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I was really hoping he was going to convert the amount of energy needed into calories, then from calories into peanuts butter sandwiches
1 calorie is interchangeable for approximately 4.1868 joules. Therefore, assuming his math was correct (many say it was not), I’m coming up with 2,687,016,337 calories needed. According to google, sourcing from the USDA, your average peanut butter sandwich has 384 calories. Therefore you’d be expending approximately 6,997,438 peanut butter sandwiches worth of energy to punt the ungrateful little shit into the sun.
In the US, the calorie used in nutrition data is actually a kilo calorie.
In the United States, in a nutritional context, the “large” unit is used almost exclusively.
384 cal or 384 kcal per sandwich?
Same.
If your leg has a mass of 2kg, 1.1×10^10 J of kinetic energy would require your leg to be moving at about
150100 km/second not faster than the speed of light.TLDR: Their math is shit.
Pretty sure you’re generating twice as much energy as needed, the required speed is only about 106km/s
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Besides, if you really needed those kinds of speed, you’d obviously have to calculate with relativistic formulas. Energy is asymptotical at the speed of light.
100 km/sec is not relativistic and even if it were, at no point would that object need to or could exceed the speed of light. Its a fundamental limit that cant be broken.
Setting aside the correctness in OP for the moment, what’s being said here is that you don’t actually need to break lightspeed. The foot would have to be moving asymptomaticly close to lightspeed, but not passing it. OP used an equasion that works classically, but we’re in territory where that model breaks down.
But if the math doesn’t work out that way, anyway, then whatever, classical equasions are fine.
Yeah, I was refering to the OP’s calculated result in that it’s incorrect not only by incorrect math, but also incorrect physics.
Yeah, their answer just intuitively seems very wrong. The ratio between the kid’s weight and your foot’s weight should be equal to the ratio between their final speed and your foot’s required speed. Ridiculous.
Not to mention the fusion reaction triggered by an FTL foot connecting with said child’s backside would annihilate both parent and child immediately.
There would be a crater where the parent and child were, and buildings would be leveled by the resulting shockwave.
Okay the math is obviously wrong, and it’s not even answering the question.
The question was, how much force. If punting the kid involves a kick, let’s say the foot makes contact with the kid for about 25 cm. Then the force required over this distance is on average 45 GN.
This is equivalent to the child experiencing roughly 180,000,000 G
Or just get another child. I know they don’t grow on trees but I’m sure they grow somewhere
The fiery pits of hell
That not a nice name for the uterus
But still accurate. That fucker tries to kill me every few weeks for some reason.
Faster than the speed of light.
Lol that is some shit maths for a checks note astrophysics major i am shit at maths and even i know its wrong.
Are you arguing that 1.12 billion m/s is NOT faster than the speed of light, or are you arguing that the speed required by the kick is not 1.12 billion m/s? Because if it’s the former, the speed of light in a vacuum is 300 million m/s (to 3 significant figures), or less than one third of that kick speed. If you’re arguing the latter, I don’t feel like checking all of the calculations this early in the morning, but you are probably right on that point.
Velocity addition and conservation of momentum don’t work like that if the speeds are close to the speed of light.
For further details, please check out special relativity theory.
The reliable way to get an answer from the internet is to provide the wrong answer, then someone will come and correct you, providing the answer you seek. (Xkcd, probably maybe?)
This is a proven fact. Expose yourself early and often, that’s my motto.
Im not allowed within 500m of the supermarket after i exposed myself last time.
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Cut the extra inch off the long side to get a 4" square, then cut the remaining 1" x 4" piece into 4 1" squares. The boy never said the squares had to be the same size.
If the triangles have already been cut, it’s a peanut butter sandwich: use peanut butter on the edges to glue it back together and cut the squares. The child gave you a challenge, think outside the box!
i’d cut squares out of the triangles.
Once the kid realizes he’s getting less because of his demands, he might change his mind about shape being important.
Edit: or make him do it. toss in a lesson about geometry, too.
If the triangles have already been cut, the kid gets a brand new sandwich fully intact, crust and all, and a knife. Let’s see you cut this sandwich better than I can brayxtyn
If that’s the child’s name, you have no one to blame but yourself, and are probably underqualified for handling a butter knife.
I’m partial to Brexit for my first childs name.
Parent’s already thinking outside the biosphere.
Thats inefficient, you dont need to cancel the angular momentum as there was no time limit on how long it takes rhe child to enter the sun and there also was not a specified required trajectory. The child can just spiral into the sun
There are no spiral orbits. Canceling the forward motion is exactly what you need to do, to bring down the next periapsis to 0. Now, you can go with a periapsis of about half a million km, because the sun is pretty big, but that is not a significant difference. Getting anywhere near the sun, is the hard part.
It’s much more efficient in this case to do a bi-elliptic transfer: raise apoapsis very far out, then lower your periapsis once you are at apoapsis. Wikipedia says you could do it with about 8.8 km/s delta v. Versus 24 or so for a basic Hohman transfer (still a bit better than 30)
Sadly the bi-elliptic transfer requires two burns so you can’t do it with a kick.
Right, I wanted to ask: is that actually the minimum energy to make the child reach the sun? What’s the minimum energy to launch something so it reaches the sun?
The minimum would be something like punting your kid to the orbit of Venus for a gravity assist that takes it to one of the outer planets where another gravity assist can push it to the edge of the solar system.
Out there, the angular momentum of the orbiting child will be very low and can be canceled out by a small thrust.
The child will then fall back into the sun. But this requires remote controlled thrusters strapped to the child. And a life support system if you want your child to actually die by burning in the sun. And then, the child will be well into their teens by the time they reach it.Give them time to think about what they did, maybe learn from it
Just tell them to give a small poot on their tenth birthday it’s all good
The speed will be related to escape speed of sun. Based on https://en.m.wikipedia.org/wiki/Escape_velocity It will be Vte which is 16.6 km/s (or Ve 42.1). So when object at earth orbit and has lower speed than 16.6 it can’t keep the orbit and will slowly fall into sun.
Right, and what force is acting on the child to make it deviate from a circular orbit into a spiral one?
You could use a gravity assist from another planet
Why cancel the angular momentum when we can utilize it?
i may be mistaken but I don’t think it can be utilized to descend toward the sun
Still this approach seems wasteful. Just making it sufficiently far from the surface travelling in the right direction should be enough… As long as you aren’t in a rush.
Yeah it just depends on whether the goal is to launch the child into the sun or just in an orbit with a low periapsis
It’s the foundational principle of “Launch Windows.” Because the earth rotates the sun and also spins on an axis, we can launch at a time of day that gives us time to accelerate and then leave earth orbit in the direction of earth’s orbit around the sun with minimum amount of energy required. The majority of energy used is simply to escape Earth orbit. Once orbiting the sun, comparatively very little energy would be required to approach it utilizing it’s own gravity.
During Perihelion the sun is 147100632 KM away, as the distance from the sun is not constant for earth’s orbit.
It requires energy to shrink an orbit just as it does to grow it. Since we’re launching from Earth, we start with Earth’s orbit around the sun, and we have to burn enough to bring the perihelion of our launched child’s solar orbit to within the sun itself. We could use Venus or Mercury for gravity assists but the angular speed of the orbit does have to be mostly cancelled to hit the sun
It’s still a lot cheaper than instantaneously cancelling angular velocity.
well the other stipulation was that it has to happen from a single kick
Yes the source of thrust is still cheaper using a launch window. Basically, you don’t need to kick directly to the sun you just need the child to land somewhere below a maintainable orbit given their velocity. The closer they get the more gravity will effect them and do more work that we don’t have to do.
Why does it have to be a kick? Could I generate that force with a car? It weighs a lot more so I assume the speed wouldn’t need to be nearly as high.
The request was to “launch”, so I’d say a railgun would also be an acceptable answer, having much lower acceleration values, as you could just accelerate the son continuously, over the distance of the barrel.
500m should be good enough, no?
“Then you’ll be fired.”
“Fine!”
“Out of a canon into the sun.”
Do. It.
I think I scared my wife and kid I laughed so hard.
Surely they’re not so mad that they need to kick their child into the sun. I’m sure a low solar orbit would suffice.
Always funny to see the memes show up here a week after I get sent them from friends who still use Facebook and Fark.
The circle of memes
Then how does Superman throw people into the sun? I think this mathematician needs to read a few comic books.
