I'm going to throw you right into the sun.
April 15, 2009 4:26 PM Subscribe
Fictional gravity: Let's say I was the complete master of all things gravity and I wanted to hurl a person into the sun.
Stipulations: I am in complete, omnipotent, magical control of gravity. I can increase and decrease gravity at will, relative to all objects, across all distances.
I want to hurl a person into the sun. Specifically, I want to increase the gravitational attraction between the sun and a specific person, such that the person is lifted off the ground, achieves escape velocity and eventually is pulled right into the sun.
When I do this, it will not have an effect on anything besides the targeted person and the sun. No other objects will have their gravity tweaked with.
Time is a factor, inasmuch as I'd like this person to be leaving the breathable atmosphere within ... oh, let's say 10 minutes after I flip the switch. Think of it like a survivable rocket launch (until he runs out of oxygen, that is). Beyond that, it could take forever for him to actually reach the sun. But he's out of eyeball range pretty damn quickly.
This will happen at noon for the targeted person, outside, so there's a straight line from him to the sun. He just goes straight up into the air and doesn't need to be pulled through walls, ceilings or the planet itself.
What is the minimum amount of gravitational increase, in percentage, to do this?
Stipulations: I am in complete, omnipotent, magical control of gravity. I can increase and decrease gravity at will, relative to all objects, across all distances.
I want to hurl a person into the sun. Specifically, I want to increase the gravitational attraction between the sun and a specific person, such that the person is lifted off the ground, achieves escape velocity and eventually is pulled right into the sun.
When I do this, it will not have an effect on anything besides the targeted person and the sun. No other objects will have their gravity tweaked with.
Time is a factor, inasmuch as I'd like this person to be leaving the breathable atmosphere within ... oh, let's say 10 minutes after I flip the switch. Think of it like a survivable rocket launch (until he runs out of oxygen, that is). Beyond that, it could take forever for him to actually reach the sun. But he's out of eyeball range pretty damn quickly.
This will happen at noon for the targeted person, outside, so there's a straight line from him to the sun. He just goes straight up into the air and doesn't need to be pulled through walls, ceilings or the planet itself.
What is the minimum amount of gravitational increase, in percentage, to do this?
Response by poster: But there's a straight line from them to the sun no matter when you fling them, except when it's night.
What I mean is that, when I flip the switch, they're not going to be pulled through the earth or anything else in order to take the shortest possible distance to the sun.
then just figuring out how fast they have to go to cross an AU in ten minutes.
No! It's how fast to cross out of the breathable atmosphere (~25,000 feet) in 10 minutes.
posted by Cool Papa Bell at 4:34 PM on April 15, 2009
What I mean is that, when I flip the switch, they're not going to be pulled through the earth or anything else in order to take the shortest possible distance to the sun.
then just figuring out how fast they have to go to cross an AU in ten minutes.
No! It's how fast to cross out of the breathable atmosphere (~25,000 feet) in 10 minutes.
posted by Cool Papa Bell at 4:34 PM on April 15, 2009
Oh.
Well, once you get out of the atmosphere, it's just creating enough thrust in the right directions to reach the sun.
How to get out of the breathable atmosphere - well, the acceleration of gravity is 9.8 m/s2. Wikipedia says escape velocity is 11.2 km/s, which translates into about 8 miles per second.
25,000 feet is about five miles.
posted by kldickson at 4:37 PM on April 15, 2009
Well, once you get out of the atmosphere, it's just creating enough thrust in the right directions to reach the sun.
How to get out of the breathable atmosphere - well, the acceleration of gravity is 9.8 m/s2. Wikipedia says escape velocity is 11.2 km/s, which translates into about 8 miles per second.
25,000 feet is about five miles.
posted by kldickson at 4:37 PM on April 15, 2009
The problem is, when standing on Earth, a person is moving around the sun at all times; even if you stopped the Earth's gravity and increased the Sun's gravity, the person at noon will begin a long spiral inwards to the Sun. Their orbital motion would have to stop as well, which means that there will need to be some force off-angle to that sun-earth line to affect the person's orbit. Think of it like when somebody on the space station drops their tools. The astronaut has negligible gravity, the Earth still has about the same amount of strength in orbit, but the bag os moving so fast in orbit that it won't fall until it can slow down enough. The Earth is the astronaut, the man is the toolbag, and he'll continue to orbit until he doesn't have enough speed to counteract gravity.
posted by AzraelBrown at 4:38 PM on April 15, 2009 [2 favorites]
posted by AzraelBrown at 4:38 PM on April 15, 2009 [2 favorites]
I'm not really sure what you even mean here. Are you talking about just changing the gravitational strength between any two objects? If that's the case I guess you could do two things.
1) Increase the gravity between the person and some other planet or maybe the moon to knock them off the planet and into space, contrary to earth's orbit around the sun.
2) simply create a negative gravity between them and the earth, ejecting them from the planet.
You can't just increase the gravitational pull between the sun and the earth because they would already be in orbit around the sun (in earth's orbit). You have to get them out of earth's around the sun. As long as they keep going the same speed as the earth, they'll orbit the sun forever. You have to slow them down enough to fall into the sun.
posted by delmoi at 4:49 PM on April 15, 2009
1) Increase the gravity between the person and some other planet or maybe the moon to knock them off the planet and into space, contrary to earth's orbit around the sun.
2) simply create a negative gravity between them and the earth, ejecting them from the planet.
You can't just increase the gravitational pull between the sun and the earth because they would already be in orbit around the sun (in earth's orbit). You have to get them out of earth's around the sun. As long as they keep going the same speed as the earth, they'll orbit the sun forever. You have to slow them down enough to fall into the sun.
posted by delmoi at 4:49 PM on April 15, 2009
When I do this, it will not have an effect on anything besides the targeted person and the sun. No other objects will have their gravity tweaked with.
Okay I missed this. Anyway, there is simply no way to do this just by modifying the gravitational pull between the sun and the object. It would have no effect at all.
When you take an empty balloon and fill it with water, you increase it's gravitational pull thousands of times, but does it fall to the sun? No, of course not, because it's already in orbit around the sun.
posted by delmoi at 4:53 PM on April 15, 2009
Okay I missed this. Anyway, there is simply no way to do this just by modifying the gravitational pull between the sun and the object. It would have no effect at all.
When you take an empty balloon and fill it with water, you increase it's gravitational pull thousands of times, but does it fall to the sun? No, of course not, because it's already in orbit around the sun.
posted by delmoi at 4:53 PM on April 15, 2009
Upon thinking of this, it is possible by adjusting just the Sun's gravity -- the problem is the speed of Earth's orbit, not Earth's gravity at all. When I stand here and drop a ball, it falls an apparently 'straight' line downwards, even though it it slightly curved because of its motion on larger scales. if I throw a ball, it is travelling faster perpendicular to the Earth's gravitational pull, so it has an apparent spiral, but the slower I throw it, the more 'vertical' it is, more apparently straight towards the Earth. The key, if you want somebody to "appear" to fly straight up, is to increase the Sun-person gravitiational pull by an unbelievably immense degree, so that the person's motion at an orbit of 1AU appears to be an negligibly slight curve. The number will be immense, because 1AU is a long ways, the person's orbital speed is extremely high, but it can be done. The math has nothing to do with Earth's escape velocity -- it has to do with increasing gravity enough to defeat the person's existing speed, which is equal to his escape velocity at an orbit of 1AU.
posted by AzraelBrown at 4:59 PM on April 15, 2009
posted by AzraelBrown at 4:59 PM on April 15, 2009
I think it's a simpler problem than that. I think you just need to increase the sun's mass until the gravitational attraction felt by the man at his current location (distance from the sun) is greater than that felt by the earth. He would have no choice but to start moving toward the sun. The astronaut dropping a tool isn't the same thing because the mass of the earth isn't changing.
The mass increase would be HUGE, though, I suspect - isn't gravitational acceleration related to distance in an inverse square relationship? It's been too long for me to remember any of the calculations.
posted by ctmf at 5:02 PM on April 15, 2009
The mass increase would be HUGE, though, I suspect - isn't gravitational acceleration related to distance in an inverse square relationship? It's been too long for me to remember any of the calculations.
posted by ctmf at 5:02 PM on April 15, 2009
Also, I'm to lazy to do the math right now but according to this the sun actually exerts a greater gravitational pull on the moon then the earth does. I would imagine that being on the surface of the earth (6.3km from the center, which is where you count from) the gravitational attraction would still be greater from an object and the sun then an object and the earth.
posted by delmoi at 5:05 PM on April 15, 2009
posted by delmoi at 5:05 PM on April 15, 2009
Rough estimate using Newton's law:
Mass of earth: 5.9742 * 10^24 kg
Mean radius of earth: 6.371 * 10^6 m
Mass of sun: 1.9891 * 10^30 kg
Distance from earth to sun: 1.496 * 10^11 m
Let's assume that the mass of your enemy is 80 kg, and that he is standing on the point on the earth's surface nearest the sun. We'll ignore the miniscule distance between his center of mass and the earth's surface, and we'll also assume that the earth-sun distance is the distance between him and the center of the sun. (I'm not sure if the given figure includes the radii of the earth and/or sun, but at that scale it won't make much difference.)
The force of attraction between your enemy and the earth is:
(G * 80 * 5.9742e24) / (6.371e6)^2 ~= 780 N
And the force between him and the sun:
(G * 80 * 1.9891e30) / (1.496e11)^2 ~= 0.47 N
...so you will need about 1660 times the present attraction between him and the sun.
posted by equalpants at 5:13 PM on April 15, 2009 [2 favorites]
Mass of earth: 5.9742 * 10^24 kg
Mean radius of earth: 6.371 * 10^6 m
Mass of sun: 1.9891 * 10^30 kg
Distance from earth to sun: 1.496 * 10^11 m
Let's assume that the mass of your enemy is 80 kg, and that he is standing on the point on the earth's surface nearest the sun. We'll ignore the miniscule distance between his center of mass and the earth's surface, and we'll also assume that the earth-sun distance is the distance between him and the center of the sun. (I'm not sure if the given figure includes the radii of the earth and/or sun, but at that scale it won't make much difference.)
The force of attraction between your enemy and the earth is:
(G * 80 * 5.9742e24) / (6.371e6)^2 ~= 780 N
And the force between him and the sun:
(G * 80 * 1.9891e30) / (1.496e11)^2 ~= 0.47 N
...so you will need about 1660 times the present attraction between him and the sun.
posted by equalpants at 5:13 PM on April 15, 2009 [2 favorites]
Mr. F thinks you could probably just reverse the effects of gravity on your target, causing him or her to fall up at terminal velocity (adjusted for lessening air resistance). He also thinks this may be faster than you want your victim to fall.
posted by fairytale of los angeles at 5:14 PM on April 15, 2009
posted by fairytale of los angeles at 5:14 PM on April 15, 2009
Essentially, all you need to worry about is attaining escape velocity and then just figuring out how fast they have to go to cross an AU in ten minutes.
By definition, it takes something with a velocity of C about eight minutes to do this.
posted by ricochet biscuit at 5:16 PM on April 15, 2009
By definition, it takes something with a velocity of C about eight minutes to do this.
posted by ricochet biscuit at 5:16 PM on April 15, 2009
...until the gravitational attraction felt by the man at his current location (distance from the sun) is greater than that felt by the earth. He would have no choice but to start moving toward the sun.
That might not be enough. He wouldn't fall directly into the sun, he'd go into an elliptical orbit with the sun at one focus.
You'd have to use a force great enough so that the perihelion of the orbit was smaller than the sun's radius. Otherwise he'd just orbit forever without ever making contact.
posted by Chocolate Pickle at 5:18 PM on April 15, 2009 [3 favorites]
That might not be enough. He wouldn't fall directly into the sun, he'd go into an elliptical orbit with the sun at one focus.
You'd have to use a force great enough so that the perihelion of the orbit was smaller than the sun's radius. Otherwise he'd just orbit forever without ever making contact.
posted by Chocolate Pickle at 5:18 PM on April 15, 2009 [3 favorites]
I think this does make sense:
He's effectively hypothesizing the creation of a fifth fundamental force, call it gravity-prime. Gravity-prime has exactly the same behavior as gravity, it obeys the inverse squared law and its strength is relative to "mass" and all, but instead of its coefficient being G it has a different coefficient G'. And it only affects two bodies in the universe, the Sun and the victim.
So you'd set up an equation such that the sum of the forces on the victim due to both gravity and gravity-prime from the Sun and the Earth results in the victim accelerating toward the Sun, crossing 25,000 feet in 10 minutes. The only unknown would be G', so you could solve for it.
His particular question, the percentage increase, could then be calculated from the ratio of the sum of G and G' to G alone.
But I don't feel like doing the actual math.
Also, I don't think the orbits would matter, would they? Any velocity from the orbits would be perpendicular to movement towards the Sun, if this happened at noon. If you were going to be picky I would expect that air resistance would be a more significant force than anything having to do with orbits.
posted by XMLicious at 5:18 PM on April 15, 2009
He's effectively hypothesizing the creation of a fifth fundamental force, call it gravity-prime. Gravity-prime has exactly the same behavior as gravity, it obeys the inverse squared law and its strength is relative to "mass" and all, but instead of its coefficient being G it has a different coefficient G'. And it only affects two bodies in the universe, the Sun and the victim.
So you'd set up an equation such that the sum of the forces on the victim due to both gravity and gravity-prime from the Sun and the Earth results in the victim accelerating toward the Sun, crossing 25,000 feet in 10 minutes. The only unknown would be G', so you could solve for it.
His particular question, the percentage increase, could then be calculated from the ratio of the sum of G and G' to G alone.
But I don't feel like doing the actual math.
Also, I don't think the orbits would matter, would they? Any velocity from the orbits would be perpendicular to movement towards the Sun, if this happened at noon. If you were going to be picky I would expect that air resistance would be a more significant force than anything having to do with orbits.
posted by XMLicious at 5:18 PM on April 15, 2009
By definition, it takes something with a velocity of C about eight minutes to do this
It's true that it's about 8 minutes (plus or minus; varies depending on the time of year) but this is not a matter of definition.
posted by Chocolate Pickle at 5:19 PM on April 15, 2009
It's true that it's about 8 minutes (plus or minus; varies depending on the time of year) but this is not a matter of definition.
posted by Chocolate Pickle at 5:19 PM on April 15, 2009
He would have no choice but to start moving toward the sun. The astronaut dropping a tool isn't the same thing because the mass of the earth isn't changing.
The requirement is that the person start heading straight into the sun; Even at large multiples of increasing the sun's gravity, the person will begin to lift up off the Earth, the Earth will continue to rotate underneath them, and they will continue in orbit around the Sun. When an astronaut drops a tool, it doesn't fall straight down towards the Earth; even if the astronaut pushed it into a lower orbit, it wouldn't fall straight down. The tool would have to be very close to the Earth to have an apparent "straight-down" motion, because it's moving so fast compared to the surface of the Earth.
Or, envision this: A bullet is fired from a gun parallel to the apparent surface of the Earth; how much would gravity have to increase for the bullet to go straight into the ground after leaving the muzzle? It's possible, but it's an extremely high multiple to cause such a large change in motion so quickly. That person destined for the sun, whether they're stuck to the surface of the Earth or Earth's gravity is no factor, is a speeding bullet passing around the sun. That motion and energy needs to be counteracted somehow.
I suppose, if the requirement is that the person apparently heads "straight" into the sun, meaning other people in the same orbit will see the person appear to head towards the sun in a more-or-less vertical direction, it can happen at any speed; if you want to make the person cross 5 miles of atmosphere in 10 minutes, the speed is 30mph, so increase the sun's gravity to cause an appropriate speed. It won't be 'straight' into the sun on a solar-system scale, but that may be the fun of the exercise.
posted by AzraelBrown at 5:22 PM on April 15, 2009
The requirement is that the person start heading straight into the sun; Even at large multiples of increasing the sun's gravity, the person will begin to lift up off the Earth, the Earth will continue to rotate underneath them, and they will continue in orbit around the Sun. When an astronaut drops a tool, it doesn't fall straight down towards the Earth; even if the astronaut pushed it into a lower orbit, it wouldn't fall straight down. The tool would have to be very close to the Earth to have an apparent "straight-down" motion, because it's moving so fast compared to the surface of the Earth.
Or, envision this: A bullet is fired from a gun parallel to the apparent surface of the Earth; how much would gravity have to increase for the bullet to go straight into the ground after leaving the muzzle? It's possible, but it's an extremely high multiple to cause such a large change in motion so quickly. That person destined for the sun, whether they're stuck to the surface of the Earth or Earth's gravity is no factor, is a speeding bullet passing around the sun. That motion and energy needs to be counteracted somehow.
I suppose, if the requirement is that the person apparently heads "straight" into the sun, meaning other people in the same orbit will see the person appear to head towards the sun in a more-or-less vertical direction, it can happen at any speed; if you want to make the person cross 5 miles of atmosphere in 10 minutes, the speed is 30mph, so increase the sun's gravity to cause an appropriate speed. It won't be 'straight' into the sun on a solar-system scale, but that may be the fun of the exercise.
posted by AzraelBrown at 5:22 PM on April 15, 2009
equalpants, didn't you just calculate what it would take to balance the force from the Sun and the Earth? But that would just result in the victim becoming weightless and we need him to accelerate towards the Sun.
posted by XMLicious at 5:24 PM on April 15, 2009 [1 favorite]
posted by XMLicious at 5:24 PM on April 15, 2009 [1 favorite]
One can have a ladder theoretically long enough to simply climb away. Therefore, escape velocity is only relevant as an idea of how fast you must be going if you jump away, without additional thrust.
I'm working with the idea that the Hurlee (more of a Driftee) is still within the gravity well of the Earth, and therefore going along with the Earth as it orbits around the Sun. It makes a bit more sense, since we're using gravity and not fixing them through an absolute point (relative to what?) in spacetime, or suddenly disconnecting them from Earth's gravity.
The acceleration due to gravity of an object upon the surface of the Earth is 9.8 m/s2 down. As soon as you increase the pull between the Hurlee and the Sun to a counter 9.8 m/s2 up (where "up" is a straight line proceeding from the center of the Earth, to the Hurlee, and on to the center of the Sun), they would be effectively weightless.
After that, they drift towards the Sun, oh so gently, perhaps helplessly, screaming. So, anything more than 9.8 m/s2 sets them on their way. But we have an additional constraint, leaving the atmosphere in 10 minutes. Now, the definition of "atmosphere" is pretty tenuous *rimshot*, so you have a lot of ways to define it, but let's get breathable in there. I'll take Mt. Everest as a fairly arbitrary point of reference, which has a height above sea level of of 29,029 feet.
Let us assume your hapless Hurlee is at sea level. Then, we must get the Hurlee from sea level to the top of Everest in ten minutes, which is 600 seconds. I'd like to neglect air resistance.
Now we have numbers:
d = 8,848 meters.
t = 600 seconds.
a = an additional acceleration, in m/s2, against our 9.8 2.
The most simplistic version of this looks like
d = 1 / 2 * a * t2.
Solving for a, we get a = 2 * d / t2.
Making our substitutions, we see
a = 2 * 8,848 / 6002
a = 0.04915 m/s2
So, the counter pull from the Sun would be 9.80665 m/s2 + 0.04915 m/s2.
Set your Master of Gravity beam for 9.855805bar m/s2. As a percentage, that would be -100.5012%. -100% to counteract the Earth's gravity, and the 0.5012% to get them out of breathable atmosphere in ten minutes.
So, should we care about air resistance? Maybe. The air resistance is more complex, and goes up proportionally both to velocity and the square of velocity.
v = a * t, so the Hurlee would be going at about 29.49 m/s at the point that they are level with the tip of Everest. That's about 65 mph, but only at the end of the trip. That's getting close to the terminal velocity of a human being plummeting through the atmosphere, arms spread wide.
So, yes, we'd probably want to look at air resistance if you cared much about that level of accuracy.
posted by adipocere at 5:24 PM on April 15, 2009
I'm working with the idea that the Hurlee (more of a Driftee) is still within the gravity well of the Earth, and therefore going along with the Earth as it orbits around the Sun. It makes a bit more sense, since we're using gravity and not fixing them through an absolute point (relative to what?) in spacetime, or suddenly disconnecting them from Earth's gravity.
The acceleration due to gravity of an object upon the surface of the Earth is 9.8 m/s2 down. As soon as you increase the pull between the Hurlee and the Sun to a counter 9.8 m/s2 up (where "up" is a straight line proceeding from the center of the Earth, to the Hurlee, and on to the center of the Sun), they would be effectively weightless.
After that, they drift towards the Sun, oh so gently, perhaps helplessly, screaming. So, anything more than 9.8 m/s2 sets them on their way. But we have an additional constraint, leaving the atmosphere in 10 minutes. Now, the definition of "atmosphere" is pretty tenuous *rimshot*, so you have a lot of ways to define it, but let's get breathable in there. I'll take Mt. Everest as a fairly arbitrary point of reference, which has a height above sea level of of 29,029 feet.
Let us assume your hapless Hurlee is at sea level. Then, we must get the Hurlee from sea level to the top of Everest in ten minutes, which is 600 seconds. I'd like to neglect air resistance.
Now we have numbers:
d = 8,848 meters.
t = 600 seconds.
a = an additional acceleration, in m/s2, against our 9.8 2.
The most simplistic version of this looks like
d = 1 / 2 * a * t2.
Solving for a, we get a = 2 * d / t2.
Making our substitutions, we see
a = 2 * 8,848 / 6002
a = 0.04915 m/s2
So, the counter pull from the Sun would be 9.80665 m/s2 + 0.04915 m/s2.
Set your Master of Gravity beam for 9.855805bar m/s2. As a percentage, that would be -100.5012%. -100% to counteract the Earth's gravity, and the 0.5012% to get them out of breathable atmosphere in ten minutes.
So, should we care about air resistance? Maybe. The air resistance is more complex, and goes up proportionally both to velocity and the square of velocity.
v = a * t, so the Hurlee would be going at about 29.49 m/s at the point that they are level with the tip of Everest. That's about 65 mph, but only at the end of the trip. That's getting close to the terminal velocity of a human being plummeting through the atmosphere, arms spread wide.
So, yes, we'd probably want to look at air resistance if you cared much about that level of accuracy.
posted by adipocere at 5:24 PM on April 15, 2009
Uh, sorry, obviously that should be: you will need at least 1660 times the present attraction.
If you want him to take off like a rocket, then let's say we want an acceleration of 5 gs ~= 49 m/s^2. Using Newton's second law (F = ma), we'll need a net force of:
80 * 49 = 3920 N
...which means that the attraction between him and the sun needs to be 3920 + 780 = 4700 N, almost exactly a 10000x increase.
posted by equalpants at 5:25 PM on April 15, 2009
If you want him to take off like a rocket, then let's say we want an acceleration of 5 gs ~= 49 m/s^2. Using Newton's second law (F = ma), we'll need a net force of:
80 * 49 = 3920 N
...which means that the attraction between him and the sun needs to be 3920 + 780 = 4700 N, almost exactly a 10000x increase.
posted by equalpants at 5:25 PM on April 15, 2009
Oh, dear. And here I did the percentage relative to the Earth.
Must have been obscuring my view of Venus.
posted by adipocere at 5:27 PM on April 15, 2009
Must have been obscuring my view of Venus.
posted by adipocere at 5:27 PM on April 15, 2009
Well, other people have given you the math needed. To deal with the problem of making the hypothetical person actually hit the sun, rather than just being in a different orbit around it, I propose the following:
1. Wait until the moon is in a position where it is following behind the earth on its journey around the sun (as happens every lunar month).
2. Wait until the hypothetical person is in a position to see the moon (e.g. the earth is not directly between them and the moon) which should take 12 hours or less.
3. Increase the gravitational pull of the moon on the hypothetical person to pull them off the surface of the earth and decrease the perihelion of their orbit around the sun to approximately zero. Quickly enough that they don't hit the moon before they have been "slowed" enough.
4. Increase the gravitational pull of the sun on the person to pull them into the sun in whatever time you would like, but fast enough that they don't hit the moon (which they'll be hurtling toward after step 3).
posted by FishBike at 5:39 PM on April 15, 2009
1. Wait until the moon is in a position where it is following behind the earth on its journey around the sun (as happens every lunar month).
2. Wait until the hypothetical person is in a position to see the moon (e.g. the earth is not directly between them and the moon) which should take 12 hours or less.
3. Increase the gravitational pull of the moon on the hypothetical person to pull them off the surface of the earth and decrease the perihelion of their orbit around the sun to approximately zero. Quickly enough that they don't hit the moon before they have been "slowed" enough.
4. Increase the gravitational pull of the sun on the person to pull them into the sun in whatever time you would like, but fast enough that they don't hit the moon (which they'll be hurtling toward after step 3).
posted by FishBike at 5:39 PM on April 15, 2009
Ooooh, how about this: if you can only change one object's gravity, find a planet that's around 90 degrees behind the Earth, and increase that planet's gravity to affect the person, thus slowing them down -- once their motion approaches zero, they'll begin to fall according to the sun's accelleration, which is huge compared to Earth's acelleration. From Earth, however, the person will seem to zip off to the West, not because the person is moving, but because the Earth moves out from underneath them.
posted by AzraelBrown at 5:41 PM on April 15, 2009
posted by AzraelBrown at 5:41 PM on April 15, 2009
It might be easiest to create a railgun effect by increasing the gravitational pull of molecules of air above the person, while dcreasing the pull of things (air, earth) under the person, creating a gravitational tube they're pulled along until they're spat out of the Earth's atmosphere toward the Sun; by angling the tube correctly you'd get the effect you desire.
I'm assuming an ability for fine-grained control here.
posted by rodgerd at 5:49 PM on April 15, 2009
I'm assuming an ability for fine-grained control here.
posted by rodgerd at 5:49 PM on April 15, 2009
Note: Ok, other people have already solved this, but god dammit I spent an hour figuring and typing this out, so I'm going to post it anyway. So there.
-------
This has nothing to do with escape velocity. The problem of escape velocity assumes that you have no propulsion beyond your initial speed. In this scenario, you have a constant force on you.
And really, that makes it fairly simple. Assume a stationary earth held in place by magic, and a person standing on it at noon -- it may help to visualize it sideways. He is held on by a force of 9.8 N/kg minus the gravitational pull of the sun at a distance of 1 AU. To make him go up, you have to make the force of the gravitational pull of the sun at 1 AU be greater than 9.8 N/kg. Let's give him a mass of 100 kg -- it won't matter in the end, but it's easier to give him one.
The formula for the force of gravity between two objects is G(M1*M2)/r2, where G is the Gravitational Constant, M1 and M2 are the masses of the objects in question, and r is the distance between them. Fortunately Google knows all of these constants, so I Google the following:
So let's start by making the sun's pull equal to the earth's, making the guy weightless. We do this by changing the mass of the sun. We want the following to be true:
G*(mass of the new sun)*(100 kg)/(1 AU)^2 = 979.982305 newtons
where "mass of the new sun" is now a variable. You know what to do when you see a variable: solve for it.
That makes Mr. Heliotrope weightless, but you want him to go 25,000 feet in ten minutes on top of that. Not too hard. Let's figure out how to make him go 25,000 feet in ten minutes first, then make the sun do it.
Acceleration from rest is calculated using y = (a * t2) / 2, where y is position (distance), a is acceleration, and t is time. y is 7620 meters (25,000 feet), and t is 600 seconds, so a = 25.4 m/s^2 -- rather more than Earth's pull.
F = ma, so F = 100kg * 25.4m/s^2 = 2540 Newtons. That is, a force of 2540 N will make a 100 kg object, in this case Mr. Heliotrope, go 25,000 feet from rest in 10 minutes. That means we need the sun's pull to be 980 newtons more than that to counter the Earth's pull that we worked out before, plus get him airborne.
So now we want G*(mass of the final sun)*(100 kg)/(1 AU)^2 = 3430 Newtons. Solve for the final sun mass and plug into Google:
Of course, I've already solved your problem as you asked it, since you asked for forces. The current force between the sun and Mr. H is 0.59304393 Newtons, and the required force of the sun on Mr. H is 3430 N. So you would have to increase that force about 5,784 times, or 578,400%.
I have not bothered to worry in the above about significant digits. Also, the probability approaches 1 that there is a stupid math error somewhere. I advise you to check it yourself if you want to be sure to have your potential victim sun-immolated properly. I'm not going to. I should be working right now.
posted by darksasami at 5:54 PM on April 15, 2009 [1 favorite]
-------
This has nothing to do with escape velocity. The problem of escape velocity assumes that you have no propulsion beyond your initial speed. In this scenario, you have a constant force on you.
And really, that makes it fairly simple. Assume a stationary earth held in place by magic, and a person standing on it at noon -- it may help to visualize it sideways. He is held on by a force of 9.8 N/kg minus the gravitational pull of the sun at a distance of 1 AU. To make him go up, you have to make the force of the gravitational pull of the sun at 1 AU be greater than 9.8 N/kg. Let's give him a mass of 100 kg -- it won't matter in the end, but it's easier to give him one.
The formula for the force of gravity between two objects is G(M1*M2)/r2, where G is the Gravitational Constant, M1 and M2 are the masses of the objects in question, and r is the distance between them. Fortunately Google knows all of these constants, so I Google the following:
G*(mass of the sun)*(100 kg)/(1 AU)^2which gets me
(G * mass of the sun * (100 kg)) / ((1 AU)^2) = 0.59304393 newtonsand
G*(mass of the earth)*(100 kg)/(radius of the earth)^2which produces
(G * mass of Earth * (100 kg)) / (radius of Earth^2) = 979.982305 newtonsAll of this is mostly to show that the force of gravity between the man and the earth is lots bigger than the force of gravity between the man and the sun, and make sure the math is sane. (I'm totally winging this.)
So let's start by making the sun's pull equal to the earth's, making the guy weightless. We do this by changing the mass of the sun. We want the following to be true:
G*(mass of the new sun)*(100 kg)/(1 AU)^2 = 979.982305 newtons
where "mass of the new sun" is now a variable. You know what to do when you see a variable: solve for it.
mass of the new sun = 979.982305 newtons * (1 AU)^2/(G*100kg)The actual mass of the sun is 1.98892 × 1030 kilograms, so we're scaling up pretty big--three orders of magnitude big.
mass of the new sun = 3.28661387 × 1033 kilograms
That makes Mr. Heliotrope weightless, but you want him to go 25,000 feet in ten minutes on top of that. Not too hard. Let's figure out how to make him go 25,000 feet in ten minutes first, then make the sun do it.
Acceleration from rest is calculated using y = (a * t2) / 2, where y is position (distance), a is acceleration, and t is time. y is 7620 meters (25,000 feet), and t is 600 seconds, so a = 25.4 m/s^2 -- rather more than Earth's pull.
F = ma, so F = 100kg * 25.4m/s^2 = 2540 Newtons. That is, a force of 2540 N will make a 100 kg object, in this case Mr. Heliotrope, go 25,000 feet from rest in 10 minutes. That means we need the sun's pull to be 980 newtons more than that to counter the Earth's pull that we worked out before, plus get him airborne.
So now we want G*(mass of the final sun)*(100 kg)/(1 AU)^2 = 3430 Newtons. Solve for the final sun mass and plug into Google:
mass of the final sun = (3430 newtons) * (1 AU)^2/(G * 100 kg)So our new and improved sun is four orders of magnitude larger than the real one.
((3430 newtons) * ((1 AU)^2)) / (G * 100 kg) = 1.15033563 × 1034 kilograms
Of course, I've already solved your problem as you asked it, since you asked for forces. The current force between the sun and Mr. H is 0.59304393 Newtons, and the required force of the sun on Mr. H is 3430 N. So you would have to increase that force about 5,784 times, or 578,400%.
I have not bothered to worry in the above about significant digits. Also, the probability approaches 1 that there is a stupid math error somewhere. I advise you to check it yourself if you want to be sure to have your potential victim sun-immolated properly. I'm not going to. I should be working right now.
posted by darksasami at 5:54 PM on April 15, 2009 [1 favorite]
It might be easiest to create a railgun effect
May I just say that on no other advice site would I read this line.
posted by ricochet biscuit at 6:01 PM on April 15, 2009 [2 favorites]
May I just say that on no other advice site would I read this line.
posted by ricochet biscuit at 6:01 PM on April 15, 2009 [2 favorites]
To answer your question of "how much do you need to increase G to make your enemy reach 10000 feet in 10 minutes" you actually need to solve a differential equation.
Your enemy's acceleration is
x(t)'' = k * G * mass_sun * mass_enemy / (distance_sun - x(t))^2 -
G * mass_earth * mass_enemy / x(t)^2
x(0)' = 0
x(0) = earth_radius
where:
x(t) = your enemy's position
earth's position is 0
distance_sun = distance between centers of sun and earth
k = multiplier for gravity between sun and enemy
So you need to first solve for x(t), and then find a value of k such that x(10 minutes) = 25000 feet + earth's radius
I don't remember how to do the diff eq... it's been ten years since I touched that stuff, and I don't want to spend hours on it.
k will be the amount you need to multiply the sun's gravity by
posted by qxntpqbbbqxl at 6:06 PM on April 15, 2009
Your enemy's acceleration is
x(t)'' = k * G * mass_sun * mass_enemy / (distance_sun - x(t))^2 -
G * mass_earth * mass_enemy / x(t)^2
x(0)' = 0
x(0) = earth_radius
where:
x(t) = your enemy's position
earth's position is 0
distance_sun = distance between centers of sun and earth
k = multiplier for gravity between sun and enemy
So you need to first solve for x(t), and then find a value of k such that x(10 minutes) = 25000 feet + earth's radius
I don't remember how to do the diff eq... it's been ten years since I touched that stuff, and I don't want to spend hours on it.
k will be the amount you need to multiply the sun's gravity by
posted by qxntpqbbbqxl at 6:06 PM on April 15, 2009
oh, and that's ignoring aerodynamics...
posted by qxntpqbbbqxl at 6:06 PM on April 15, 2009
posted by qxntpqbbbqxl at 6:06 PM on April 15, 2009
Also, the whole concept of escape velocity isn't relevant here... with gravity changed, this person is going to fall towards the sun.
posted by qxntpqbbbqxl at 6:07 PM on April 15, 2009
posted by qxntpqbbbqxl at 6:07 PM on April 15, 2009
It depends a lot on how you're affecting gravity...
Go read Wells' "The First Men in the Moon" and Asimov's "The Billiard Ball" for two different takes on how you might go about that.
posted by Pinback at 7:03 PM on April 15, 2009
Go read Wells' "The First Men in the Moon" and Asimov's "The Billiard Ball" for two different takes on how you might go about that.
posted by Pinback at 7:03 PM on April 15, 2009
Short answer: use the stars. You can "increase and decrease gravity at will, relative to all objects, across all distances"? Increase the pull between the person and some star directly overhead by, oh, let's call it a bajillion times, until they're above the atmosphere. Then "pan" across the sky to the west, switching to one star after the other, killing their angular velocity relative to the sun and getting them well clear of the Earth. Then switch to using the sun's gravity and drop 'em straight down in a blaze of superheated solar wind.
posted by The Tensor at 7:38 PM on April 15, 2009
posted by The Tensor at 7:38 PM on April 15, 2009
He wouldn't fall directly into the sun, he'd go into an elliptical orbit with the sun at one focus.
You'd have to use a force great enough so that the perihelion of the orbit was smaller than the sun's radius. Otherwise he'd just orbit forever without ever making contact.
This.
The Earth is totally irrelevant in this discussion. The only thing that matters is Earth's orbit. If the planet Earth suddenly disappeared, I would still be in orbit around the sun--I already have the velocity.
So, to suck the target into the sun, you would need to change the gravitational constant for the sun-target interaction such that the target's current Earth-orbit velocity was woefully insufficient to maintain orbit. Well, specifically, insufficient to maintain an orbit such that the length of the minor elliptical axis is smaller than the diameter of the sun.
posted by Netzapper at 7:40 PM on April 15, 2009
You'd have to use a force great enough so that the perihelion of the orbit was smaller than the sun's radius. Otherwise he'd just orbit forever without ever making contact.
This.
The Earth is totally irrelevant in this discussion. The only thing that matters is Earth's orbit. If the planet Earth suddenly disappeared, I would still be in orbit around the sun--I already have the velocity.
So, to suck the target into the sun, you would need to change the gravitational constant for the sun-target interaction such that the target's current Earth-orbit velocity was woefully insufficient to maintain orbit. Well, specifically, insufficient to maintain an orbit such that the length of the minor elliptical axis is smaller than the diameter of the sun.
posted by Netzapper at 7:40 PM on April 15, 2009
The earth's current speed, perpendicular to the direction of the sun (approx):
(1 AU) * pi * 2/(1 year) = 29 785.9163 m / s
The speed of your opponent, perpendicular to the direction of the center of the earth, if he's on the equator:
(radius of the earth)*pi*2/(1 day) = 463.828521 m / s
If you cancel the pull of the earth, the pull of the sun still has to be quite massive to get him to fall straight down.
posted by Monday, stony Monday at 8:14 PM on April 15, 2009
(1 AU) * pi * 2/(1 year) = 29 785.9163 m / s
The speed of your opponent, perpendicular to the direction of the center of the earth, if he's on the equator:
(radius of the earth)*pi*2/(1 day) = 463.828521 m / s
If you cancel the pull of the earth, the pull of the sun still has to be quite massive to get him to fall straight down.
posted by Monday, stony Monday at 8:14 PM on April 15, 2009
You'd have to use a force great enough so that the perihelion of the orbit was smaller than the sun's radius.
...and at this point you might also have to worry about tidal effects*. You want your enemy to sail up into the sky kicking and screaming, not instantly turn into a column of pulped gore streaking into the sky.
Unless turning him into ascending gore is more appealing, anyway.
*Or not. I don't know from the math.
posted by ROU_Xenophobe at 8:43 PM on April 15, 2009
...and at this point you might also have to worry about tidal effects*. You want your enemy to sail up into the sky kicking and screaming, not instantly turn into a column of pulped gore streaking into the sky.
Unless turning him into ascending gore is more appealing, anyway.
*Or not. I don't know from the math.
posted by ROU_Xenophobe at 8:43 PM on April 15, 2009
To condense the above a bit, using the sun's gravity to pull the victim off the face of the earth is not a problem.
The problem comes from the fact that the pull vector to the victim is perpendicular to the victim's velocity vector (which is/was in near-circular orbit around the sun).
It's a somewhat trivial problem to separate the victim from the earth, turn off gravity to him to let his (formerly orbital) momentum put him in a position further away from the Sun and with an altered velocity such that reestablishing sufficient gravity to him will suck him into the sun without having to turn up the gravity to force a pure elliptical orbit to tighten into the sun as described above.
posted by mrt at 9:05 PM on April 15, 2009
The problem comes from the fact that the pull vector to the victim is perpendicular to the victim's velocity vector (which is/was in near-circular orbit around the sun).
It's a somewhat trivial problem to separate the victim from the earth, turn off gravity to him to let his (formerly orbital) momentum put him in a position further away from the Sun and with an altered velocity such that reestablishing sufficient gravity to him will suck him into the sun without having to turn up the gravity to force a pure elliptical orbit to tighten into the sun as described above.
posted by mrt at 9:05 PM on April 15, 2009
So I'll try to answer your specific numerical question first. Wikipedia has the formulas for periapsis and apoapsis of an orbit -- the lowest and highest points.
Given:
rap = 1.52x1011m
vap = 2.98x104 m/s
rper = 6.96x108 m
I get:
a = 7.64x1010 m
e = 0.991 (ridiculously elliptical)
μ = 1.48x1013 km3/s2
The end result is that the value for μ, the gravitational parameter, would have to be about 110x stronger than it currently is. That's ignoring the effects of earth's gravity; even with that much of an increase, the sun would only be exerting an extra 0.06g of acceleration at the range of the earth's orbit. You'd need to either turn off earth's gravity at the same time, or else temporarily boost the sun's pull by another factor of 20 or so until the hapless victim reaches escape velocity.
That said, this has the disadvantage that you need to keep up the acceleration for half an orbit until they end up in the sun, which would be about 6 days. A more elegant method would be to wait for a first quarter moon, and increase the moon's gravity enough that the person slingshots past, canceling out their orbital velocity relative to the sun. At 1g, that would take an hour or so, after which you can return the universe to its regularly scheduled laws of physics.
Yet another way, which only requies changing the sun's gravity, is to give the person a brief boost of speed that gets them as far as Jupiter's orbit, wait a few year for them to get there, and then execute a series of slingshots to drop their periapsis.
posted by teraflop at 9:15 PM on April 15, 2009
Given:
rap = 1.52x1011m
vap = 2.98x104 m/s
rper = 6.96x108 m
I get:
a = 7.64x1010 m
e = 0.991 (ridiculously elliptical)
μ = 1.48x1013 km3/s2
The end result is that the value for μ, the gravitational parameter, would have to be about 110x stronger than it currently is. That's ignoring the effects of earth's gravity; even with that much of an increase, the sun would only be exerting an extra 0.06g of acceleration at the range of the earth's orbit. You'd need to either turn off earth's gravity at the same time, or else temporarily boost the sun's pull by another factor of 20 or so until the hapless victim reaches escape velocity.
That said, this has the disadvantage that you need to keep up the acceleration for half an orbit until they end up in the sun, which would be about 6 days. A more elegant method would be to wait for a first quarter moon, and increase the moon's gravity enough that the person slingshots past, canceling out their orbital velocity relative to the sun. At 1g, that would take an hour or so, after which you can return the universe to its regularly scheduled laws of physics.
Yet another way, which only requies changing the sun's gravity, is to give the person a brief boost of speed that gets them as far as Jupiter's orbit, wait a few year for them to get there, and then execute a series of slingshots to drop their periapsis.
posted by teraflop at 9:15 PM on April 15, 2009
2nd rail gun for the big buy: Rather than using the sun to suck why not stick to launching from earth? If you are controlling gravity wouldn't a simple answer be to just inverse gravity? Like falling, they would increasingly zip away and its got the bonus of being simple on the math end; -200%. And it would be super freaky to fall up, especially as you inner ear would totally lose it. Once in space it would be simply a matter of pointing (or timing) the body in the path of the sun and letting them rip - again, an oversimplification but you are in control of a rather powerful force and while spiraling someone to the fiery end might take a little while, this fits with the intention of the OP's nefarious plan.
Here's some info from those Kittinger, who jumped out of high altitude balloons: 95% of the atmosphere is an altitude of 82,200 feet (25,055 meters). I think I originally saw this on mefi.
And more info from a 19 1/2 mile jump:
.... and the tropopause barrier.
My -200% earth gravity estimation using these nuts jumping to earth as a reference: It took Kittinger approximately 14 minutes with a somewhat controlled descent from basically space and his reported maximum speed was 614 miles per hour and we know Kittinger survived (I mean the whole point is to really smite this character right?). Of course our little fella would moving a bit quicker than that but its a rough starting point. Keeping it as simple as possible let's say we launch Mr. Heliotrope into space at 614 m/h, and with the sun being 93,000,000 miles away it would take 151466 hours or 6311 days, or about 24 years, to reach the sun, assuming (impossibly) a direct path. Of course maintaining a repulsive gravitation force away from earth would speed the trip up, but with everything moving it would become quite complicated past my meager skills, and the OP was fine with a timely journey.
More info regarding maximum velocity of regular sky divers with a great little diagram of a falling human cartoon with the math included.
posted by zenon at 9:20 PM on April 15, 2009
Here's some info from those Kittinger, who jumped out of high altitude balloons: 95% of the atmosphere is an altitude of 82,200 feet (25,055 meters). I think I originally saw this on mefi.
And more info from a 19 1/2 mile jump:
Aerodynamically, space begins about 120 miles from earth. Physiologically and psychologically, however, it starts only 12 miles up, where survival requires elaborate protection against an actual space environment.Apparently spinning and temperature are also big issues for human shaped objects:
Flat spin is a characteristic of any falling object that is aerodynamically unstable. Dummies dropped from balloons up to 100,000 feet have attained 200 revolutions per minute, whereas tests show that 140 r.p.m. would be harmful, possibly fatal.
.... and the tropopause barrier.
The temperature drops steadily until it reaches -94° F. at 50,000 feet, then starts to rise.So Mr. Heliotrope would, in addition to running out of oxygen, likely also die of exposure and spinning, and old Helio hasn't even gotten to space. Getting launched into space quickly or slowly without protection sounds like it would be rather no fun, and that's long before getting cooked in the sun. Also, my understanding that somewhere around Venus, which is halve the distance to the Sun, the radiation level is already increased by four and so somewhere between there and Mercury you would additionally get very very radiated. It would be epic way to finish someone.
My -200% earth gravity estimation using these nuts jumping to earth as a reference: It took Kittinger approximately 14 minutes with a somewhat controlled descent from basically space and his reported maximum speed was 614 miles per hour and we know Kittinger survived (I mean the whole point is to really smite this character right?). Of course our little fella would moving a bit quicker than that but its a rough starting point. Keeping it as simple as possible let's say we launch Mr. Heliotrope into space at 614 m/h, and with the sun being 93,000,000 miles away it would take 151466 hours or 6311 days, or about 24 years, to reach the sun, assuming (impossibly) a direct path. Of course maintaining a repulsive gravitation force away from earth would speed the trip up, but with everything moving it would become quite complicated past my meager skills, and the OP was fine with a timely journey.
More info regarding maximum velocity of regular sky divers with a great little diagram of a falling human cartoon with the math included.
posted by zenon at 9:20 PM on April 15, 2009
Those saying that he would go into a spiral around the sun are correct, except that really he'd be in a spiral around the sun's center of mass, and the sun isn't a point mass (that is, it has a finite, non-zero size), so eventually the spiral around the sun would become tight enough that he would run into the surface of the sun. You can't just pull him right into the sun, because he's already moving in a direction 90 degrees to the surface of the sun. If you stopped his movement in that direction first, so that he was no longer revolving around the sun, you wouldn't have to do anything at more, he would just start slowly falling towards the sun.
Discussions of air resistance are minute details, in my opinion.
posted by !Jim at 12:30 AM on April 16, 2009
Discussions of air resistance are minute details, in my opinion.
posted by !Jim at 12:30 AM on April 16, 2009
That might not be enough. He wouldn't fall directly into the sun, he'd go into an elliptical orbit with the sun at one focus. You'd have to use a force great enough so that the perihelion of the orbit was smaller than the sun's radius. Otherwise he'd just orbit forever without ever making contact.
Perfect! Then you could have a sequel when the guy shows back up one orbit around, looking for revenge on the math-challenged gravity-changing smart-ass.
posted by ctmf at 12:55 AM on April 16, 2009 [2 favorites]
Perfect! Then you could have a sequel when the guy shows back up one orbit around, looking for revenge on the math-challenged gravity-changing smart-ass.
posted by ctmf at 12:55 AM on April 16, 2009 [2 favorites]
Without having done out the numbers, I think that the air resistance is going to be quite significant and that its end effect will be at least several hundred percent of G. The figures above look to me like they'll involve exceeding the terminal velocity for a human body which would mean the force of air resistance would get pretty high.
posted by XMLicious at 1:07 AM on April 16, 2009
posted by XMLicious at 1:07 AM on April 16, 2009
I think that the air resistance is going to be quite significant
nah, the poster stipulated 10 minutes to get through the atmosphere. 12 miles in 10 minutes is only freeway speeds.
posted by mrt at 8:07 AM on April 16, 2009
nah, the poster stipulated 10 minutes to get through the atmosphere. 12 miles in 10 minutes is only freeway speeds.
posted by mrt at 8:07 AM on April 16, 2009
This thread is closed to new comments.
Essentially, all you need to worry about is attaining escape velocity and then just figuring out how fast they have to go to cross an AU in ten minutes.
posted by kldickson at 4:29 PM on April 15, 2009