 8*10^20 meters, or 90 000 light years. That's about the diameter of the Milky Way, and way closer than Andromeda, which is our nearest neighbour galaxy.
posted by Herr Zebrurka at 8:42 PM on January 6, 2015 [5 favorites]
The gravitational acceleration due to an object of mass M is given by g = G*M/r^2, where G is the gravitational constant and r is the distance from M. The mass of the observable universe is estimated to 10^53 kg according to Wikipedia and at the surface of the earth, we have g_earth = 9.8m/s^2.
So, if all the matter in the universe was gathered in one point, it would exert a stronger gravitational pull on you than the earth if it was any closer to us than r = sqrt(G*M/g_earth) = 8*10^20 meters, or 90 000 light years. That's about the diameter of the Milky Way, and way closer than Andromeda, which is our nearest neighbour galaxy.
posted by Herr Zebrurka at 8:42 PM on January 6, 2015 [5 favorites]
It seems like what you are imagining is some amount of mass that could be brought within a certain distance of the earth to make everything float. That wouldn't work because the earth is (roughly) spherical. That means only the side of the earth facing the mass would experience lower net gravity; the opposite half of the earth would experience stronger net gravity.
Gravity is an inverse cube law that depends on mass and distance. How much mass is on "one side" of Earth, how tightly is it compacted, and how far is it from the Earth's surface? Also, the Earth itself would be affected. It is not fixed in position.
That said, if you wanted an "overhead" mass to make a person float away, you would just need to mass to be of sufficient mass and distance from the earth that the acceleration of gravity in its field was 9.8 m/s^2 at the point of the earth's surface. Since gravity is an inverse square law, this is easy to do with mental math. Let's pretend the mass is ultra dense so it can be relatively small, like the size of a city block (no such material exists outside the cores of certain stars). The radius of the earth is approximately 4,000 miles, so if the mass had the same mass as the Earth, it would need to be 4,000 miles "overhead" to neutralize gravity. If we double the distance to 8,000 miles, we now need 4 earth masses. Triple it to 12,000 miles and you need nine earth masses. Tem times the distance at 40,000 miles would need 100 earth masses. Please keep in mind, by the way, that 40,000 miles from the earth's surface is much closer to the earth than to the moon.
SO, you probably get the idea. The farther away you get, the more mass you need, but no substance in the universe is dense enough to create the effect you are imagining. The Earth is already quite dense. A planet twice as dense as Earth would be the density of lead. Four times the density of Earth is greater than the density of uranium. At that point, you're done. No mass is going to exist at that density.
posted by Tanizaki at 8:46 PM on January 6, 2015 [1 favorite]
Gravity is an inverse cube law that depends on mass and distance. How much mass is on "one side" of Earth, how tightly is it compacted, and how far is it from the Earth's surface? Also, the Earth itself would be affected. It is not fixed in position.
That said, if you wanted an "overhead" mass to make a person float away, you would just need to mass to be of sufficient mass and distance from the earth that the acceleration of gravity in its field was 9.8 m/s^2 at the point of the earth's surface. Since gravity is an inverse square law, this is easy to do with mental math. Let's pretend the mass is ultra dense so it can be relatively small, like the size of a city block (no such material exists outside the cores of certain stars). The radius of the earth is approximately 4,000 miles, so if the mass had the same mass as the Earth, it would need to be 4,000 miles "overhead" to neutralize gravity. If we double the distance to 8,000 miles, we now need 4 earth masses. Triple it to 12,000 miles and you need nine earth masses. Tem times the distance at 40,000 miles would need 100 earth masses. Please keep in mind, by the way, that 40,000 miles from the earth's surface is much closer to the earth than to the moon.
SO, you probably get the idea. The farther away you get, the more mass you need, but no substance in the universe is dense enough to create the effect you are imagining. The Earth is already quite dense. A planet twice as dense as Earth would be the density of lead. Four times the density of Earth is greater than the density of uranium. At that point, you're done. No mass is going to exist at that density.
posted by Tanizaki at 8:46 PM on January 6, 2015 [1 favorite]
For fun, we can do the same math on smaller scales too: If all the mass of the Milky Way was assembled in a point, its gravitational pull would match that of the earth at a distance of 0.4 light years. Our nearest star after the sun is 4 light years away.
For the sun to pull as hard on us as the earth, it would have to be 4 million kilometers away, which is ~1/40 of where it actually is.
posted by Herr Zebrurka at 8:51 PM on January 6, 2015
For the sun to pull as hard on us as the earth, it would have to be 4 million kilometers away, which is ~1/40 of where it actually is.
posted by Herr Zebrurka at 8:51 PM on January 6, 2015
Response by poster: Thanks for the responses. To clarity, I'm not imagining a single point, but instead I'm imagining every celestial body lined up single file with the Earth at the head of line, with each object the same distance it is from us. They'd all be directly overhead, appearing as a single point with the moon in the foreground (like the lunar eclipse of all lunar eclipses!). For arguments sake I'm standing on a bathroom scale. If I normally weigh 200 lbs, what would the scale show? 199.9?
To simplify the math, can you take the total mass of the universe, and half the diameter of the universe, and calculate the gravity based on that? Is a single point the same as a single file line?
posted by karst at 9:02 PM on January 6, 2015
To simplify the math, can you take the total mass of the universe, and half the diameter of the universe, and calculate the gravity based on that? Is a single point the same as a single file line?
posted by karst at 9:02 PM on January 6, 2015
To simplify the math, can you take the total mass of the universe, and half the diameter of the universe, and calculate the gravity based on that? Is a single point the same as a single file line?
No, because the gravity falls off by the square of the distance. For example, imagine if the entire universe consisted of two Earth-like planets, one light year apart.
The force you would feel standing on one of the planets would be much greater than what you would feel a half light year away from a planet twice the Earth's in mass.
posted by justkevin at 9:17 PM on January 6, 2015
No, because the gravity falls off by the square of the distance. For example, imagine if the entire universe consisted of two Earth-like planets, one light year apart.
The force you would feel standing on one of the planets would be much greater than what you would feel a half light year away from a planet twice the Earth's in mass.
posted by justkevin at 9:17 PM on January 6, 2015
The Earth is already quite dense. A planet twice as dense as Earth would be the density of lead. Four times the density of Earth is greater than the density of uranium. At that point, you're done. No mass is going to exist at that density.
Both degenerate matter and neutronium exceed that density by a considerable margin.
Is a single point the same as a single file line?
No - distance is important, because of the inverse square law.
posted by HiroProtagonist at 9:18 PM on January 6, 2015 [1 favorite]
Both degenerate matter and neutronium exceed that density by a considerable margin.
Is a single point the same as a single file line?
No - distance is important, because of the inverse square law.
posted by HiroProtagonist at 9:18 PM on January 6, 2015 [1 favorite]
No, because the gravity falls off by the square of the distance.
But mass increases by the cube of the distance assuming uniform density. If the universe were large enough, the effect of the accumulated mass would eventually overtake the effect of the distance. However the universe is not of uniform density due to expansion, so it would ultimately depend on the rate of expansion of the universe.
posted by JackFlash at 2:10 AM on January 7, 2015
But mass increases by the cube of the distance assuming uniform density. If the universe were large enough, the effect of the accumulated mass would eventually overtake the effect of the distance. However the universe is not of uniform density due to expansion, so it would ultimately depend on the rate of expansion of the universe.
posted by JackFlash at 2:10 AM on January 7, 2015
I'm not imagining a single point, but instead I'm imagining every celestial body lined up single file with the Earth at the head of line, with each object the same distance it is from us. They'd all be directly overhead, appearing as a single point with the moon in the foreground (like the lunar eclipse of all lunar eclipses!). For arguments sake I'm standing on a bathroom scale. If I normally weigh 200 lbs, what would the scale show? 199.9?
To simplify the math, can you take the total mass of the universe, and half the diameter of the universe, and calculate the gravity based on that?
No, because the inverse square law means that the result will be completely dominated by the closest few masses.
The order you pick to put your celestial bodies in therefore becomes critically important, as does the lower limit of size at which you're willing to declare something to be a celestial body (because that affects the number of bodies you're going to have in your 14 billion light year string of celestial beads and therefore the bead spacing).
posted by flabdablet at 4:36 AM on January 7, 2015
To simplify the math, can you take the total mass of the universe, and half the diameter of the universe, and calculate the gravity based on that?
No, because the inverse square law means that the result will be completely dominated by the closest few masses.
The order you pick to put your celestial bodies in therefore becomes critically important, as does the lower limit of size at which you're willing to declare something to be a celestial body (because that affects the number of bodies you're going to have in your 14 billion light year string of celestial beads and therefore the bead spacing).
posted by flabdablet at 4:36 AM on January 7, 2015
Oh hang on. Sorry, I'd misinterpreted "with each object the same distance it is from us". So you're talking about leaving all existing objects at their existing distances from here, but moving them around to make them all lie on the same bearing as viewed from here?
If that's what you're doing, the Moon is still so much closer to us than any other significant chunk of mass that it will utterly dominate the result. The next nearest object both big enough and near enough to matter is the Sun, and even though it's twenty-seven million times as massive as the Moon, its gravitational influence is only enough to account for the difference between spring tides and neap tides.
The Moon is about 1.3 light-seconds away. The Sun is eight light-minutes away. The next nearest object of comparable mass is over four light-years away. Its gravitational influence is completely swamped by variations in the distance to the Sun during the year's orbit.
posted by flabdablet at 4:50 AM on January 7, 2015
If that's what you're doing, the Moon is still so much closer to us than any other significant chunk of mass that it will utterly dominate the result. The next nearest object both big enough and near enough to matter is the Sun, and even though it's twenty-seven million times as massive as the Moon, its gravitational influence is only enough to account for the difference between spring tides and neap tides.
The Moon is about 1.3 light-seconds away. The Sun is eight light-minutes away. The next nearest object of comparable mass is over four light-years away. Its gravitational influence is completely swamped by variations in the distance to the Sun during the year's orbit.
posted by flabdablet at 4:50 AM on January 7, 2015
I'm imagining every celestial body lined up single file with the Earth at the head of line
This would never happen. You may have noticed that objects of a certain size and density such as planets, stars, and the like are (roughly) spheres. This is because when enough mass comes together, the force of gravity is strong enough to overcome the structural forces of the mass to squish it into a (rough) sphere.
Also, it is helpful to think of gravity as emanating omnidirectional from an object's center of mass. So, lining objects up does not make a super gravity gun that pulls in a given direction like a sci fi tractor beam. The most efficient way to make a strong gravity field is to have a sphere that is as dense as possible. (this was the concept of To Crush The Moon, but that book ignored energy issues that made that concept impossible)
Both degenerate matter and neutronium exceed that density by a considerable margin.
Neutronium is a type of degenerate matter. You may have noticed my previous disclaimer that "no such material exists outside the cores of certain stars". Science books for kids talk about how "heavy" a teaspoon full of neutronium would be, but that's impossible. There is not going to be a "planet" of neutronium floating around.
posted by Tanizaki at 7:26 AM on January 7, 2015
This would never happen. You may have noticed that objects of a certain size and density such as planets, stars, and the like are (roughly) spheres. This is because when enough mass comes together, the force of gravity is strong enough to overcome the structural forces of the mass to squish it into a (rough) sphere.
Also, it is helpful to think of gravity as emanating omnidirectional from an object's center of mass. So, lining objects up does not make a super gravity gun that pulls in a given direction like a sci fi tractor beam. The most efficient way to make a strong gravity field is to have a sphere that is as dense as possible. (this was the concept of To Crush The Moon, but that book ignored energy issues that made that concept impossible)
Both degenerate matter and neutronium exceed that density by a considerable margin.
Neutronium is a type of degenerate matter. You may have noticed my previous disclaimer that "no such material exists outside the cores of certain stars". Science books for kids talk about how "heavy" a teaspoon full of neutronium would be, but that's impossible. There is not going to be a "planet" of neutronium floating around.
posted by Tanizaki at 7:26 AM on January 7, 2015
Response by poster: The point is was trying to get to was that the alignment of Jupiter and Pluto obviously isnt enough to influence gravity. It's so absurd that the alignment of everything that is isn't enough to effect gravity here on Earth. I've seen dozens of friends post a link to that article and it pains me to see that. Though floating weightless would be cool...
posted by karst at 7:49 AM on January 7, 2015
posted by karst at 7:49 AM on January 7, 2015
the alignment of Jupiter and Pluto obviously isnt enough to influence gravity
Absolutely not. I haven't run the numbers, but I'd expect to experience more gravitational anomaly by standing close to a large building than anything Jupiter or Pluto or both could ever be responsible for on Earth.
posted by flabdablet at 8:34 AM on January 7, 2015 [1 favorite]
Absolutely not. I haven't run the numbers, but I'd expect to experience more gravitational anomaly by standing close to a large building than anything Jupiter or Pluto or both could ever be responsible for on Earth.
posted by flabdablet at 8:34 AM on January 7, 2015 [1 favorite]
Here are some numbers:
Acceleration due to gravity at Earth's surface: 9.8m/s2
Mass of you: Y kg
Gravitational force of Earth on you, at surface (i.e. your weight):
Y kg × 9.8m/s2 = 9.8Y kg m/s2 = 9.8Y N
Mass of Jupiter: 1.9×1027kg
Distance to Jupiter at closest approach: 6.3×1011m
Universal gravitational constant G: 6.673×10-11Nm2/kg2
Gravitational attractive force of Jupiter at closest approach to you:
6.673×10-11Nm2/kg2 × Y kg × 1.9×1027kg ÷ (6.3×1011m)2 = 3.2×10-7Y N
As a percentage of your weight:
3.2×10-7Y N ÷ 9.8Y N × 100% = 0.0000033%
Not even going to bother with Pluto.
posted by flabdablet at 9:17 AM on January 7, 2015
Acceleration due to gravity at Earth's surface: 9.8m/s2
Mass of you: Y kg
Gravitational force of Earth on you, at surface (i.e. your weight):
Y kg × 9.8m/s2 = 9.8Y kg m/s2 = 9.8Y N
Mass of Jupiter: 1.9×1027kg
Distance to Jupiter at closest approach: 6.3×1011m
Universal gravitational constant G: 6.673×10-11Nm2/kg2
Gravitational attractive force of Jupiter at closest approach to you:
6.673×10-11Nm2/kg2 × Y kg × 1.9×1027kg ÷ (6.3×1011m)2 = 3.2×10-7Y N
As a percentage of your weight:
3.2×10-7Y N ÷ 9.8Y N × 100% = 0.0000033%
Not even going to bother with Pluto.
posted by flabdablet at 9:17 AM on January 7, 2015
In the hypothetical proposed, mass increases as the cube of the distance in the numerator, which dominates the inverse square term in the denominator if the universe is big enough.
If you cancel out terms in the universal gravitation law, the effect of distant objects, while initially tiny, increases linearly with distance in this scenario.
posted by JackFlash at 9:21 AM on January 7, 2015
If you cancel out terms in the universal gravitation law, the effect of distant objects, while initially tiny, increases linearly with distance in this scenario.
posted by JackFlash at 9:21 AM on January 7, 2015
Fair point, though I think the cube thing is a bit wonky.
For any given distance R you would have a spherical shell, with surface area proportional to R2, from which to gather the objects to be placed at distance R along your hypothetical straight bearing. So once you get more than a few tens of light years out and the crowding has a chance to set in properly, you can approximate most of your lined-up objects as a long thin continuous body whose density at any given distance R is proportional to R2.
That would put its center of mass much closer to the far end than the near end, and the far end is 14 billion light years distant.
Which means, it seems to me, that provided Herr Zebrurka's result is correct, it would still only exert a relatively tiny gravitational force on an object on the Earth's surface.
posted by flabdablet at 9:48 AM on January 7, 2015
For any given distance R you would have a spherical shell, with surface area proportional to R2, from which to gather the objects to be placed at distance R along your hypothetical straight bearing. So once you get more than a few tens of light years out and the crowding has a chance to set in properly, you can approximate most of your lined-up objects as a long thin continuous body whose density at any given distance R is proportional to R2.
That would put its center of mass much closer to the far end than the near end, and the far end is 14 billion light years distant.
Which means, it seems to me, that provided Herr Zebrurka's result is correct, it would still only exert a relatively tiny gravitational force on an object on the Earth's surface.
posted by flabdablet at 9:48 AM on January 7, 2015
For arguments sake I'm standing on a bathroom scale. If I normally weigh 200 lbs, what would the scale show? 199.9?
The important thing to consider here is that the bathroom scale is also being pulled by the alignment.
So is the Earth.
Since things fall at the same rate (acceleration, I know, I'm trying to keep this simple), you, the scale, and the Earth are all being pulled toward this super-alignment together. You would get a miniscule advantage for being closest, but the difference in distances between you and the Moon and Earth and the Moon is small compared to the total Earth/Moon distance.
posted by miguelcervantes at 10:17 AM on January 7, 2015 [1 favorite]
The important thing to consider here is that the bathroom scale is also being pulled by the alignment.
So is the Earth.
Since things fall at the same rate (acceleration, I know, I'm trying to keep this simple), you, the scale, and the Earth are all being pulled toward this super-alignment together. You would get a miniscule advantage for being closest, but the difference in distances between you and the Moon and Earth and the Moon is small compared to the total Earth/Moon distance.
posted by miguelcervantes at 10:17 AM on January 7, 2015 [1 favorite]
The point is was trying to get to was that the alignment of Jupiter and Pluto obviously isnt enough to influence gravity
There's the lede. If that was the point, this is best shown by simply doing the math. The pull of Jupiter on Earth is 34 million times less than Earth's gravity on objects at the surface. I calculated the pull of Pluto on the Earth as being 20 quadrillion times less than the pull of gravity on the surface. It is utterly negligible.
(I ended up doing the Pluto math even though you could just double Jupiter's pull and still have an utterly negligible force)
posted by Tanizaki at 10:18 AM on January 7, 2015
There's the lede. If that was the point, this is best shown by simply doing the math. The pull of Jupiter on Earth is 34 million times less than Earth's gravity on objects at the surface. I calculated the pull of Pluto on the Earth as being 20 quadrillion times less than the pull of gravity on the surface. It is utterly negligible.
(I ended up doing the Pluto math even though you could just double Jupiter's pull and still have an utterly negligible force)
posted by Tanizaki at 10:18 AM on January 7, 2015
you, the scale, and the Earth are all being pulled toward this super-alignment together.
This is kind of irrelevant, since the force between you and the Earth is what the scale is measuring, and that force does depend on the prevailing gravitational field strength. With a sufficiently large mass sufficiently close overhead, you could certainly reduce that local field strength to zero.
The point is that even shoving all the mass in the Universe around so that it lines up directly overhead does not a sufficiently large mass sufficiently close overhead make.
Now, if you could arrange for Jupiter to crash into the Earth and you stood in the impact zone, you'd probably get somewhere. You'd need a planet at least as massive as the Earth to get fully weightless a perceptible time before being crushed, though; the Moon wouldn't do.
posted by flabdablet at 10:27 AM on January 7, 2015
This is kind of irrelevant, since the force between you and the Earth is what the scale is measuring, and that force does depend on the prevailing gravitational field strength. With a sufficiently large mass sufficiently close overhead, you could certainly reduce that local field strength to zero.
The point is that even shoving all the mass in the Universe around so that it lines up directly overhead does not a sufficiently large mass sufficiently close overhead make.
Now, if you could arrange for Jupiter to crash into the Earth and you stood in the impact zone, you'd probably get somewhere. You'd need a planet at least as massive as the Earth to get fully weightless a perceptible time before being crushed, though; the Moon wouldn't do.
posted by flabdablet at 10:27 AM on January 7, 2015
If you're talking about taking everything in the universe and aligning it in a straight line, where everything stays the same distance from earth, you have to consider that as you move outward, the surface area that you're shuffling around to a single point gets larger, and larger, along with the amount of mass that you're moving around. At some point, you're going to start creating an infinite line of black holes, which will of course, all immediately start join together. So that's going to be a problem.
I think the answer is that the entire universe will almost immediately get sucked into a giant black hole.
posted by empath at 10:42 AM on January 7, 2015 [1 favorite]
I think the answer is that the entire universe will almost immediately get sucked into a giant black hole.
posted by empath at 10:42 AM on January 7, 2015 [1 favorite]
This is kind of irrelevant, since the force between you and the Earth is what the scale is measuring, and that force does depend on the prevailing gravitational field strength. With a sufficiently large mass sufficiently close overhead, you could certainly reduce that local field strength to zero.
Yes but no.
In our current Universe, when you stand on a scale, you have two forces acting on you -- the gravitational pull from the Earth and the repulsive force (normal force) of the Earth itself preventing you from falling any further. The scale is measuring the second of the two.
Now, to be weightless, you need the total force acting on you to be 0. This can be caused by a node in the gravitational potential, or by orbiting the planet, or by strapping magnets to yourself. But the key factor is F_net = 0. So, let's try this -- we put a point mass some distance away, such that it is exerting a gravitational force of 1 g on us. We pull ourselves a few rungs up a ladder, push off, and viola -- we're weightless. This object pulls us up as much as the Earth pulls us down.
But, there's a catch. The gravitational pull of that object is also going to affect Earth. It is now going to be falling in toward that object, accelerating at 1 g.* And our free-floating days are quickly over as the Earth slams into us. After dusting ourselves off, we walk over to the scale. The total gravitational force on us is now 0; however, we have a normal force caused by the Earth trying to fly through us. Since the Earth is accelerating at 1 g, we have to be accelerating at 1 g, meaning the normal force from the Earth makes the scale read -- 200 lbs.
Now, you're right, it is irrelevant, since there is no way to arrange the Universe like how the poster suggested and produce any sort of perceptible shift. However, the actual maths behind it are not the same as the standard physics of gravitation problems.
*(if we want the effective gravitational force to drop off rapidly by the time that it reaches the core of the Earth, then we ourselves are going to be sitting on an unstable point in the combined gravitational fields; if we float a bit too high, the gravity from the second object is going to quickly become stronger than that of the Earth, and we fly off into the heavens. Such a case would be interesting, but beyond the scope of this question.)
posted by miguelcervantes at 12:22 PM on January 7, 2015 [1 favorite]
Yes but no.
In our current Universe, when you stand on a scale, you have two forces acting on you -- the gravitational pull from the Earth and the repulsive force (normal force) of the Earth itself preventing you from falling any further. The scale is measuring the second of the two.
Now, to be weightless, you need the total force acting on you to be 0. This can be caused by a node in the gravitational potential, or by orbiting the planet, or by strapping magnets to yourself. But the key factor is F_net = 0. So, let's try this -- we put a point mass some distance away, such that it is exerting a gravitational force of 1 g on us. We pull ourselves a few rungs up a ladder, push off, and viola -- we're weightless. This object pulls us up as much as the Earth pulls us down.
But, there's a catch. The gravitational pull of that object is also going to affect Earth. It is now going to be falling in toward that object, accelerating at 1 g.* And our free-floating days are quickly over as the Earth slams into us. After dusting ourselves off, we walk over to the scale. The total gravitational force on us is now 0; however, we have a normal force caused by the Earth trying to fly through us. Since the Earth is accelerating at 1 g, we have to be accelerating at 1 g, meaning the normal force from the Earth makes the scale read -- 200 lbs.
Now, you're right, it is irrelevant, since there is no way to arrange the Universe like how the poster suggested and produce any sort of perceptible shift. However, the actual maths behind it are not the same as the standard physics of gravitation problems.
*(if we want the effective gravitational force to drop off rapidly by the time that it reaches the core of the Earth, then we ourselves are going to be sitting on an unstable point in the combined gravitational fields; if we float a bit too high, the gravity from the second object is going to quickly become stronger than that of the Earth, and we fly off into the heavens. Such a case would be interesting, but beyond the scope of this question.)
posted by miguelcervantes at 12:22 PM on January 7, 2015 [1 favorite]
In addition to what miguelcevantes said, keep in mind that the Earth is held together by gravity, too, so if you are floating, so is the massive amount of rock that you are standing on. And since that means both you and the rock now are accelerating away from the Earth's center rather quickly and things are pretty much literally about to go all pear-shaped.
posted by Zalzidrax at 1:05 PM on January 7, 2015
posted by Zalzidrax at 1:05 PM on January 7, 2015
Forget the numbers. Forget the fancy physics equations.
Gravity is what pulls you in some direction. You are being pulled in a direction right now. If it is towards the center of the Earth, then the odds that anything other than the center of the Earth is seriously affecting the gravity you are in is incredibly small.
So, either we happen to be located, right now, at the dead center of mass of the universe itself - and what are the odds of that miraculous coincidence? (Hint: we aren't), or the Earth is 99.99% of the gravity we're experiencing, and the rest barely makes any difference at all.
In fact, the next-biggest source of gravity, the Moon, only has enough effect to move all the water in all the oceans on Earth a few feet. Remember that the oceans range up to thousands of feet deep; that's not a lot of movement.
I'm doing some hand-waving here with physical principles ("center of mass of the Universe" would make Heisenberg throw things at me, and gravitational force is proportional to M1*M2...) - but the basic idea is sound: if all the rest of the mass would pull us noticeably, it would already be noticeable.
posted by IAmBroom at 1:18 PM on January 7, 2015
Gravity is what pulls you in some direction. You are being pulled in a direction right now. If it is towards the center of the Earth, then the odds that anything other than the center of the Earth is seriously affecting the gravity you are in is incredibly small.
So, either we happen to be located, right now, at the dead center of mass of the universe itself - and what are the odds of that miraculous coincidence? (Hint: we aren't), or the Earth is 99.99% of the gravity we're experiencing, and the rest barely makes any difference at all.
In fact, the next-biggest source of gravity, the Moon, only has enough effect to move all the water in all the oceans on Earth a few feet. Remember that the oceans range up to thousands of feet deep; that's not a lot of movement.
I'm doing some hand-waving here with physical principles ("center of mass of the Universe" would make Heisenberg throw things at me, and gravitational force is proportional to M1*M2...) - but the basic idea is sound: if all the rest of the mass would pull us noticeably, it would already be noticeable.
posted by IAmBroom at 1:18 PM on January 7, 2015
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posted by DoubleLune at 8:37 PM on January 6, 2015