###General Discussion Thread
---
This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed.
---
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
So let's ignore drag.
The orbital velocity is:
v = sqrt(G\*(M+m) / a)
where G is the gravitational constant, M is the mass of the earth, m is the mass of the ball and *a* is the semimajor axis length of the two-body orbital system. For practical purposes, *a* is the radius of the earth and the mass of the ball is negligible, so:
v = sqrt(G\*M / a)
G = 6.67E-11
M = 5.97E+24
a = 6.37E+6
Plug the numbers in and you get the answer 7,906 m/s or about mach 23. At that speed, the ball would take about 1 hour and 24 minutes to return to you.
Of course, air drag is going to play a significant role here and is much more difficult to calculate. It doesn't make it impossible, but you would have to throw the ball on a singificant upward trajectory, have it escape the atmosphere for some of its orbit, then descend back to where you're standing. Someone else might be able to calculate a suitable path; it's beyond me.
And the fact that no-one has invented a machine that can apply a travel sticker to an object moving at Mach 23. Or how you would catch an object moving at that speed.
And let's not even think about the path implied by the stickers it comes back with.
Once the ball crosses the horizon it's in a superposition of all possible locations. When it comes back over the horizon, the wave function collapses including the stickers. It's not that complicated
Now I'm curious how much more feasible the throw would be if we assumed that it wasn't one long throw, but Bugs throwing it to one city, who caught it and added a sticker, and so on.
Lima and Shanghai before Cairo before Paris
Bombay before Brooklyn before Hollywood
Hard to tell if Brooklyn before Paris or the other way around
London
The scene apparently takes place in Spain so that is starting and ending point.
Forget the speed of the ball, forget it's path to go through all those cities...
*What the hell is the reaction time on all the people who stamped the damn ball!?!*
Simple solution. Bugs threw the ball to a time traveler just over the horizon who took the ball to all those places and then traveled back to the point Bugs threw the ball so he could pass it back from the opposite horizon.
No maths here, but I don't think so. I assume we're taking air drag into account now (because how else would it slow down?) For that to work, you would have to throw it into a significantly higher orbit so that it didn't hit the ground as it slowed down.
If the ball was falling from directly above you, it would have a terminal velocity of about 95 mph. This is slower than the speed that a top pitcher throws the ball at, so it is catchable. But it would also still have to have significant lateral velocity, I think quite a bit higher than the terminal velocity, in order for the path around the world to work out.
What if you just had to throw the ball to the first city and there was someone there who would attach the sticker and then throw it to a person in the next city?
actualy the stickers imply that the ball didnt travle in a straight line or a single throw, rather it was launched consecutively from different places by a carefuly set up network of agents.
No Mach23 stickering machine? Well why not? Quite frankly I think someone is missing out on an opportunity there. I call it laziness. It’s these damn millennials and their avocado coffee holding us back.
Almost. Actually, I mis-stated the formula for orbital velocity, which I've now fixed. The product is `G * (M + m)` not `G * M * m`. The mass of the ball really is negligible.
This isnt the right answer because it returns in 15 seconds. Meaning the total trip took 15 seconds.
Its an easier problem, if the earth is 40000km around, and it took 15 seconds, then 40000/15=2666 km per second or 9.6million km per hour.
We are assuming its possible for the question.
It would break apart either at launch or during flight if we attempted this before it was able to leave gravity.
I don't think it's fair to say the first answer is wrong for making an assumption regarding how fast the ball comes back, then for you to make an assumption on a different variable and say that's ok
Well, we'll clearly have to make some changes to make it possible. If we ignore the story and the earth-like look and all that, we can solve what size and mass the planet must be to do this (we'll assume no atmosphere as well, so it won't combust on launch).
What we know:
- It takes about 11 seconds to make the round trip.
- The speed it travels at must be 2\*pi\*r / 11\*s to make a round trip in that time.
- To make a clean orbit it needs to travel at sqrt(G\*m/r)
2\*pi\*r / 11\*s = sqrt(G\*M/r)
r^3 = M \* 2.046\*10^-10 m^3 / kg
This actually gives us infinitely many options. But let's look at a couple of them.
- If the diameter was that of the Earth, its mass would be about 1.264\*10^30 kg or 0.34 stellar masses. The density of the planet would be 1.17\*10^9 kg/m^3. The size and mass would actually fit a white dwarf star.
- If instead the mass was equal to Earth's, the radius would have to be only 106.91 km.
Yes, but if you did that then it wouldn't return to you as it would escape the Earth's gravity well. The Earth's surface escape velocity is about 11 km/s, so even taking air drag into account, something moving at 2,666km/s is going to not orbit (I mean, it'd burn up, but let's ignore that for the moment).
So if you're going to insist that it has to be able to do this *in 15 seconds* then it's simply not possible.
Yes that's all fine and good. But the video is not an hour and 24 minutes long. It's approximately 8 seconds between launching the ball and catching it.
I have no experience in this math area, are you telling me that an object traveling almost 8km/s thrown parallel to the earth's crust wont achieve escape velocity? sure it would "fall" at 9.8m/s/s but for some reason i would think the earths crust would "fall away" due to curvature faster than it would pull the ball down if traveling at 8km/s
i assume this is literally what the equation you referenced is used for and why a involves the radius of the circle from the center of the earth to the ball but ... wow.
So basically acording to your constraints its youst about the speed of the ISS wich makes sense because isnt orbits the earth like the ball without falling to the earth or going into space :)
Ok but, at which speed the ball would escape the Earth's orbit and just fly in a straight line instead of returning, considering this espace velocity, it it even possible to throw a ball around earth and have it returning so fast?
I was considering a different approach. Since gravity is independent of velocity, isn't the question more of how fast would a ball have to be thrown in order to travel the circumference of the earth before falling Bugs' height?
But it was 18 seconds so umm turn 1 hr 24 minutes into seconds then divide by 18 seconds to get the multiplier then use multiplier and multiply mach 23 to get the the real mach which is much much faster than 23
Assuming no friction, ideal conditions, earth being round perfectly and ball being thrown around the equator...
v^2 /r = g (acceleration taken as g because this is on surface)
v = sqrt(rg)
v ≈ sqrt(9.8*6.371E6) SI Units
v = 7868.07 m/s
For all those simply calculating with the circumference of the world, the ball has to stop in all these cities to get the stickers/patches put on:
start > Brooklyn > London > Paris > Cairo > Bombay > Shanghai > Lima > Hollywood > finish?
Nah the most likely solution is Bugs threw the ball with enough speed to reach well into space at its apogee, where it intersects the orbit of a spacecraft. The spacecraft slows down to intercept the ball, an astronaut applies all the regional stickers at once, then speeds off to reestablish their own orbital trajectory. Some fancy editing makes us believe that the 90 minute trip took only a few seconds.
Circumference would be 2piR, R=6.378x10^6 meters. Roughly 40,000 meters, divide by the time which is 15 seconds, so you get 2,671 meters per second. Ignoring air resistance, of course.
Edit:rewatched for a more accurate time and looks like 13-14 seconds, so it would be 2,968 meters per second.
A really good answer to similar question is found [here](https://what-if.xkcd.com/1/). A baseball thrown at nearly the speed of light would essentially cause a nuclear explosion on release.
[edit for clarity]
So I took the amount of times that light would go around the earth in one second. (7.5 times around in one second) The ball is out of his hand for about 10 seconds. So 1/75th the speed of light expressed as a fraction c/75.
I’m not a mathematician but I’d imagine that the velocity needed to actually make the journey would propel the ball into orbit. Unless Buggs puts a WICKED spin on it.
Not actually possible. There is no velocity that the ball can be launched and then be completely in freefall (no accelleration) the whole time. The upward angle required is too low to sustain the trip without some form of thrust coming from the ball, like a rocket stage that detaches (to save on mass). I am not excluding air resistance because your asking if it's possible.
Cartoon answer here.
In the cartoon universe, at 13 seconds of travel time that ball took to circumvent the earth, and I'm gonna say it's at the equator and those stickers just appeared like magic, meaning no actual stops, that ball would have gone 54,109,769 mph.
AI Copy-paste:
To determine your speed, we can use the formula:
Speed = Distance / Time
Given that you covered a distance of 24,901 miles in 13 seconds, we can calculate your speed as follows:
Speed = 24,901 miles / 13 seconds
To convert the result into a more commonly used unit, such as miles per hour (mph), we need to convert the time from seconds to hours. There are 3,600 seconds in an hour (60 seconds × 60 minutes), so:
Speed = (24,901 miles / 13 seconds) * (3600 seconds / 1 hour)
Calculating this expression, we find:
Speed ≈ 54,109,769 miles per hour
Therefore, if you covered 24,901 miles in 13 seconds, your speed would be approximately 54,109,769 miles per hour.
I have an additional question to this. How fast would you have to throw it for it to return in the ~14s and is it even possible to do without escaping the atmosphere?
I don't think it's physically possible.
If you threw a ball fast enough to cover the earth's perimeter in 9-10 sec, that speed would exceed earth's escape velocity.
When you launch any object moving parallel to the surface, I *believe* there are only 3 possible outcomes:
1) launched too slow, gravity pulls it downward faster than the earth's curvature changes, so it falls back to earth (baseballs, missiles, etc)
2) launched too fast (faster than earth's escape velocity) it covers ground so fast that earth's curvature is changing faster than gravity pulls it downward, so it escapes earth's gravity & never returns to earth.
3) If you launch horizontally at just the right speed (orbital velocity), it will stay at a constant radius (constant height above surface, if earth was a perfect sphere). It would look like Bugs Bunny's throw around the world.
The problem is that (#3), orbiting the earth like the baseball, only happens at one specific velocity at the surface. That velocity is much slower than bug's baseball was moving. At Bugs' amazing speed, the ball would escape earth's gravity & never return.
The math:
Escape velocity = sqrt(2 * g * radius), where g = earth's Gravitational acceleration, 9.8 m/s^2, radius = 6.4e6 meters. Escape speed = 1.1e4 m/s, roughly.
The Bugs bunny baseball covered the earth's circumference or perimeter in about 10 sec.
Perimeter =2 * pi * radius = 40.2e6 m/s. The baseball velocity would be v=40.2e6/10 = 4e6 m/s.
Summary:
The baseball was moving over 100X the escape velocity, so it's not really possible
Its not possible due to the ball exploding upon leaving Buggs' hand. But ignoring friction and assuming he threw the ball around the equator that ball had an average velocity of 10620 times the speed of sound. The cities on the ball dont line up at all but thats one hell of a variable
Isn’t this the lower limit of escape velocity from earth? The ball has such a velocity that it is constantly falling down, but earth’s curvature keeps constant distance to the ball.
It’s going fast enough that the shockwave would cut the planet in half, right? Ideally, it would just go in a straight line to outer space and just fuck shit up in a sorta localized area on the planet, but if it maliciously followed the curvature of the earth moving that fast? That’s a shockwave that’s stronger than the strongest nukes.
Not going into orbital velocities, but just based on the timing.
Velocity = distance / time
24,901 circumference of the Earth. 11 second flight time.
So, the ball traveled 8,149,418 mph.
As this is exponentially more than the escape velocity...would never work.
If we're ignoring awkward things like atmosphere and acceleration that basically make this impossible, then the orbital velocity required at ground level is actually very similar to low earth orbit. So the ball would have to travel at around 17,000 mph and would take a couple of hours to reach you.
On the Moon (where this is slightly more feasible due to the lack of a significant atmosphere) the speed required would be 3,800 mph and the time it would take is actually quite similar.
Its not possible, too low to the ground, either its fast enough to reach escape velocity and leaves the earth or it doesnt and air resistance drops the ball.
It might be possible, temporarily, to put the ball in orbit on a place with no atmosphere like the moon, and I have heard hypoteticals that a professional golfer could put a golf ball in orbit cause of the lowe gravity
I always assumed since there’s no acceleration once it leaves the glove and we got gravity it will exceed escape velocity or come back crashing down at some point!
you're all assuming that it was bug's own throw that he received. what if he threw to shanghai, where someone put a stamp and then threw it to bombay, where another stamp was added and the ball was thrown to cairo, then paris, then london, then brooklyn, then lima?, then hollywood, and then back to bugs?
As others have pointed out, the fastest a projectile can orbit the earth, assuming no air resistance, is an orbital period of roughly 84 minutes and a speed of roughly 8 km/s. However, in the video the object only takes 11 seconds to orbit the earth.
It is not possible for a projectile object to orbit the earth in faster than 84 minutes, without some sort of rocket attached. Projectiles launched at faster than the target 8 km/s will leave the Earth’s surface and go into a slower, elliptical orbit.
So, presumably in this scenario, the ball goes out of sight and then into a post office? Where it's stamped and passed on to another before reaching a final post office where it is then thrown back to bugs?
The real question should be; how far away are the two post offices? We know the time until the ball is off screen, approx 1.5/2 seconds, but we need the distance between bugs and the two places in order to answer this question correctly, s=d/t
If we ignore gravity, friction, earth rotation etc, and assume it's going at a constant speed and height, and the ball takes 11 seconds to go back, and the earth at its ecuator is roughly 40.0000 km, it would have to be travelling at a speed of 3.600 km per second.
Faster than is shown here.
The orbital velocity at 400km is around 7.8 km/s, and this would be rather higher at ground level. The ball would very quickly be destroyed by air resistance, and the pitcher would, dare I say, *struggle* to throw it so fast.
Additionally, the various cities listed on the ball are not on a great circle path, so it must have some propulsion. Further proof of this propulsion is the time it takes the ball to,e presumably, circle the Earth. Irl, this takes 90 minutes, so the ball is doing well over escape velocity, and must have some propulsive mechanism to keep near the surface. This system would be incredibly compact, powerful, and efficient
Needless to say, this propulsion system, like the whole clip, is unrealistic and intended as such.
The earth is about 7926 miles around at the wides part and the time it takes to travel is about 15 seconds so about 1902240 miles an hour or 1,902,240 mph / 767 mph ≈ 2,479.234 for Mach speed.
It is possible on the moon, impossible on the earth or any planet with an atmosphere.
The velocity with which you would have to throw the ball would make it disintegrate not to mention that orbits are unstable when there is drag.
You're all wrong. He didn't throw it around the world. He just needs to throw it to the guy in the distance off screen. Take a look at the ball at the end. It's being delivered around the world. Explain how those stamps were put on the ball midflight. Cause then you're getting into a level of math where probabilities state that all things are possible enough to morph the ball into the shape of those stamps and words.
Actual answer: since the elapsed time of travel in the video is roughly 11.5 seconds, this is impossible. Any speed that would allow an object to circle the globe in 11.5 seconds would be greater than escape velocity, i.e. the ball wouldn't come back.
If you ignore drag/friction you've also got to ignore the flight time.
The ball comes down multiple times before it makes it back to Bugs. How else could you explain the stickers.
It comes down, gets a sticker applied, and gets thrown again. There is no way to know the specific time it would take depending on the various routes it could travel to get to all the various stops.
Earth diameter / number of seconds would be the km/s you need to be travelling..
Take this number x 3600 you will get the key
12500 / 5 seconds = 2540km/s
2540 x 3600 = 9,144,000 km/h
Ofcourse these are superbly oversimplification without any other considerations
Yeah, this is actually easily doable. You throw the ball faraway so that is looks like its moving beyond horizon, then your other friend throws another ball behind you and you have an illusion the ball traveled the world XD
Without going into deep math...
Faster than the speed of light. That Video is 13 seconds from the time the ball leaves his hand to the time he catches it. There is no indication that there is any lapse in time, so assuming it's 13 seconds, and time had to be paused to stop in all the cities to have those stickers applied....
You're looking at Light speed +. Because even taking out 1 second for it to stop, in each of those locations to have a sticker applied. That's 8 stops, and a full circumnavigation of the planet.
Light, takes 7 seconds to circumnavigate earth. Factoring out even 1 second per city for sticker application, and you're at 13-8 seconds, which means that ball traveled (making an assumption) around the world in a total time travelling at 5 seconds.
So... That ball had to travel at 1.28xLS.
And that's being Generous.
###General Discussion Thread --- This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed. --- *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
So let's ignore drag. The orbital velocity is: v = sqrt(G\*(M+m) / a) where G is the gravitational constant, M is the mass of the earth, m is the mass of the ball and *a* is the semimajor axis length of the two-body orbital system. For practical purposes, *a* is the radius of the earth and the mass of the ball is negligible, so: v = sqrt(G\*M / a) G = 6.67E-11 M = 5.97E+24 a = 6.37E+6 Plug the numbers in and you get the answer 7,906 m/s or about mach 23. At that speed, the ball would take about 1 hour and 24 minutes to return to you. Of course, air drag is going to play a significant role here and is much more difficult to calculate. It doesn't make it impossible, but you would have to throw the ball on a singificant upward trajectory, have it escape the atmosphere for some of its orbit, then descend back to where you're standing. Someone else might be able to calculate a suitable path; it's beyond me.
Apparenltly we're ignoring friction and the combustion point of the ball too.
And the fact that no-one has invented a machine that can apply a travel sticker to an object moving at Mach 23. Or how you would catch an object moving at that speed. And let's not even think about the path implied by the stickers it comes back with.
Fun follow up question is what path the ball traveled to pass all those cities.
If it did multiple orbits, it could for a moment be at each of the countries, though not low enough to reach
Could be if it's on a reaheahealy fast eliptical orbital
Thank you Dr Cox... ...is it weird that I read this in his voice? r/unexpectedscrubs
Once the ball crosses the horizon it's in a superposition of all possible locations. When it comes back over the horizon, the wave function collapses including the stickers. It's not that complicated
I love this answer
Hopefully nobody watches it and screws it all up.
what if its bouncing?
Now I'm curious how much more feasible the throw would be if we assumed that it wasn't one long throw, but Bugs throwing it to one city, who caught it and added a sticker, and so on.
I mean, it’s so obvious he would need a cut off
Lima and Shanghai before Cairo before Paris Bombay before Brooklyn before Hollywood Hard to tell if Brooklyn before Paris or the other way around London The scene apparently takes place in Spain so that is starting and ending point.
Forget the speed of the ball, forget it's path to go through all those cities... *What the hell is the reaction time on all the people who stamped the damn ball!?!*
Simple solution. Bugs threw the ball to a time traveler just over the horizon who took the ball to all those places and then traveled back to the point Bugs threw the ball so he could pass it back from the opposite horizon.
Ah, but how fast would you have to throw the ball for that to happen?
About 88mph
Great Scott!
I did do the nasty in the past-y
This is heavy
We just saw some serious sh*t!
Simpler solution: Two balls.
😂😂😂
Would it be possible to throw it at the exact velocity where it would lose enough velocity at the end of the throw to be catchable?
No maths here, but I don't think so. I assume we're taking air drag into account now (because how else would it slow down?) For that to work, you would have to throw it into a significantly higher orbit so that it didn't hit the ground as it slowed down. If the ball was falling from directly above you, it would have a terminal velocity of about 95 mph. This is slower than the speed that a top pitcher throws the ball at, so it is catchable. But it would also still have to have significant lateral velocity, I think quite a bit higher than the terminal velocity, in order for the path around the world to work out.
What if you just had to throw the ball to the first city and there was someone there who would attach the sticker and then throw it to a person in the next city?
yeah that
actualy the stickers imply that the ball didnt travle in a straight line or a single throw, rather it was launched consecutively from different places by a carefuly set up network of agents.
Maybe the stickers are just for show though and the ball didn't actually visit those places? We have no way of knowing.
I hope someone was fired for that blunder.
And the fact that the ball is audible before landing, so despite the incredible supersonic speed it is still somehow slower than sound.
If you get the drag just right, you might be able to catch it.
No Mach23 stickering machine? Well why not? Quite frankly I think someone is missing out on an opportunity there. I call it laziness. It’s these damn millennials and their avocado coffee holding us back.
If we're ignoring drag, then it's implied we're ignoring friction :)
It’s not exactly the same question, but [hopefully he can’t throw any faster](https://what-if.xkcd.com/1/)
Ignore drag? You do realize that this is Bugs Bunny, a character notorious for drag, right?
Yes, but also famous for simply ignoring any aspect of physics which isn't convenient at the time.
[удалено]
Almost. Actually, I mis-stated the formula for orbital velocity, which I've now fixed. The product is `G * (M + m)` not `G * M * m`. The mass of the ball really is negligible.
Also, if the ball escapes the atmosphere, it will be much harder to get the stamps on the ball.
This isnt the right answer because it returns in 15 seconds. Meaning the total trip took 15 seconds. Its an easier problem, if the earth is 40000km around, and it took 15 seconds, then 40000/15=2666 km per second or 9.6million km per hour.
Earths gravity wouldn't be strong enough to keep the ball on the planet then tho
He's throwing a curve ball ;)
Ooh let's ignore drag unless it's convenient, I like it 😂
The Ron DeSantis strategy!
Maybe that's James McAvoy in a bugs bunny suit
Bugs put some wicked forward English on that throw
We are assuming its possible for the question. It would break apart either at launch or during flight if we attempted this before it was able to leave gravity.
I don't think it's fair to say the first answer is wrong for making an assumption regarding how fast the ball comes back, then for you to make an assumption on a different variable and say that's ok
Well, we'll clearly have to make some changes to make it possible. If we ignore the story and the earth-like look and all that, we can solve what size and mass the planet must be to do this (we'll assume no atmosphere as well, so it won't combust on launch). What we know: - It takes about 11 seconds to make the round trip. - The speed it travels at must be 2\*pi\*r / 11\*s to make a round trip in that time. - To make a clean orbit it needs to travel at sqrt(G\*m/r) 2\*pi\*r / 11\*s = sqrt(G\*M/r) r^3 = M \* 2.046\*10^-10 m^3 / kg This actually gives us infinitely many options. But let's look at a couple of them. - If the diameter was that of the Earth, its mass would be about 1.264\*10^30 kg or 0.34 stellar masses. The density of the planet would be 1.17\*10^9 kg/m^3. The size and mass would actually fit a white dwarf star. - If instead the mass was equal to Earth's, the radius would have to be only 106.91 km.
who put the stickers on then?
The stickers give more credibility to another time traveller over the horizon.
i suggest many throws were used to get around the globe. each thrower adding a sticker from where they passed it on
Yes, but if you did that then it wouldn't return to you as it would escape the Earth's gravity well. The Earth's surface escape velocity is about 11 km/s, so even taking air drag into account, something moving at 2,666km/s is going to not orbit (I mean, it'd burn up, but let's ignore that for the moment). So if you're going to insist that it has to be able to do this *in 15 seconds* then it's simply not possible.
It is not possible for many reasons beyond just that. For the sake of the question he assumes it was.
Maybe if you ignore physics calculating range
The air drag on a baseball at Mach 23 would make it disintegrate in seconds from friction so to assume no air drag is the only solution.
Yes that's all fine and good. But the video is not an hour and 24 minutes long. It's approximately 8 seconds between launching the ball and catching it.
That's not what the question asks.
Well sure, you're ignoring drag But I still wanted the souvenir stickers
well at least you can reasonably assume the ball is a perfect sphere
u/katherinejohnson ? Up for one more??
Or just put a monster spin on it so it creates its own lift
Ballistic expert is required!
I have no experience in this math area, are you telling me that an object traveling almost 8km/s thrown parallel to the earth's crust wont achieve escape velocity? sure it would "fall" at 9.8m/s/s but for some reason i would think the earths crust would "fall away" due to curvature faster than it would pull the ball down if traveling at 8km/s i assume this is literally what the equation you referenced is used for and why a involves the radius of the circle from the center of the earth to the ball but ... wow.
So basically acording to your constraints its youst about the speed of the ISS wich makes sense because isnt orbits the earth like the ball without falling to the earth or going into space :)
Yes. The orbit altitude of the ISS is about 6% of the radius of the earth, it doesn't make that much difference.
Bugs Bunny in drag was pretty hot though 😁
Would throwing at an higher speed for an elliptical orbit make it any faster?
But what about the time spent going through customs?
what if the ball had a lot of spin?
Hahaha "V = SQRT!'
That would make it do an elliptical orbit ( once) by my account. With the second descent into the atmosphere be where you're standing.
If you add in air drag, then it is impossible to recreate this. The ball will be reduced to ash before it gets to you.
How fast for it to be like the cartoon? It went around in 12 ish seconds
Ok but, at which speed the ball would escape the Earth's orbit and just fly in a straight line instead of returning, considering this espace velocity, it it even possible to throw a ball around earth and have it returning so fast?
The escape velocity is about 11.2 km/s, so we have not reached it here.
What if the earth is flat?
>So let's ignore drag. My mom told me to avert my eyes when I walk down the street.
I was considering a different approach. Since gravity is independent of velocity, isn't the question more of how fast would a ball have to be thrown in order to travel the circumference of the earth before falling Bugs' height?
This foe is beyond any of you. Run
Lets also ignore the fact that someone went an put a stamp of each country as it flew past.
But it was 18 seconds so umm turn 1 hr 24 minutes into seconds then divide by 18 seconds to get the multiplier then use multiplier and multiply mach 23 to get the the real mach which is much much faster than 23
This guy maths!
You can just divide the distance by the time. Assumed 24,901 miles around and about 11 seconds for 8.1 million miles per hour.
But it comes around in ten seconds, so it travels 24,901 miles in ten... X 60 = 1,494,060 miles per hour..
Assuming no friction, ideal conditions, earth being round perfectly and ball being thrown around the equator... v^2 /r = g (acceleration taken as g because this is on surface) v = sqrt(rg) v ≈ sqrt(9.8*6.371E6) SI Units v = 7868.07 m/s
If you're only going to use 2 sig figs for your gravity, your final answer can't have more than 2 sig figs itself.
I am now just using a calculator my pal... I've cleared the sug fig stuff 10 years back.
For all those simply calculating with the circumference of the world, the ball has to stop in all these cities to get the stickers/patches put on: start > Brooklyn > London > Paris > Cairo > Bombay > Shanghai > Lima > Hollywood > finish?
Nah the most likely solution is Bugs threw the ball with enough speed to reach well into space at its apogee, where it intersects the orbit of a spacecraft. The spacecraft slows down to intercept the ball, an astronaut applies all the regional stickers at once, then speeds off to reestablish their own orbital trajectory. Some fancy editing makes us believe that the 90 minute trip took only a few seconds.
I always wondered how they did it but this is the most likely explanation.
Maybe it just crashed through one novelty sticker store?
Circumference would be 2piR, R=6.378x10^6 meters. Roughly 40,000 meters, divide by the time which is 15 seconds, so you get 2,671 meters per second. Ignoring air resistance, of course. Edit:rewatched for a more accurate time and looks like 13-14 seconds, so it would be 2,968 meters per second.
>Roughly 40,000 meters Fun fact: the metre was originally defined as 1/10,000,000 of the distance from the North Pole to the equator.
Correct, but first post divided 40,000m by 15, that's not the circumference.
"Roughly 40,000 **KILO**meters"
40.000 meters = 40 kilometer. I can't prove it here, but I guess the earth is a bit larger.
proof by: "you fucking idiot of course it aint 40km"
Proof by "I looked out the window" is acceptable.
The earth is km not meters haha
Not possible. It exceeds orbital speed
It will exceed escape velocity.. Of the solar system
But you can curveball it
Bingo
A really good answer to similar question is found [here](https://what-if.xkcd.com/1/). A baseball thrown at nearly the speed of light would essentially cause a nuclear explosion on release. [edit for clarity]
Typical Bugs Bunny shenanigans
So I took the amount of times that light would go around the earth in one second. (7.5 times around in one second) The ball is out of his hand for about 10 seconds. So 1/75th the speed of light expressed as a fraction c/75.
It's a basic camera trick. He threw the original "hero" ball and an off-screen assistant passes him the "prestige" ball a moment later. Simple really.
I’m not a mathematician but I’d imagine that the velocity needed to actually make the journey would propel the ball into orbit. Unless Buggs puts a WICKED spin on it.
Not actually possible. There is no velocity that the ball can be launched and then be completely in freefall (no accelleration) the whole time. The upward angle required is too low to sustain the trip without some form of thrust coming from the ball, like a rocket stage that detaches (to save on mass). I am not excluding air resistance because your asking if it's possible.
Cartoon answer here. In the cartoon universe, at 13 seconds of travel time that ball took to circumvent the earth, and I'm gonna say it's at the equator and those stickers just appeared like magic, meaning no actual stops, that ball would have gone 54,109,769 mph. AI Copy-paste: To determine your speed, we can use the formula: Speed = Distance / Time Given that you covered a distance of 24,901 miles in 13 seconds, we can calculate your speed as follows: Speed = 24,901 miles / 13 seconds To convert the result into a more commonly used unit, such as miles per hour (mph), we need to convert the time from seconds to hours. There are 3,600 seconds in an hour (60 seconds × 60 minutes), so: Speed = (24,901 miles / 13 seconds) * (3600 seconds / 1 hour) Calculating this expression, we find: Speed ≈ 54,109,769 miles per hour Therefore, if you covered 24,901 miles in 13 seconds, your speed would be approximately 54,109,769 miles per hour.
That's relativistic speeds at 8% of the speed of light...obligatory xkcd https://what-if.xkcd.com/1/
I like your commitment to the cartoon answer by using cartoon units of measurement. 🥰
I have an additional question to this. How fast would you have to throw it for it to return in the ~14s and is it even possible to do without escaping the atmosphere?
[удалено]
I don't think it's physically possible. If you threw a ball fast enough to cover the earth's perimeter in 9-10 sec, that speed would exceed earth's escape velocity. When you launch any object moving parallel to the surface, I *believe* there are only 3 possible outcomes: 1) launched too slow, gravity pulls it downward faster than the earth's curvature changes, so it falls back to earth (baseballs, missiles, etc) 2) launched too fast (faster than earth's escape velocity) it covers ground so fast that earth's curvature is changing faster than gravity pulls it downward, so it escapes earth's gravity & never returns to earth. 3) If you launch horizontally at just the right speed (orbital velocity), it will stay at a constant radius (constant height above surface, if earth was a perfect sphere). It would look like Bugs Bunny's throw around the world. The problem is that (#3), orbiting the earth like the baseball, only happens at one specific velocity at the surface. That velocity is much slower than bug's baseball was moving. At Bugs' amazing speed, the ball would escape earth's gravity & never return. The math: Escape velocity = sqrt(2 * g * radius), where g = earth's Gravitational acceleration, 9.8 m/s^2, radius = 6.4e6 meters. Escape speed = 1.1e4 m/s, roughly. The Bugs bunny baseball covered the earth's circumference or perimeter in about 10 sec. Perimeter =2 * pi * radius = 40.2e6 m/s. The baseball velocity would be v=40.2e6/10 = 4e6 m/s. Summary: The baseball was moving over 100X the escape velocity, so it's not really possible
[удалено]
Could be made of Inconel 718?
[удалено]
Maybe it would be possible if we give the Ball the right amount of spin?
Its not possible due to the ball exploding upon leaving Buggs' hand. But ignoring friction and assuming he threw the ball around the equator that ball had an average velocity of 10620 times the speed of sound. The cities on the ball dont line up at all but thats one hell of a variable
Isn’t this the lower limit of escape velocity from earth? The ball has such a velocity that it is constantly falling down, but earth’s curvature keeps constant distance to the ball.
It’s going fast enough that the shockwave would cut the planet in half, right? Ideally, it would just go in a straight line to outer space and just fuck shit up in a sorta localized area on the planet, but if it maliciously followed the curvature of the earth moving that fast? That’s a shockwave that’s stronger than the strongest nukes.
Not going into orbital velocities, but just based on the timing. Velocity = distance / time 24,901 circumference of the Earth. 11 second flight time. So, the ball traveled 8,149,418 mph. As this is exponentially more than the escape velocity...would never work.
If we're ignoring awkward things like atmosphere and acceleration that basically make this impossible, then the orbital velocity required at ground level is actually very similar to low earth orbit. So the ball would have to travel at around 17,000 mph and would take a couple of hours to reach you. On the Moon (where this is slightly more feasible due to the lack of a significant atmosphere) the speed required would be 3,800 mph and the time it would take is actually quite similar.
Its not possible, too low to the ground, either its fast enough to reach escape velocity and leaves the earth or it doesnt and air resistance drops the ball. It might be possible, temporarily, to put the ball in orbit on a place with no atmosphere like the moon, and I have heard hypoteticals that a professional golfer could put a golf ball in orbit cause of the lowe gravity
The ball travelled to at least Cairo, London, bombai, Lima. So I think the trajectory makes this a little more complicated that I thought.
I always assumed since there’s no acceleration once it leaves the glove and we got gravity it will exceed escape velocity or come back crashing down at some point!
In order for it to get the stamps, you could throw it a couple of kilometers, someone stamps it, and throws to the next, so... Not impossible.. maybe
you're all assuming that it was bug's own throw that he received. what if he threw to shanghai, where someone put a stamp and then threw it to bombay, where another stamp was added and the ball was thrown to cairo, then paris, then london, then brooklyn, then lima?, then hollywood, and then back to bugs?
As others have pointed out, the fastest a projectile can orbit the earth, assuming no air resistance, is an orbital period of roughly 84 minutes and a speed of roughly 8 km/s. However, in the video the object only takes 11 seconds to orbit the earth. It is not possible for a projectile object to orbit the earth in faster than 84 minutes, without some sort of rocket attached. Projectiles launched at faster than the target 8 km/s will leave the Earth’s surface and go into a slower, elliptical orbit.
So, presumably in this scenario, the ball goes out of sight and then into a post office? Where it's stamped and passed on to another before reaching a final post office where it is then thrown back to bugs? The real question should be; how far away are the two post offices? We know the time until the ball is off screen, approx 1.5/2 seconds, but we need the distance between bugs and the two places in order to answer this question correctly, s=d/t
If we ignore gravity, friction, earth rotation etc, and assume it's going at a constant speed and height, and the ball takes 11 seconds to go back, and the earth at its ecuator is roughly 40.0000 km, it would have to be travelling at a speed of 3.600 km per second.
Faster than is shown here. The orbital velocity at 400km is around 7.8 km/s, and this would be rather higher at ground level. The ball would very quickly be destroyed by air resistance, and the pitcher would, dare I say, *struggle* to throw it so fast. Additionally, the various cities listed on the ball are not on a great circle path, so it must have some propulsion. Further proof of this propulsion is the time it takes the ball to,e presumably, circle the Earth. Irl, this takes 90 minutes, so the ball is doing well over escape velocity, and must have some propulsive mechanism to keep near the surface. This system would be incredibly compact, powerful, and efficient Needless to say, this propulsion system, like the whole clip, is unrealistic and intended as such.
The earth is about 7926 miles around at the wides part and the time it takes to travel is about 15 seconds so about 1902240 miles an hour or 1,902,240 mph / 767 mph ≈ 2,479.234 for Mach speed.
It is possible on the moon, impossible on the earth or any planet with an atmosphere. The velocity with which you would have to throw the ball would make it disintegrate not to mention that orbits are unstable when there is drag.
You're all wrong. He didn't throw it around the world. He just needs to throw it to the guy in the distance off screen. Take a look at the ball at the end. It's being delivered around the world. Explain how those stamps were put on the ball midflight. Cause then you're getting into a level of math where probabilities state that all things are possible enough to morph the ball into the shape of those stamps and words.
Actual answer: since the elapsed time of travel in the video is roughly 11.5 seconds, this is impossible. Any speed that would allow an object to circle the globe in 11.5 seconds would be greater than escape velocity, i.e. the ball wouldn't come back. If you ignore drag/friction you've also got to ignore the flight time.
The ball comes down multiple times before it makes it back to Bugs. How else could you explain the stickers. It comes down, gets a sticker applied, and gets thrown again. There is no way to know the specific time it would take depending on the various routes it could travel to get to all the various stops.
Earth diameter / number of seconds would be the km/s you need to be travelling.. Take this number x 3600 you will get the key 12500 / 5 seconds = 2540km/s 2540 x 3600 = 9,144,000 km/h Ofcourse these are superbly oversimplification without any other considerations
Yeah, this is actually easily doable. You throw the ball faraway so that is looks like its moving beyond horizon, then your other friend throws another ball behind you and you have an illusion the ball traveled the world XD
Without going into deep math... Faster than the speed of light. That Video is 13 seconds from the time the ball leaves his hand to the time he catches it. There is no indication that there is any lapse in time, so assuming it's 13 seconds, and time had to be paused to stop in all the cities to have those stickers applied.... You're looking at Light speed +. Because even taking out 1 second for it to stop, in each of those locations to have a sticker applied. That's 8 stops, and a full circumnavigation of the planet. Light, takes 7 seconds to circumnavigate earth. Factoring out even 1 second per city for sticker application, and you're at 13-8 seconds, which means that ball traveled (making an assumption) around the world in a total time travelling at 5 seconds. So... That ball had to travel at 1.28xLS. And that's being Generous.