Problem #9 - Tunnel Through Earth

Problem #9 - Tunnel Through Earth

[@kallmekrissarchivetiktoks8012]

I love your enthusiasm. It always brings a smile on my face 😊😊😊

[@lecturesbywalterlewin.they9259]

:)

[@stephenfoster7149]

I really don't know physics, but it would seem to me that we'd only have to calculate the time to the halfway mark. Given the assumptions, I tend to imagine that the speed/energy gained heading toward the middle point would be equivalent to the required energy to make it up the second half. The time might be derived by calculating the velocity that gravity can pull an object at the given angle, and then using that velocity determine how long it would then take to travel half the distance of the tunnel. Last, multiplying by two. The reason for the halfway calculation is in case it takes more distance than the halfway mark to reach terminal velocity. Love watching your videos Walter. Even though the physics are beyond me, your enthusiasm is contagious. Thanks for everything you do!

[@bernhartschmieder1632]

its super awesome that i just had calculated this a couple of days ago. Listening to you talk about this problem brought a smile to my face :)

[@ramonedgardosarmientomatut2851]

Hi professor Lewin, mi name is Ramon Sarmiento , I am from Honduras , Central America. I am mechanical engineer I used to see your videos to learn about physics and to teach also, I taught Physics for more than 12 years in one of the universities here in Honduras , your lectures inspired me to teach it in a different way , Its Amazing how students like physics when teachers demonstrate its principles in the classroom using equipment . My best regards professor , you are one of the best in the world- if not the best!!

[@lecturesbywalterlewin.they9259]

:)

[@ramonedgardosarmientomatut2851]

Have a nice day. God bless you!!

[@milanguzvic8457]

You sir are one of my heroes, I've never seen a better teacher

[@lecturesbywalterlewin.they9259]

:)

[@andreathecat100]

In summer is so hot! So hard to think! IF only it was " spring"! the problem would be more easy to face!

[@shrinivasaprabhu8931]

you are looking really colourful today sir!

[@lecturesbywalterlewin.they9259]

:)

[@Bibibosh]

shrinivas prabhu amazing shirt

[@gildemann]

Simple words - Thank You Professor!

[@DaviFeliciano]

Assuming that the density of the Earth is constant, the gravitational force on a point inside of the sphere of the planet grows linearly as you get near to the center. The resultant force, for r < R, in which R is the radius of the Earth, would be given by the formula F(r) = GMmr/R³. That is true only because the gravitational force is proportional to 1/r², what makes possible to consider the entire mass of the planet as if it was on its center, and because the Earth has a simetric distribution of mass. So, a good way to solve this problem is to consider that the force which is pulling the falling object is a force that complies the Hooke's Law. Then, it is possible to consider the system as a harmonic oscillator, and after some calculations, which include the solution of a second-linear ordinary differential equation, find the angullar frequency ω. So, if the found value is ω = √[(GM)/R³], and the period is T = 2π/ω, the corresponding value of T is approximately 5073 seconds.. That was the way i found to resolve this problem.. :)

[@rihow6071]

beautiful shirt

[@lecturesbywalterlewin.they9259]

:)

[@Bibibosh]

Ri How? amazing right?

[@hareendraguruge1375]

T^2=4pi^2Ro^3/GMo, this yields the time T = 84.52 minutes (the period from Oslo to Oslo back) so the answer is half this time 42.26 min. This is as same as the time it takes if the tunnel passes through the centre of the earth. This means any tunnel that gets through earth will take the same time to get from one side to the other.

[@MarkoTManninen]

First time here, I like the problem set up. My estimation is that I would probably end up to Honolulu faster than I can solve the problem for now.

[@lecturesbywalterlewin.they9259]

:)

[@lera2794]

thanks to you I fell in love with physics, thank you very much!

[@peeyushgoyal9587]

84 minutes from Oslo to Oslo back again

[@trevorteolis3691]

Writing the problem in the description would be really convenient for reference

[@koenth2359]

Data: R=6.4E6m, G = 6.7E-11 m^3/kgs^2, M= 6.0E24 kg, pi =3.14, Oslo=60N11E, Honolulu 21N158 W In this model, with a somewhat idealized earth, gravity is always directed towards its center. And the gravity at a point p is just as if only the mass that is closer to the earths center than p, would be located in the center of the earth. So for height r

[@AfonsoVale177]

Thanks for the challenge! I got about 100min, but I disregarded the effect from the "smaller piece of earth" as I calculated them to be negligible.

[@greenmanreddog]

I suspect the time taken is the same regardless of the entry and exit points on the surface as the acceleration vector along the tunnel is proportional to the angle of the tunnel to the angle of the normal to the radius. So we can simplify the calculation by going through the center of the Earth. The time taken will be twice the time to get to the centre of the Earth as getting back to the surface is just the reverse of getting to the centre. As I recall from Physics classes: s=ut+1/2(at^2) So initial speed u = 0, acceleration a = 9.8ms^-2 and distance s = 6.4*10^6 (to the center of the Earth) Rearranging all this I get: total time t = 2 * sqrt(2*s/a) Substituting in: t = 2 * sqrt(2 * 6.4*10^6 / 9.8) Total time = 2285.71 seconds.

[@deyvrules1809]

You are awesome !

[@AMotoVlogger]

The (spectrum) frequency of your sense of fashion is unparalleled!

[@lecturesbywalterlewin.they9259]

:)

[@AndrewDotsonvideos]

I hope I can explain physics this well one day. I started my own channel just because of how well he does it

[@lecturesbywalterlewin.they9259]

:)

[@non-inertialobserver946]

Hello there tensor boi

[@tanmayramani5491]

required time= 42.26 mins as the body just fall in tunnel(friction less) it will execute SHM between Oslo and Honolulu with time period = 2pi√(R/g) as there is no external force on the body. Required time will be half of the total time period = pi√(R/g) = 42.260 mins

[@lovejazzbass]

Thank you professor for this wonderful challenge. My dad my sister and I are trying to figure this problem out I am a 7th grader my sister is a 4th grader my dad allowed us to use his account. Thank you once again.

[@lecturesbywalterlewin.they9259]

:)

[@russchadwell]

air is let in from above and ends up with an atmospheric ambient pressure so high that vehicle occupants get the bends and die as the near Hawaii? or, the ascend slow enough to prevent this?

[@tusharmaurya1668]

Good one sir!

[@SongSeeker7]

So If I were to pitch a baseball from Oslo at 21600 Km/hr to Honolulu, It would arrive at this magic time. ...well at least I think? :)

[@MuhammadAbdulrahim-cl3wk]

As the gravitational field is a conservative field, we can derive a formula for potential energy as a function of position; >> delta Pe = integral (GMm)/R^2 from R2 >> R1 , M(R) = D*4/3*pi*R1^3 = (GmM(R1^2-R2^2))/2*R1^3 >> delta Pe = - 1/2*mv^2 D is the density of the earth Pe is the potential energy difference v is the speed at the midpoint of the tunnel R1 is the radius of the earth at its surface R2 is the direct radial distance between the vehicle at the midpoint of the tunnel and the center of the earth so, v = sqrt(GM*(R1^2-R2^2)/R1^3) and x = sqrt(GM*(R1^2-R2^2)/R1^3) * t >> the total length of the tunnel is about 9660 km, and the entire trip will take approximately 27 minutes >> note that the external spherical shell outside has zero net force on the vehicle at every point of the tunnel and only the internal sphere does such force on it

[@ernestschoenmakers8181]

I had about the same answer cause i had an OH distance of 9698 km and a time t= 26.93 min.

[@alleneverhart4141]

reminds me of 'alice in wonderland' down the rabbit hole - a not insignificant problem in the day of lewis carrol was what would happen if there was a hole straight through the center of the earth and one jumped into it. it is not as simple as it sounds. if i recall correctly, one can show, with a little calculus, that the pull of gravity from parts of the earth at greater radii than your position cancel out. so that at the center, there is no pull at all because everything cancels out. then as one ascends, under momemtum, back toward the surface, more and more gravity is felt - the effective mass of the earth actually increases. i mention this because there is this matter of dark matter and how that could affect the rotational speeds of stars in the galaxy. it turns out that if there is a spherical distribution of unseen stuff, that the stars furthest from the center of the galaxy, "see" a larger center of mass than the stars closer to the center and, thus, move faster than one would anticipate without dark matter.

[@Bibibosh]

I love your t-shirt!! IT IS AMAZING! Wow it is actually really nice. Were did you get that tshirt? i want to buy one for my dad!

[@lecturesbywalterlewin.they9259]

ask google where to buy one

[@shreeswarajpanchal4362]

Sir today is Indian national festival teacher's day. So I wish you HAPPY TEACHER'S DAY as your lectures taught me many things

[@lecturesbywalterlewin.they9259]

:)

[@psodq]

Please excuse me, when I don't even try to figure out the problem, as I have no mastery of these physics. I just started to ponder how a passenger might experience gravity in such a ride. Along the entire journey a gravitational force would be directed towards the center of the sphere. At the start of the journey there'd be perhaps free-fall or weightlessness, then at the mid-point a slight pull towards the center, which gradually "turns around" in the direction of travel as the journey continues. The feel of gravitation would increase gradually, and at the other end there'd be a 1 G pull perpendicular to that local surface. I'd think it would make for an interesting sci-fi movie.

[@robrechtv6868]

Hello. It's obvious that g is not constant due to the changing depth in the Earth. Determining variable g uposed to its location in Earth, using that M below the position is a function of the volume of a sphere through the position V(Earth) and the density of the Earth. Then make kinematic calculations with given distance |OH| and the place dependant g to calculate time t. Does this make sense Professor?

[@pm1248]

Mass M below the position r is a function of the volume, that is right. So it is a function of r^3 assuming constant density. Gravitational force on the other hand decreases with r^-2. In total gravitational force is proportional with distance r. That gives a harmonic oscillation! Use the law of the graviational force at the surface of earth to calculate the constants for this oscillation and that gives you the period of the oscillation T=2*pi*sqrt(m/D). By the way: the value of T is independent of the way through the earth.

[@pm1248]

Addition: the value of T is independent of the way through the earth - when assuming a straight way ... ... other shapes of tunnels will give longer or shorter travelling times. You may travel faster if you come closer to the center of the earth compared to the straight way.

[@mguebel]

G is an universal constant called « Gravitational constant ». G is not the gravity on earth surface. en.wikipedia.org/wiki/Gravitational_constant

[@Yeatzy]

I love your attitude, can I get a smiley? ;)

[@lecturesbywalterlewin.they9259]

:)

[@PC_YouTube_Channel]

This was my self-assigned final project for Differential Equations a few months ago!

[@togethervines2653]

Beautiful bracelet sir n your lectures .....damn!!

[@lecturesbywalterlewin.they9259]

yup!

[@mixer8774]

Now this one must involve calculus because of all the changing variables right? Or at least it would be much easier with calculus, right?

[@ganymedkallisto5561]

Hello, my answer is 42.26 minutes. Surprisingly it is the same time it would take to fall through a hole from the northpole to the southpole under the same restrictions (no friction etc.). That is because the force which accelerates the mass is the gravitational force times sin(phi), when phi is the angle between the distance from the mass and the center of earth and the distance from the mass to the middle of the tunnel. But at the same time, the distance between the mass and the middle of the tunnel is the distance between the mass and the center of the earth times sin(phi). So, the sines cancel out when writing down the equation of motion. The result is a force on the object, that is proportional to the distance from the middle of the tunnel. That means, the motion is a simple harmonic. And since we know the period- time T of a simple harmonic, we know the time it would take an object to fall through the tunnel (T/2). Best Regards

[@xpyrox5462]

Mr Walter, I have also had a problem that will keeps me awake at night trying to figure out.. If we managed to drilled a tunnel, perfectly vertical ( ignoring earth rotation ) and one jumped into said tunnel... i am assuming you would be stuck in a sort of infinite falling never being able to fall out the other side - but my question is, Would you ever stop ? but i believe you would actually live from the fall and just be stuck in a never ending freefall. = these are the things that keep me up at night... please tell me i'm not the only crazy person who tries to figure out these weird quirky theory's ! Also, love your videos and will continue to learn from you. it is quite an honour.

[@lecturesbywalterlewin.they9259]

all tunnels you dig through the Earth (thus also one going through the center of the Earth) will act the same way. SHM and the period for one way trip is the same for all tunnels. Watch my solution, I make that very clear.

[@HenkJanBakker]

I'm pretty sure the distance is near irrelevant when sizable enough Assuming it is safe to go straight through the core you'd take the same time as going at an angle and only travel 1/4 of the earth. Pendulum and mass made famous by you would apply here. So I don't need Pi. But I also don't need the actual location coordinates as any two points will yield the same result. My approach would be to calculate how long it would take me to get to the center of the earth and double that. Instinct says it will take me about 30 minutes. Ok now the tough part. The actual math. 6.4 .10^3 km to accelerate at 6.7E-11km h2 takes...... I will need to get back on that one.

[@ABaumstumpf]

The hard part is just that the gravity is not constant - a good approximation is making it linear with the distance towards the centre.

[@maximedelboo9021]

is there an easy way to convert the geographical coordinates to an angle on a circle?

[@Zxymr]

The latitude and longitude ARE angles on the sphere. Look up spherical coordinates.

[@stuartyellow1679]

I used vektors and the law of cosines to get the angle between OCH (Oslo, Centre of the Earth, Honolulu). But it meight turn out that the angle is not really important ;) To answere your question: No xD

[@alleneverhart4141]

Haversine's formula www.intmath.com/vectors/3d-earth-geometry.php

[@panaroyaltea]

My calculation comes to 3656.4 Seconds. If correct, please respond. Will love to share how I arrived at this calculation. Thank you for the challenge. I love your book and your lectures.

[@riks7256]

isn't the same time ou spend to go from the north pole to the south pole?so i think because the force of gravity will be splitted in 2, one to the center and the aother to the line you have to follow..then about 40 minutes or so? maybe i am wrong

[@lecturesbywalterlewin.they9259]

I miss your derivations

[@forloop7713]

I see this as a complex calculus problem that I cannot solve with my current mathematical knowledge, But I still really really like it

[@gotperp66]

Martin S its not complex at all

[@pm1248]

Walter, "km" is of course an SI unit! Using the prefix "kilo" is allowed within the SI units. See chapter 3.1 "SI prefixes" of the SI Brochure of the BIPM.

[@lecturesbywalterlewin.they9259]

yes you are right

[@jetman551]

If you fell into the hole or tunnel you would have to travel 9,660 km to the other side. The initial gravity at the beginning in either city would be 9.8 m/s^2 but as soon as you begin to fall would reduce to 7.4 m/s^2 because of the angle of 41 degrees off the perpendicular to the earth's center. I'm suggesting the KE would peak at 4,830 km, which is halfway to the other side and the resultant gravity would equal the first half but in a negative manner slowing you down to 0 KE & 0 PE at the other end. I see ~ 27 minutes for the journey.

[@ernestschoenmakers8181]

I had about the same answer cause the OH distance = 9698 km and time t= 26.93 min.

[@ayman1966100]

ACTUALLY THE SLOPE OF THE TUNNEL IS NOT CONSTANT IT IS Parabola WE SHOULD CONSIDER THAT THIS MEAN THAT THE ANGLE OF SLOPE ALFA IS NOT CONSTANT IT WILL BE MAXIMUM IN THE BEGINNING AND STARTED TO BE LESS UP TO THE MEDIUM POINT THEN IT WILL STARTED TO INCREASED BUT IN OPPOSITE SIDE the car or the body which will pass from O to H will be free fall . if we do not consider that the gravity constant but it is not constant another factor to be added to the equation

[@mikethom6155]

hi, I'd be interested if you could do a video about how you have used the physics in these problems in your own research or how colleagues have used it to solve in problems in their own fields. Or how the problem is related to an interesting topic, I'd find that interesting and point me to areas of interest I'm ignorant of. thanks

[@lecturesbywalterlewin.they9259]

:)

[@ybaggi]

is there an isomorphism with the swing time of a pendulum?

[@andreathecat100]

you could say so, I also would day it is also easier couse you don't have to resort to small angle approximation.

[@ThomasHaberkorn]

Considering the case that the earth tunnel goes through the center of the earth, how high would the air pressure be in the tunnel at the center of the earth (r=0) ?

[@lecturesbywalterlewin.they9259]

good question - try to calculate it

[@bluegender2005]

Should we ignore the fact, that the gravitational acceleration goes towards the center of the Earth, and in the middle we would crash on the side of the tunnel?

[@Quasilobo]

There's no gravity at all at the center of the earth. Everything balances out. Same for the center of a black hole. Zero gravity--everything balances out, also. Consider the Van De Graaff generator: There is no electrical field inside the dome-everything balances out.

[@nitinranacluster1737]

fantastic sir

[@samyuen8234]

if the person didn't grab the soil(could be anything as long as to make him stop falling) after come out from the other side. will him keep on falling towards the center of the earth over and over again?

[@theextremist4829]

I don't know if we have to account for the fact that the force of gravity toward the earth's center changes as we got closer to the center and mass accumulates above us.

[@hotszho9331]

Let x be the displacement from the midpoint of Oslo and Honolulu and r be the distance between the object and the centre of the earth. Let m=M(r^3/R^3). The acceleration on the particle x"=-(Gm/r^2)(x/r)=-(GM/R^3)(x). From this result, the object will be performing simple harmonic motion. The period T=2pi*sqrt(R^3/GM). But the time needed to slide from Oslo to Honolulu is only half of the period. Therefore, time needed = pi*sqrt(R^3/GM) = 2536 sec, which is around 42.3 mins

[@VikasSharma-nv3pu]

hi sir ,sir how should I watch your videos on KZbin 10 video per week is ok to clear the concept or any suggestions for me because sir I love physics after see the video of 8.02

[@diptanshude2525]

land vehicles will not be able to complete the entire journey from h to o , without friction. somewhere in the midpoint of the tunnel the vehicle would stop due to the least gravitational potential energy over there . Now, to complete the rest of the journey, form the midpoint to o, i.e, from a region of lower gravitational potential to a region of higher, it would have to move against gravity, and without friction it would be impossible to do so.It would be like a car climbing a hill without friction. It might overshoot due to its high inertia at that point but im sure it wouldnt be enough to complete the journey. A rocket might serve for the purpose.

[@vikrantshrimali2287]

prof. Lewin i have a question about light. when a light passes through a polarizer, the electric wave component which is perpendicular to the direction of polarization get obsorbed by the polarizer and light doesnt pass through it. but its magnetic component is still there. wouldn't that magnetic component inturn create electric component regenerating the wave...

[@lecturesbywalterlewin.they9259]

watch my lectures or use google

[@lecturesbywalterlewin.they9259]

Use my lectures and/or google

[@minutemanhomestead7214]

-hey Mr Lewin i have a question thats a little off topic but not sure who else to ask.... can a ship's changing buoyancy be used as potential energy?

[@lecturesbywalterlewin.they9259]

use google, or quora.

[@amolauliyan5326]

GOOD LECTURES SIR, BUT PLEASE TRY JEE ADVANCED QUESTION PAPERS FOR MORE INTERESTING STUFF.IT WILL SURELY MAKE YOU LOVE PHYSICS EVEN MORE.

[@andreathecat100]

Oh my Gauss! This is harder than I thought!

[@manishprasad1298]

andreathecat100 Nice one man, I don't have the talent to make puns😅

[@rajpanchal9226]

andreathecat100 yes you need to use guass

[@burpleson]

2.5 x 10^3 seconds (Unless I made a silly mistake while entering numbers, which is always the most treacherous part.)

[@desimejdani7533]

Hi Professor! I guess the acceleration due to gravity changes while in the tunnel. Its absolute value would decrease in the middle of the tunnel because there is less mass pulling us toward the center. Hope I am right :)

[@farahunn]

i also think that

[@biswaroopjoarder8135]

Basically a simple proof. just take the component of gravitational force of the earth along the tunnel and we will see it takes the form of an SHM where restoring force is of form F=-kx. So time period is 2π√(m/k) and since we're going one way, it's just the half of the time period. Sir, tell me if I'm correct.

[@saumya942]

Professor I have a question which I am sure many others have too. We all know about the classical physics that we're taught in highschool and your lectures prove to be of great help. Then there's books on quantum physics and relativistic mechanics which we're all still familiar with. But where we all face some problem is regarding the mathematics that is demanded by physics. Could you please tell us about the topics that a physicist must know (just a name list would do) and if possible then could you list a few books for those topics?

[@saumya942]

Btw that's an amazingly colourful shirt

[@lecturesbywalterlewin.they9259]

use google

[@saumya942]

I visited a website called how to be a good physicist by gerard t hooft, do you think following that path should be good enough?

[@lecturesbywalterlewin.they9259]

do what you think is best

[@saumya942]

Lectures by Walter Lewin. They will make you ♥ Physics. Thank you professor. I would like to someday meet you in person after getting my degree in physics. I'd owe you a lot.

[@esa062]

Of course we also need to ignore that drilling into the thousands of degrees hot molten rock of Earth's mantle under enormous pressure is a bit more than just an engineering challenge :-) But sadly I don't have the math to form this integral, so I have to wait for the solution.

[@PeperMintification]

Dear prof.Walter Lewin, I came to know in one of your physics lectures that the astronauts that went to the moon left reflective mirrors there. However, I have met a few scientists that are quite sceptical of this. One of the reasons being the enormous precision required to successfully detect the reflected laser beam, which they claim is not achievable. It would be great if you made a video about this topic explaining how this is possible and the level of precision that is required! Thank you!

[@lecturesbywalterlewin.they9259]

>>>I have met a few scientists that are quite sceptical of this. One of the reasons being the enormous precision required to successfully detect the reflected laser beam>>> use google - all details are there

[@PeperMintification]

will do. On wikipedia it says " This is one of the most precise distance measurements ever made, and is equivalent in accuracy to determining the distance between Los Angeles and New York to 0.25 mm". It is certainly extra-ordinary accuracy, since according to the same source 0.25mm is around the length of a fine grain of sand!

[@lecturesbywalterlewin.they9259]

:)

[@icystrangers5482]

A laser beam aimed at the Moon diverges to hundreds of metres wide by time it hits the Moon and the beam reflected back to Earth is so weak that it is measured in photons per second. The beam also wobbles significantly due to a number of factors, like refraction. You do not need a mirror to detect reflection of a laser beam off the Moon. The Moon's surface is sufficiently reflective, given that it captures a much, much larger area of the 200+ metre wide beam. Bear in mind that many factors are changing over time, including laser and detector technology and weather conditions. So it all boils down to interpreting statistical data for a cocktail of factors and attributing it to presence of mirrors. Recall that it took quite some time for lasers to "locate the mirrors" on the Moon, during which the aforementioned factors were all changing. So what lay folk often cite as categorical evidence of mirrors on the Moon is not so indisputable.

[@jetman551]

BTW the tunnel would reach a maximum depth of 2,202 km from the surface at midpoint, which would be 4,198 km from the center of the earth.

[@jetman551]

Let me correct "initial gravity" to "initial acceleration" due to gravity. The midpoint of the tunnel would be directly under the northern end of the Brooks Range in Alaska on the coast of the Beaufort Sea.

[@josephf151]

Well, I got it down to: 9.8 times the integral from 0 to 4815KM of arctan ((X+dX)/4171.7KM)dX =sum of accelerative forces for one half of the trip. But I am unsure of how on earth to continue. somehow I feel like I did it all wrong anyway, as I tried to sum up all the forces due to acceleration through one half pass through the tunnel by looking at how the angle coming outwards from the earths core changes relative to the distance dX traveled. thats what I get for trying calc whilst tired...

[@lecturesbywalterlewin.they9259]

:)

[@josephf151]

I am sorry I did not notice you had replied! It seems I was taking a far overcomplicated approach.

[@mettlertoledo2]

~27 minutes for the 9700 km trip. I am looking forward to see how Dr. Lewin solves the problem and his answer. :)

[@ernestschoenmakers8181]

I had the same answer.

[@MindCrime550]

You use a vacuum, a really really long one.

[@atalazs]

My Solution: 42 min 17 s How I did it: -convert coordinates to kartesian and determine distance by phythagoras -determine gravitational acceleration as a function of distance along the tunnel (which I named x) Here I used the Law, that there is no gravitational Force in a hollow sphere. So I just have to calculate the force of the sphere inside the momentary position. -determine component of acceleration along x as function of x -solve ODE a = -cx -apply all the numbers and calculate

[@atalazs]

actually, the distance between the cities cancelled out - so it wasnt necessary to calculate that.

[@boborulllz]

Great problem sir, but I have a question: what do you mean by the fact that you will post/not post out solutions? As in where will you post them? In the next video, in the comment section or somewhere else? I just did not understand what you meant by that, thank you for your time.

[@micahnightwolf]

So let me get this straight. Someone actually wants to build a tunnel where I'm supposed to slide on my ass from Oslo to Honolulu for several days, through thousands of miles of Earth's mantle, while the oxygen supply in the tunnel is low, and the air inside is boiling from all the magma around it? Don't those people know that airplanes are a thing that exists?

[@arvildasgupta5938]

thank you sir

[@sivasandilya366]

hi professor why does menthol has a cooling effect when applied on skin ??

[@lecturesbywalterlewin.they9259]

ask google

[@carultch]

Due to the fact that the air has A LOT less alcohol in gas phase as part of its composition when compared to water vapor, alcohol tends to evaporate a lot quicker than water. The air is nowhere near saturation, and the forces behind evaporation are a lot greater per unit of the substance. Also contributing to this fact is the fact that it takes less energy per gram to boil alcohol, than it does to boil water. In order for the alcohol to change phase, it must absorb energy from somewhere, to convert it from liquid to vapor. The energy comes from your body, which is why you feel cold as a result.

[@mohammadkhalid9288]

55.08 min ,I guess kind of weird, I applied conservation of energy and periodic motion principles ,can any one help with the length of the tunnel I got it as 9649.22 km just having trouble at this part

[@zafarparkar98]

professor u r really gr8. however i wanted to ask, why are satellites not perpetual motion machines? i would be overjoyed if you answered by a video :-) :-) :-)

[@lecturesbywalterlewin.they9259]

use google for the precise definition of Perpetual motion machines. There is no friction on planets. Thus they continue rotating. Energy is conserved. Anything you build on Earth will be subject to friction. Thus they will not be able to continue. If they did, you would be able to use the energy released in friction (heat) and they would still keep going. Thus you would get free energy.

[@biglift1]

Walter ik heb een verzoek : zou jij een lecture op youtube willen zetten over rendementen en verliezen bij opwekking van elektriciteit transport en verliezen die daarbij optreden en de verliezen die optreden als je deze elektrische energie in een accu laad en daarna weer omzet naar mechanische energie. Heb vaak heftige discussie met mensen die geen jota snappen van rendementen en verliezen die optreden. Jij lijkt me de perfecte teacher om dat eens goed uit te leggen aan de leken.

[@lecturesbywalterlewin.they9259]

Heb je enig idee hoe veel tijd het mij kost om een lecture voor te bereiden? ongeveer 60 uur. Dat deed ik met liefde toen ik 600 studenten had op MIT die mijn 8.01 cursus en later mijn 8.02 cursus volgden. Totaal 71 lectures. Dan ook nog 23 lectures voor 8.03 (ieder ongeveer 80 min lang). Ik ben nu gepensioeneerd. Doe met liefde de bi-weekly problems. een 15 minuten video vereist ongeveer 4 uur voorbereiding.

[@biglift1]

hoi walter nee ik had daar totaal geen idee van . sorry . i had zo het idee van dat zon geweldige vakman als jij dat zo uit je losse pols schud ( weet je meteen hoe hoog ik je heb *hihi* ) maar goed ik leg mijn idee in je idee bakje . dus als je ooit zin en tijd hebt :) cheers mate en bedankt voor het toevoegen op facebook .

[@AManKumar-gh1vt]

can u tell me if a partcle travel in cirle and completed one circle so displacement is 0 but angular disp =2pie why

[@lecturesbywalterlewin.they9259]

yes that's possible, ask google for details

[@AManKumar-gh1vt]

sir give me explanation i can't find it on Google please please sir.

[@lecturesbywalterlewin.they9259]

google "charge moving in constant B-field" q(vXB) force will produce a circular motion if v is perpendicular to B.

[@lecturesbywalterlewin.they9259]

also google "satellite in circular orbit".

[@pradysadhu6890]

Not related to this video but I have a doubt, sir... Consider three sheets arranged like this | | | (From the Left) Let the first one be +vely charged and the remaining two -vely charged (From left) The first 2 act as a capacitor.. We say that interaction between two charges doesn't depend on the surrounding charges... So, there is a possibility that the plate 1(+vely charged) and plate 3(-vely charged) act a capacitor.. The capacity of 'capacitor 1-3' should be C* as well.. Possible? If not, why? *C is the capacity b/w sheets 1&2

[@lecturesbywalterlewin.they9259]

use google

[@AlqGo]

Will you collaborate with Minute Physics please? Would be awesome!

[@lecturesbywalterlewin.they9259]

I prefer they collaborate with me.

[@AlqGo]

Of course :D

[@DesEResources]

If there was a hole in the ground from one side of the earth to the opposite side of the earth, and you dropped ball in the hole, will the ball come out the other end traveling upwards??

[@lecturesbywalterlewin.they9259]

It will be a simple harmonic motion, Watch the videos with my solution.

[@arjunp1394]

at the nearest point to the centre of the earth ,the gravitational pull will be very high .so how can the body reach honol.there is no other force acting expect the gravity.so in my opinion the body will get stuck in a point nearest to the centre of earth.

[@lecturesbywalterlewin.they9259]

watch the video of the solution to this problem. It's simple physics.

[@spacetime3121]

Approximately, 84.5 minutes.

[@spacetime3121]

Yess sorry

[@russchadwell]

45 minute total fall and ascent if no friction

[@ThePanicButton2]

Dear sir why the earth rotate in anticlock wise direction ?

[@lecturesbywalterlewin.they9259]

ccw if earth i viewed from above North Pole, cw if viewed from the axis below South Pole. use google.

[@carultch]

Most objects in the solar system rotate/orbit in the counterclockwise direction, as viewed from constellations such as Draco (pole star of Earth's orbital plane) and Ursa Minor (pole star of Earth's rotation). The exceptions that require an explanation, are those that rotate clockwise when viewed from this direction, such as the planet Venus.

[@NHLHogeschool]

Hallo Walter! Leuke video weer, dit is dus ook een van de redenen waarom we jou graag zien op ons Opening Studiejaar :-) Zie het berichtje dat we je vorige week via walterlewin.com hebben gestuurd. We horen graag van je!

[@lecturesbywalterlewin.they9259]

Hallo NHLH. Ik heb geen uitnodiging van jullie ontvangen. Ik ben niet op de hoogte met "waterlewin.com". Stuur de uitnodiging alsnog. Het probleem met Europa is dat ik minstens 4 dagen nodig heb om aan de jetlag te wennen en dat komt boven op business class flights. Het wordt daardoor erg kostbaar om mijn kosten te vergoeden.

[@NHLHogeschool]

We kunnen altijd kijken wat de mogelijkheden zijn? ;-) Heb je ook een e-mailadres waar op we je direct een mailtje kunnen sturen?

[@deanfloyd8931]

I understand how you fall in, but how do you fall out of a hole?

[@aisains983]

You jump in and get velocity. You get the highhest velocity at the centre of the tunel and after this point you start to decrase your velovity but you still fly to Honolulu.

[@quentin2867]

80 seconds! And we don't care about the two connected cities

[@lecturesbywalterlewin.they9259]

11*10^3 km in 80 sec, that's an average of 137 km/s.

[@quentin2867]

+Lectures by Walter Lewin. They will make you ♥ Physics. That is the penalty for falling inside earth :)

[@guardiangamer2695]

an object will only accelerate until it reaches the mid point of the slanted tunnel since the acceleration due to gravity will be effective for the mass under the tunnel. find the new radius r' work out new gravity g' find time t using sqrt (2 ×distance /g') abd then double the time. this is messing with my mind

[@Bibibosh]

THIS IS ONE OF MY FAVORITE IDEAS.... IMAGINE IF A PLANE COULD FLY INTO THE TUNNEL! sorry for caps..

[@Evilope]

This was on an AP Physics C test one year

[@Evilope]

Something similar*

[@vgoj]

Are you retired from MIT?

[@gaetanle6200]

Eratum to my previous comment. Thank @andreathecat100 for pointing out my mistake ! Hello, here is my answer -37min 50s- 43min 14s see part of the demo here after Let's determine the (straight) distance *OH*. Let be a Oxyz Cartesian system. _find O and H coordinates then OH_ We have *OH=1.515 R* Let's work now on a (Ox) axis where x is the abscissa of a point M on [OH]. Let C the center of the earth and C' the middle of [OH] . Let a vect(u) unit vector in the direction MC. Let θ the angle (C'MC), (equal to (vect(u);vect(MC')) In absence of friction, the object is only submitted to its weight *Fg* . If the object has a masse m, (1) -vect(Fg) = G.M.m/(d²) vect(u)- vect(Fg) = G.M_e.m.d/R_e^3 vect(u) Afer the second law of Newton : (2) ∑vect(Fext) = m.vect(a) After (1) and (2) G.M_e.m.d/R_e^3 vect(u) = m.vect(a) we call ax the acceleration in the x axis, we have : *ax = G.M_e.d/R_e^3 cos(θ)* As the problem is symmetrical, we can determine the time that takes to travel the distance OC', that is the half of the distance OH and multiply the result by 2. Let determine d = f(x), x € [0; OH/2] In MC'C right triangle *d²=[(OH/2-x)²+(CC')²]* Let determine θ = f(x), x € [0; OH/2] In MC'C right triangle sin(θ) = CC'/d *θ=asin(CC'/d)* Here are the initial conditions : x0 =0 v0 = 0 a0 = 7,102 We already have a = f(d(x),θ(x)) v(t+1) = v(t) + a(t+1) x(t+1) = x(t) + v(t+1) We use the Euler method to solve it with a spreadsheet. There must be a better way but can't find it for the moment...

[@andreathecat100]

what is the weight of a person of 70 kg placed at the center of the planet Earth????

[@gaetanle6200]

It is zero... so it does not fit with my proposition (1) M is not constant either arf !

[@yash29210]

acceleration of body inside tunnel = (G*M_e)/(sqrt2*R^2).....................where M_E is the mass of earth. and a = dv/dt so we get, v = (GMt)/(sqrt2*R^2), ----------------EQUATION-1 where 't' is the time required to reach mid-point of the tunnel.......... and from the Work-Energy theorem, v of the body at the mid-point of the tunnel is- v = sqrt(G*M_e*((2-sqrt2))/R) --------------------EQUATION-2 so from equation-1 and equation-2 we get that- sqrt(G*M_e*((2-sqrt2))/R) = (GMt)/(sqrt2*R^2) . . . . . the time required to travel trough whole tunnel is 29.4 mins

[@yash29210]

Oh sorry!! I forgot to consider the changing mass of Earth as I did when I was asking you questions regarding Potential Energy Curve................... my bad!! :(

[@thawekpl]

How did you get this V? From integral, I believe, as by digging into the Earth, we change the distance between center of masses, so the gravity won't be the same at whole distance. But how :D ?

[@dudelmer4594]

Well assuming gravity as "defined" Infinity ...object (me?) won't over come Gravity to center ...need more inertia to make it through. At least give me a plasma thruster :)

[@dudelmer4594]

And if "I" have to walk the rest of the way again "Infinity" :)

[@robertoziano]

I missed a PI so: 77819 sec, that is 21.6 hours... maybe

[@andreathecat100]

I can't see my comments! :(