I need to be able to turn individual people's gravity off

Question

I'm writing a book to redefine the horsemen of the apocalypse. My first book concentrated on a new "Death", who reveals that there are in fact 7 "horsemen", not 4 - war, death, pestilence, famine, mania, nature and rot. I have War as an vibration which travels through vibrations, sound waves etc. I'm slowly introducing nature, and want one of her abilities to be the control or manipulation of carbon into other forms. I also want her to be able to control natural disasters (or start them) and add to that a couple of new things. To that end, is it possible to effectively negate a person's gravity, to make them "fall" off the planet? The how of it is important because I will be detailing the perspective of the first victim as they experience what will happen to them. Thank you for any thoughts on this. I need the planet not to explode, I have been told that messing with the gravity of the planet, in specific areas, causes a planet wide catastrophe - so that won't work. Thanks for your time!

Answer 1

To that end, is it possible to effectively negate a person's gravity, to make them "fall" off the planet?

Using magic? Sure, anything is possible.

Without using magic? No.

The how of it is important because I will be detailing the perspective of the first victim as they experience what will happen to them.

In the real world inertial mass and gravitational mass are indistinguishable and inseparable, but that doesn't need to be a problem fdor you.

Magically set their gravitational mass to zero, so that gravitational forces can't act upon them. Their inertial mass must remain the same, or they'll instantaneously accelerate to lightspeed with all the badness that entails. When their inertia stays the same, all other forces act upon them as normal. Some magical material that shields from the effect of gravity would also work (the classic example would be cavorite, of course), but being put in a death-box is perhaps less dramatic than simply flying up into space. Once gravity has ceased holding them to the Earth, there are various things that will fling them into space: the rotation of the Earth about its axis, the rotation of the Earth about the Sun, and the victim's bouyancy in air. Obviously they can also push themselves off the ground (eg. by trying to run away) which will speed things up a bit, but I'll ignore that factor.

The effect will be to ultimately fling them into interstellar space, but initially they'll just appear to rise slowly from the ground, and accelerate slowly into the sky where they'll freeze to death or asphyxiate after a short period of time.

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Centrifugal acceleration (which I'd originally and mistakenly called "centripetal", thanks for the correction by Nuclear Hoagie) is a fictitious force that appears to push an object in a rotating reference frame (eg. on the surface of the Earth) away from the centre of rotation (eg. the Earth's axis).

Centrifugal acceleration can be calculated using $a_c = \omega^2r$ where $\omega$ is the angular velocity of the thing in question. At Earth's equator, $\omega$ is 2π radians per day, or ~7.27 x 10^-5 radians per second. $r$ is ~6378.1km, giving an apparent upward acceleration of 0.034m/s², or about 3 milligees relative to your point of departure. I say "apparent" because they're not really accelerating upwards, but shooting off the surface of the Earth at a tangent at about 460m/s, but because their point of departure is curving away from them as Earth rotates it will appear that they're accelerating upwards. There's a worked example at the end showing the effect and the difference.

As you approach a pole, the apparent acceleration will reduce (proportional to the cosine of the latitude) and the direction will be more towards the horizon. At my latitude, ~52°N, the force will be reduced to ~62%, or ~2 milligees, and the victim would appear to be taking off at an angle of ~48° from vertical, pointing towards the south. Their tangential velocity will also change... $v_t = \omega r \cos{\lambda}$ where $\lambda$ is the latitude, 0° being the equator.

There's an additional centrifugal effect caused by the Earth's rotation about the Sun, where $r$ is 1AU and $\omega$ is 2π radians per year. This will ultimately manifest as a 26km/s trajectory out of the solar system and into interstellar space, but its immediate effect is small: at midnight on the equator it gives you an apparent upward boost of 0.006m/s², and at midday it will reduce your apparent acceleration by the same amount, but in either case its contribution at the beginning is small. It will have the same magnitude regardless of latitude, though, and add or remove about half a milligee. At other times you'll need to do a vector sum, but for a first approximation you can ignore it.

The final effect will be bouyancy (well spotted Qami!) as although the volume of the person has not changed, there's no longer any opposition to their bouyancy in air, so they will rise as if they were a vacuum balloon. The acceleration exerted by bouyancy in the air is $a_b = g\frac{\rho_f}{\rho_v}$ where $g$ is the acceleration due to gravity, $\rho_f$ is the density of air around the victim (~1.2754 kg/m³ at STP) and $\rho_v$ is the density of the victim (say, 1000 kg/m³). This gives an initial ~0.013m/s² vertical acceleration regardless of time of day or latitude, and decreases with altitude as air density decreases. That's another milligee and a bit.

Lets assume then that this is occuring on the equator (for maximum centrifugal force) and we're ignoring the effects of Earth's orbit. You'd get an apparent initial acceleration of about 4 milligees, which is pretty sedate. Bouyancy forces will accelerate them straight up, and their tangential velocity will carry them up-ish. The victim will rise up, and can reasonably be assumed to be dead by the time they've passed 8000m altitude, the conveniently named Death Zone, where they will asphyxiate if they haven't already succumbed to hypothermia (and if they took off from the equator, they might not be dressed for freezing temperatures).

The exactly rate of climb is awkward to compute, but you can reasonably assume that in about 10 minutes (give or take), they'll be dead or dying. In the immediate short term though, the acceleration is surprisingly slow... after 10 seconds, they'll have risen less than 2m, and so could be caught by a friend.

They'll be experiencing free-fall as soon as their gravitational mass goes missing, so they might be sick. The vomit will experience the same forces, so unless it was expelled quite forcefully it'll stay with them on their way up.

Peter LeFanu Lumsdaine suggested that it might be possible to swim back down to the surface, which I suspect would be impossible for someone not equipped with suitable things to flap... the possibility exists though that someone with a couple of big fans or suitably stiff, lightweight boards might be able to fly back down and grab on for safety. Interesting thought though, certainly!

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For anyone who cares, you can see that the centrifugal force, whilst fictitious, isn't wrong, and you can approximate your flight path by using the intial centrifugal acceleration. Here's a slice though the Earth at the equator. The Earth's axis of rotation is the centre of the circle, and the Earth is rotating clockwise.

$T$ is the trajectory that the victim will be travelling on, neglecting bouyancy effects. $R$ is the radius of the earth, and $A$ is the altitude of the victim, who is assumed to be at the far apex of the triangle. The length of $T$ will be the tangential velocity $v$ of the victim (~460m/s at the equator) multiplied by the flight time, $t$. You can compute $A = \sqrt{R^2 + v^2t^2} - R$. After 700 seconds, the victim's actual altitude will be ~8123m above where they started, but using the initial centrifugal acceleration $c_a$ of 0.034m/s² and $A = \frac{c_at^2}{2}$ you'd get more like 8263m which is close enough for government work. As $t$ increases, the less this lazy approximation works, but the victim will be dead by then so it doesn't really matter!

Similarly, the rate of altitude gain (or vertical speed) done using $\frac{dA}{dt} = c_at$ gives you a speed at 700 seconds of ~23.6m/s, whereas the correct derivative $\frac{dA}{dt} = \frac{v^2t}{R^2+v^2t^2}$ gives you ~23.2m/s.

Including the bouyancy acceleration vector (which always points away from the centre of the Earth and reduces in magnitude with altitude) and drag (which you can't really neglect after a couple of minutes, and acts in opposition to the velocity vector whose direction continually changes with time and also reduces with altitude) is left as an exercise for the reader.

Answer 2

If you completely negate someone's mass, then the only thing keeping them on the ground is air resistance. And not for long.

For the sake of argument, imagine not someone that has zero mass, but someone who has so little mass that it approaches zero. That person will have very high buoyancy, even in air. Remember when Lawnchair Larry gained altitude "as if shot by a cannon"? That person would go up faster than that. Might already become a meteor in the way up to the stratosphere. And I say meteor in the meteorological sense, i.e.: something burning up in the atmosphere as it moves really fast.

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Now let's really nerd this up. A consequence of special relativity is that massless particles must move at the speed of light. As soon as Nature takes away someone's mass, they will move at $c$. That is five orders of magnitude greater than both the Earth's escape velocity at sea level, and the Sun's escape velocity at Earth's orbit. The person will immediately start leaving the loving solar system. If they don't move up straight away they will just bounce on the ground and then go up anyway.

Also all the particles in the person's body will each go their own way, so you don't really have a person anymore.

Regardless, the only visible effect for regular humans is that the victim seems to disappear instantly. Along with everyone and everything around them. And an instant flash many times brighter than the Sun. Also if done in an atmosphere, you will see a mushroom cloud. As some colleagues mention in comments, turning mass into photons would likely release amounts of energy measured in the way we measure the output of nukes. If you want something less fatal, have Nature reduce someone's mass instead of removing it completely.

Answer 3

Replace "fall" with "sink" or transition to another realm.

To that end, is it possible to effectively negate a person's gravity, to make them "fall" off the planet? The how of it is important because I will be detailing the perspective of the first victim as they experience what will happen to them

The effects on the person of "falling" off the Earth will be dependent on what exactly "falling" means. I think you mean "lifting", whereas I would suggest another approach is sinking.

The energy required to lift someone is not enormous - we're talking flying here - it's hot air balloon territory. So, yes, you can make them fly. The problem is getting them to keep going past the atmosphere. You are now moving into escape velocity territory which require a lot of energy.

Also it is not clear to me why you need that. At about the 10,000 feet mark the air density drops to the point that humans die, pretty quickly (minutes or seconds) from hypoxia (no air is bad !). They will not experience anything after that, whatever happens their bodies. Planes have rather a lot of equipment dedicated to preventing this, but an unprotected human without an air supply is dead quickly.

So after 10,000 feet (or a bit before if you want them alive and alert !) you can have them transition to another realm (Hades, Disneyland, whatever). They are now out of the picture. Pushing them up to 10,000 feet is relatively small potatoes on the energy front and can be done gradually so they do not suffer injury from rapid acceleration (so just terror).

Another possibility is burial. Let's open a hole in the ground and just drop them in and cover them up. They won't be coming back. What happens when they are out of view is up to you.

is it possible to effectively negate a person's gravity

Not strictly speaking, but we routinely fly so finding a method to apply force to someone and push (lift) them up is easy. Your basic creature from beyond should have no problem magicing this without actually doing much damage to local physics on the way. No magic (well, modest amounts), no weird breaking the laws of physics (kind of), just a few writhing tentacles and a portal up there at about the 10,000 feet mark. Practically the same as any day in the Bermuda Triangle. :-)

Maybe the realm "above" (or an entrance to it) drops down tentacles to drag the lucky traveler up. Can't go without tentacles in a story like this, IMO.

Answer 4

Just use this warp drive I have lying around (some assembly needed)

You probably think I'm kidding, but here's the paper! It actually passed peer review and was published in Classical and Quantum Gravity. You just expand space under the mark and contract it on top of him and before long he has free fallen right out into space.

True, it takes an amount of energy equal to a tenth of the mass of the sun, but he thinks he can cut it down by 10^30 (yes, to the 30th!). He just needs some computer simulations ... and maybe some security protection against hostile aliens from the 27th century.

Answer 5

For a pseudo-science and handwavium answer without outright calling it magic, you could have the power's target no longer interact with gravitons from any distance over 10 feet or so.

So, only being attracted to the portion of Earth within 10 feet, gravitational attraction is now minimal, allowing the target to push themselves away from the ground with their next footstep.

I'll leave it as an exercise to the reader how this person would interact with air, wind and jetstreams, and whether he would be able to rise above the atmosphere.

Answer 6

Anti-gravity is hard, but LIFT is quite easy

Birds, planes, balloons, and rockets fall upward all the time.

As far a physics is concerned, the sum of two forces is essentially the same as a single force. So, instead of making G=-9.8m/s^2 you can leave G=9.8m/s^2 and add a second force that is 19.6m/s^2 in the opposite direction. To send an average sized man falling upwards at the speed of gravity, you simply need to exert a constant force of 1,215.2 N.

The hard part is doing this with your mind, but you already seem prepared to handwave this part anyway. So while your horseman appears to be reversing gravity, he/she is really just pushing the victims up into space using telekinesis. They will accelerate upwards very quickly at first until they hit their terminal velocity, and then as they get higher and higher air resistance decreases allowing their terminal velocity to increase. Once they hit about 10km in altitude, the air with start to thin out very quickly causing them to die of hyperthermia and/or asphyxiation pretty quickly. Once they get high enough, you can stop lifting them and let them start falling again. The atmosphere will make sure nothing is left of them to return to Earth.

Answer 7

Gravity is a distortion of the fabric of spacetime. Spacetime is distorted by all objects with mass, especially dense objects with a lot of mass, such as planets. Gravity as we experience it is a result of our planet’s immense mass tugging on our own bodies’ masses. So no, it is not possible to negate an individual’s gravity, as the tug of the gravity we all feel is that of our planet’s immense mass and the gravity well it creates. The only thing that wouldn’t feel gravity is something with zero mass. Humans can’t have zero mass without some sort of magic that turns them into pure light or something, and somehow keeps them alive.

tl;dr nope, you need magic.

Answer 8

This is something I only looked at very briefly in undergrad, so, if you care about scientific accuracy, double check everything. However, there is something called anti-matter which *I think" has negative mass. This means it would do exactly as you say - fall away from a planet. Another way to say it would be that the gravitational interaction between anti-matter and (normal)-matter is a push rather than a pull.

This is convenient in that you can get the effect you're looking for but (1) you don't have to change anything about earth's gravitational field and (2) if you want to weaken the affect, you could just stick a person with some anti-matter that has less mass than them. If the anti-matter is 1/2 the target's mass, but is, say, stuck to them, then they will only be accelerated upward 1-1/2=1/2 times as fast as we normally fall down, for instance.

As far as how this would be experience - the matter/anti-matter relationship would be almost identical to the interaction between positive/negative ends of a magnet except that opposites repel in this case. In fact, the mathematical descriptions of (Newtonian) gravity and electromagnetic forces are strikingly similar. The actually motion would be like any other acceleration - think of falling or accelerating in a sports car.

Answer 9

Swap the victim's gravitational mass to negative.

A person's weight is the result of Newton's formula for gravity: f = GMm/R^2. If one of the masses has a negative value then the force will be negative also. If the person's inertial mass remains positive then this will act to repel them from the planet, accelerating them upwards.

This is of course impossible under known physics, but is at least a logical way for you magic to work.