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.
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 150 100 km/second not faster than the speed of light.
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.
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.
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
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.
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.
I don't think you can achieve a spiral orbit in an area with so little friction, mostly devoid of dust and gas, else the earth would be on one of those too...
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.
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?)
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!
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
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.
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.
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?
What if you could kick him into space, making an orbital transfer to Jupiter, from which the kid gets a gravity assist that bounces the kid into a more elliptical orbit that then sends him into the sun?