I wish I can become a professor almost like you becausw no one can teach like you. Teachers like you and Richard Feynman are responsible for inspiring so many young students to pursue physics. Thank you for your brilliant lectures. I see you as the feynman of our age.
Sir here is your proud student . Sir I'm selected for International Scientific Physics Olympiad from Team India which is gonna held in MIPT Russia. Thank you so much sir for making me capable for this huge glory 😢
I would like to try: (a) Given that, The mass, M = 7*10^22 kg Radius, R = 1500 km = 1.5*10^6 m Radius of the hollow portion, r = (R - thickness of the shell) = (1.5*10^6 - 30000) m = 1.47*10^6 m The net volume, V = (4π/3)R^3 - (4π/3)r^3 = (4π/3)*(R^3 - r^3) which gives 8.3138*10^17 m^3. Thus the density, ρ = M/V = (7*10^22 kg)/(8.3138*10^17 m^3) =84197 kg/m^3. (Answer) (b) The density of water is 1000 kg/m^3. Thus the density of the hollow sphere is "more than 84 times" the density of water. (c) As stated in Newton's shell theorem, a spherically symmetric shell of an uniform distribution of mass exerts no net force on an object located inside it, regardless of its position. Thus the net gravitational force on the object is 0 N. (Answer) Have a nice day, Professor.❤ - Symoum Syfullah Priyo
Glad to see you're still alive, 😊 Dear sir I would like to confess I had learnt a lot from your lecture during the preparation of IIT JEE exams, unfortunately I was not in the top 2 lac students, in despite of that I had started from 2 tier College and today I am placed through campus with enough package. Thank you because there is your contribution in it too to teach.
Did you work out this problem though? (a) & (b) are easy but many make simple errors... (c) he said was easy, which it is if you have the knowledge, but that's only if you don't care about the "proof".
Hey professor! Although I don’t have a teacher who makes me love physics, I absolutely love the subject and I hope to become a physicist in the future :) thank you for your inspiring videos!
a) ρ=M/V = 7x10^22 / V eq(1). where V is the volumen of the spherical shell given by V= (4/3)π[Routside^3- Rinside^3]= (4/3)π[(1.5 x10^6)^3-(1.47 x10^6)^3] (where Routside and Rinside are the outside and inside radii of the shell) Plugging this into eq(1) and simplifying we get ρ= (21 x10^4 )/[4 π(1.5^3-1.47^3)]=84197.509 Kg/m3 (which is SI units). b) The density of water is 1000 Kg/m3. Therefore our shell is 84 times denser that water. c) The gravitational force on any object inside the shell is ZERO. There is more than one way to prove this but the easiest is the gravitational analog of Gauss’ Law (see Lecture 2 of 8.02 where Walter Lewin derives that the Electric Field anywhere inside a uniform-charged shell is zero). That lecture deals with electrostatics (static electric fields). In our case we have static gravitational field. We can apply the same mathematics: pick a close surface inside the hollow shell; and for convenience we chose a sphere of radius r=250km. Using symmetry arguments we would find that, for any point that lies on that sphere, the gravitational field is zero and thus the gravitational force too.
Namsta sir, I am from India.. studying in class 12..sir your experiment give me a engry or interest studying physics..i am enrolled in science to saw your video... thank you sir
a) the density of the sprere is 4.95 gr/cm^3. (The density of the shell is 84.24gr/cm^3) b) The density of the spere is 5 times the density of the water. c) 0. There is no gravitational force in the hollow part of the sphere.
my bet: Vsph=3/4*pi*r^3; Vext = (3/4)*pi*(1500E3)^3 = 7.952E18 cbm; Vint = (3/4)*pi*(1500E3-30E3)^3 = 7.4845E18 cbm; Vshell = Vext - Vint = 7.952E18 - 7.4845E18 = 4.6765E17 cbm; Density of the shell = Mshell/Vshell = 7.0E22/4.6765E17 = 149684.461 kg/cbm a) Density of the shell = Mshell/Vshell = 7.0E22/4.6765E17 = 149684.461 kg/cbm b) assuming density of water = 1000 kg/cbm we can conclude that density of the core is 149684.461/1000 = 149.684 times greater than density of water. c) According to Newton's Shell Theorem gravitational F=0.00 N (Newtons).
1) wrong Vsph = 3/4pir3. it is 4/3pir3 2) your rounding in V calculations introduces a few percent error into Vshel value - yo need more precision there. My result is 84,2 times more dense than water.
a) Mass(m)= 7.0 × 10²² kg Outside radius(R)= 1500km = 15 × 10^5 meter Thickness= 30.0km = 3 × 10^4 meter Therefore: inside radius(r)= outside radius minus thickness = 12 × 10^5 meter Volume=4/3 × π × (R³ - r³) Putting values Volume= 4π × 10^5 (meter)³ Density=mass/volume => 7.0 × 10²² kg ÷ 4π × 10^5 => 5.568 × 10^16 kg/m³ b) Density calculated= 5.568 × 10^6 kg/m³ Density of water = 1000kg/m³ So, 5.568 × 10^6 - 1000 = 5.567 × 10^6 So. Density of shell is 5.567 times 10^6 greater than earth...... I have doubt on option b because i still confused how to compare things in physics..
Sir, pls upload a video directing instructions for downloading PDFs of the those 3 books that you use for your lectures❤❤❤.. Me and my friends find difficult to find them on internet❤❤
Solutions: A: 84,200 kg/m^3 B: 84.2 times the density of water C: 0.00 Newtons Supporting calcs & reasoning: The volume of a spherical shell is outer sphere volume, minus inner sphere volume. V_shell = Vo - Vi V_shell = 4*pi/3 * [Ro^3 - Ri^3] Inner radius: Ri = Ro - t Combine equations & expand: V_shell = 4*pi/3 * [3*Ro^2*t - 3*Ro*t^2 + t^3] Plug in data, and translate to m^3: 4*pi/3 * [3*1500^2*30 - 3*1500*30^2 + 30^3] * (1000 m/km)^3 V_shell = 8.31*10^17 m^3 Compute density: rho = M/V_shell rho = 84,200 kg/m^3, which is 84.2 times water's density For part C, Newton's shell theorem states that a spherically symmetric body, produces zero net gravitational field at every point inside it. This is what we have in this situation, and thus zero gravitational force on this mass.
The Maxwell-Boltzmann distribution describes the distribution of speeds among the particles in a sample of gas at a given temperature. The distribution is often represented graphically, with particle speed on the x-axis and relative number of particles on the y-axis.
@@shwetasharma154 that is from a research and facilities point of view, but if you wanna go for a job later or lose interest in research it gets a little tough compared to iits.
In kinematics motion under gravity,if we take upward direction as positive and downward as negative..... A ball is falling downward....so g(acceleration due to gravity) will be positive or negative?( for downward direction)
Dear Walter Lewin Sir, I am Utkarsh from India an aspirant of JEE and i am in 9th standard and i want to study as much as science as possible for these competitive entrance exam. What do you recommend me to learn first or complete basics before I am able to solve tough questions such as these and for your knowledge we in India nowadays are starting to prepare from class 7 in which we have NSEJS and NTSE type Olympiad Examinations which have questions of Kinematics at the level of JEE Advanced and much difficult students are waking up at 3 in morning to study and they study 10+ hours even tho they arent in 11 or 12th Grade. So please help me out as im not late but also not to early to start. Thanks, 🙏🏼
Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, SolutionsY & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
a) What is the density of the shell (SI units 3 digit precision)? M = mass of the shell = 7.0×10²² kg R = outside radius = 1500 km d = thickness of the shell = 30 km V = (4/3)π[R³ - (R-d)³] => V ≈ 8.31379×10¹⁷ m³ ρ = M/V ≈ 7.0×10²²/ 8.31379×10¹⁷ Answer a): ρ ≈ 8.42×10⁴ kg/m³ b) Compare this with the density of water. Density of water is about 1000 kg /m³
Answer b): Relative density is about 84
Inside the shell at a distance of 250 km from its center is a mass of 3500 kg. c) What is the gravitational force on that mass (SI units 3 digit precision)? Answer c): Gravitational force is zero due to symmetry.
Sir I have a question but not related to this scenario...That is can the graph which shows maxwell boltsnman's distribution intersect at zero. Kinetic energy 0 molecules are 0, but kinetic energy is zero when in 0Kelvin. So There can be particles with 0 kinetic energy in 0Kelvin. So some people draw graph intersect at zero some shouldn't...What's the correct one sir ? Can you please explain this sir by a reply...😭
@@GayathriLokuge The Maxwell-Boltzmann distribution describes the distribution of speeds among the particles in a sample of gas at a given temperature. The distribution is often represented graphically, with particle speed on the x-axis and relative number of particles on the y-axis. google *MB-distribution - shows nice plots*
many aricles have been publised about bursting pulsars. I suggest you read them first. If you then think you can still make a major contributing - submit your article to a refereed journal.
Hello sir, I am 'Akash' I am preparing for the Jee exam as I am watching your lecture 8.02 and Solving the Question that you provided But it is difficult to solve the Question. So, I need your help and guidance please sir help me out ...
Hello Sir: Could you pls tell how would the total time of flight change if we consider air resistance in projectiles? If wind blows horizontally, there would be some air resistance which would be tangential to the projectile and its vertical component would point downwards and add to gravity due to which time of ascent would decrease, and likewise in the other half the vertical component would be opposite to gravity so time of descent would increase, but what about the total time of flight?
You have to work out the forces on the aircraft first. You need the characteristic lengths, drag coefficient, the mass of the plane, the velocity v, the altitudes throughout the flight, weather, pressure, wind ...etc. If you have these relevant details, you can work out the drag force (usually proportional to velocity squared), then put it all together in a differential equation and solve for v. If you have v, you can easily find the flight time t. 🌬️ ✈️
@@The_Green_Man_OAP thanks, so to sum up - the total time of flight will depend upon a lot of factors and will keep changing as these factors change, right?
"A visual guide to Bayesian thinking" - Julia Galef : "I use pictures to illustrate the mechanics of "Bayes' rule," a mathematical theorem about how to update your beliefs as you encounter new evidence. Then I tell three stories from my life that show how I use Bayes' rule to improve my thinking."
Read books, watch videos like Prof. Lewin's lectures, interact with others who are interested in the sciences, use AI, use search engines, use libraries and most importantly pay attention to your teachers or tutors if you are lucky enough to have any.
@@lecturesbywalterlewin.they9259 First of all Thank you for your response sir.. I will practice it until my neet exam sir ,after my result I tell you how it had worked for me sir😃...
yes there are 2 Lewin channels none of them are by me as I already died 4 years ago. The other 2 are run by my twin brothers; 1 was 3 yr younger than I was and one 2 yr older.
@@lecturesbywalterlewin.they9259 Well, I know it's zero but I want to prove it my way. It'll take me a while to do that. I need to find the center of mass for a hollow spherical cap which I call ”object A" and then for the rest, "object B". I already have the masses for these hypothetical objects. Then I will just work out the difference between the fields. Hopefully it'll end up being zero. 😊 - It's tricky figuring out the cog for B. I need a way of combining/averaging the cog positions of the ring and the hollow hemisphere in B.
@@carultch Feel free to carry on with what I started for part (c), if you think it's going somewhere. Maybe we'll come up with something interesting...❔I want more than just "zero" as an answer.
@@lecturesbywalterlewin.they9259 Icymi, I did write something else but I decided to change my strategy. It was to do with a comparing a contiguous model to a discrete model of the mass distribution in the shell. But a discrete model would help explain why g is roughly proportional to π² at the surface... I believe it comes from π²≈6(1+1/2²+1/3²+..+1/N²), for large N and the way the meter was defined by the French. If this is true, then GM is something to do with the internal geometry of large masses. -I worked it out in detail a while back, so maybe I'd have to dig that up. I think that different results would be obtained for each model, but of course the discrete model would limit to contiguous model as the number particles increases and their size relative to the shell decreases.
[INCORRECT❌] Density = 6.391 * 10^12 kg/m³; Relative density = 6.391*10^9.... I thought the last one was going to be just as easy but no 🤣 . A 30-min video later , turns out it is 0 . Which only somewhat makes sense , I need time to take that in , I did hear this in one of those what if you dig a hole through earth videos . I really need to learn more math. ALR I'll do that .
@@synchronium24All the perpendicular forces cancel out via symmetry wrt the axis formed by connecting the 250km point to the centre of the sphere. The parallel components add up as if there were 2 opposing masses either side of the 3500kg mass, call them A and B. Say A was the closer mass. It's closer but consists of less mass than B which is a further distance out. The masses of A & B and their distances from the 3500kg mass conspire so that the net force of A & B on the test mass is ZERO.
a) 8,42 10^4 kg/m3 b) a bit more than 84 times more dense than water - there is no such material. c) zero according to en.wikipedia.org/wiki/Shell_theorem
I wish I can become a professor almost like you becausw no one can teach like you. Teachers like you and Richard Feynman are responsible for inspiring so many young students to pursue physics. Thank you for your brilliant lectures. I see you as the feynman of our age.
Wow, thank you!
Sir here is your proud student . Sir I'm selected for International Scientific Physics Olympiad from Team India which is gonna held in MIPT Russia. Thank you so much sir for making me capable for this huge glory 😢
Wow ! Really
No you are my inspiration...
Can u tell me how should I train myself for it...
I am in 11th grade and had qualified nsejs in 9th grade
@@akshitshivhare9107 Just 2 sentences
Belive yourself and keep hardworking
I would like to try:
(a) Given that,
The mass, M = 7*10^22 kg
Radius, R = 1500 km = 1.5*10^6 m
Radius of the hollow portion, r = (R - thickness of the shell)
= (1.5*10^6 - 30000) m = 1.47*10^6 m
The net volume,
V = (4π/3)R^3 - (4π/3)r^3
= (4π/3)*(R^3 - r^3)
which gives 8.3138*10^17 m^3.
Thus the density,
ρ = M/V
= (7*10^22 kg)/(8.3138*10^17 m^3)
=84197 kg/m^3. (Answer)
(b) The density of water is 1000 kg/m^3.
Thus the density of the hollow sphere is "more than 84 times" the density of water.
(c) As stated in Newton's shell theorem, a spherically symmetric shell of an uniform distribution of mass exerts no net force on an object located inside it, regardless of its position.
Thus the net gravitational force on the object is 0 N. (Answer)
Have a nice day, Professor.❤
- Symoum Syfullah Priyo
Glad to see you're still alive, 😊
Dear sir I would like to confess I had learnt a lot from your lecture during the preparation of IIT JEE exams, unfortunately I was not in the top 2 lac students, in despite of that I had started from 2 tier College and today I am placed through campus with enough package.
Thank you because there is your contribution in it too to teach.
Did you work out this problem though? (a) & (b) are easy but many make simple errors... (c) he said was easy, which it is if you have the knowledge, but that's only if you don't care about the "proof".
Btw, he died 4 years ago.
You are seeing his twin brother.
Great professor Lewin 👏👏👏. I am a former m.d. but i love physics and your lessons and yours problems are awesome 👏👏👏
thanks ive got my alevel physics tomorrow and this problem was useful
Hey professor! Although I don’t have a teacher who makes me love physics, I absolutely love the subject and I hope to become a physicist in the future :) thank you for your inspiring videos!
I'm from India
I too
🇮🇳
a) ρ=M/V = 7x10^22 / V eq(1).
where V is the volumen of the spherical shell given by
V= (4/3)π[Routside^3- Rinside^3]= (4/3)π[(1.5 x10^6)^3-(1.47 x10^6)^3]
(where Routside and Rinside are the outside and inside radii of the shell) Plugging this into eq(1) and simplifying we get
ρ= (21 x10^4 )/[4 π(1.5^3-1.47^3)]=84197.509 Kg/m3 (which is SI units).
b) The density of water is 1000 Kg/m3. Therefore our shell is 84 times denser that water.
c) The gravitational force on any object inside the shell is ZERO. There is more than one way to prove this but the easiest is the gravitational analog of Gauss’ Law (see Lecture 2 of 8.02 where Walter Lewin derives that the Electric Field anywhere inside a uniform-charged shell is zero).
That lecture deals with electrostatics (static electric fields). In our case we have static gravitational field. We can apply the same mathematics: pick a close surface inside the hollow shell; and for convenience we chose a sphere of radius r=250km. Using symmetry arguments we would find that, for any point that lies on that sphere, the gravitational field is zero and thus the gravitational force too.
Namsta sir, I am from India.. studying in class 12..sir your experiment give me a engry or interest studying physics..i am enrolled in science to saw your video... thank you sir
b) The relative density 4.95
First coment mine❤️❤️
a) the density of the sprere is 4.95 gr/cm^3. (The density of the shell is 84.24gr/cm^3)
b) The density of the spere is 5 times the density of the water.
c) 0. There is no gravitational force in the hollow part of the sphere.
Thanks and Regards
Sir, I'm in 12th i have done some calculations in quantum mechanics. How can I share with you. I want to know what corrections are required.
I do not want to get involved as that is too time-consuming
my bet:
Vsph=3/4*pi*r^3;
Vext = (3/4)*pi*(1500E3)^3 = 7.952E18 cbm;
Vint = (3/4)*pi*(1500E3-30E3)^3 = 7.4845E18 cbm;
Vshell = Vext - Vint = 7.952E18 - 7.4845E18 = 4.6765E17 cbm;
Density of the shell = Mshell/Vshell = 7.0E22/4.6765E17 = 149684.461 kg/cbm
a) Density of the shell = Mshell/Vshell = 7.0E22/4.6765E17 = 149684.461 kg/cbm
b) assuming density of water = 1000 kg/cbm we can conclude that density of the core is
149684.461/1000 = 149.684 times greater than density of water.
c) According to Newton's Shell Theorem gravitational F=0.00 N (Newtons).
It's 4/3 not 3/4
1) wrong Vsph = 3/4pir3. it is 4/3pir3
2) your rounding in V calculations introduces a few percent error into Vshel value - yo need more precision there.
My result is 84,2 times more dense than water.
Shame on me: Vsph=(4/3) * pi * r^3 , indeed!
Sir pls make video on correct way of studying physics.....ure an renowed astrophysicist so u may know better
a)
Mass(m)= 7.0 × 10²² kg
Outside radius(R)= 1500km = 15 × 10^5 meter
Thickness= 30.0km = 3 × 10^4 meter
Therefore: inside radius(r)= outside radius minus thickness = 12 × 10^5 meter
Volume=4/3 × π × (R³ - r³)
Putting values
Volume= 4π × 10^5 (meter)³
Density=mass/volume
=> 7.0 × 10²² kg ÷ 4π × 10^5
=> 5.568 × 10^16 kg/m³
b)
Density calculated= 5.568 × 10^6 kg/m³
Density of water = 1000kg/m³
So, 5.568 × 10^6 - 1000 = 5.567 × 10^6
So. Density of shell is 5.567 times 10^6 greater than earth......
I have doubt on option b because i still confused how to compare things in physics..
Sir,have you ever presented from when and with what reason you are sending these problems and more informations about these till 202 questions
Sir, pls upload a video directing instructions for downloading PDFs of the those 3 books that you use for your lectures❤❤❤.. Me and my friends find difficult to find them on internet❤❤
a. 84.2e3 kg/m3
b. 84.2 times the density of water.
c. No force by Newton's shell theorem.
a) 84200 kg/m³
b) 84 times more dense than water
c) 0 N
Solutions:
A: 84,200 kg/m^3
B: 84.2 times the density of water
C: 0.00 Newtons
Supporting calcs & reasoning:
The volume of a spherical shell is outer sphere volume, minus inner sphere volume.
V_shell = Vo - Vi
V_shell = 4*pi/3 * [Ro^3 - Ri^3]
Inner radius:
Ri = Ro - t
Combine equations & expand:
V_shell = 4*pi/3 * [3*Ro^2*t - 3*Ro*t^2 + t^3]
Plug in data, and translate to m^3:
4*pi/3 * [3*1500^2*30 - 3*1500*30^2 + 30^3] * (1000 m/km)^3
V_shell = 8.31*10^17 m^3
Compute density:
rho = M/V_shell
rho = 84,200 kg/m^3, which is 84.2 times water's density
For part C, Newton's shell theorem states that a spherically symmetric body, produces zero net gravitational field at every point inside it. This is what we have in this situation, and thus zero gravitational force on this mass.
If both the bike and the fourwheel are accelerated at the same time and with the same speed, will they arrive at the same time?
Sir..Plz give solution
Provide answers also
I want to be like all of IITians
I looked it up and found the answer, but their explanation for how it is 0 is too brief.
The Maxwell-Boltzmann distribution describes the distribution of speeds among the particles in a sample of gas at a given temperature. The distribution is often represented graphically, with particle speed on the x-axis and relative number of particles on the y-axis.
google Maxw-B distribution
I love you Sir❤❤❤❤❤
I have my JEE Advanced exam tomorrow sir 😄 , been watching you since the last 4 years
Hope u go to IIsc indian institute of science or IIST(under ISRO) rather than iit......my advice....iisc ranks better than iit
Even walter lewin knows.....iisc better
@@shwetasharma154 that is from a research and facilities point of view, but if you wanna go for a job later or lose interest in research it gets a little tough compared to iits.
I wanna learn physics to
you
In kinematics motion under gravity,if we take upward direction as positive and downward as negative.....
A ball is falling downward....so g(acceleration due to gravity) will be positive or negative?( for downward direction)
if upwards is positive and if upwards speed is voy at t=0 then y=yo + voy*t - 0.5gt^2 here g = 10 m/s^2
Dear Walter Lewin Sir,
I am Utkarsh from India an aspirant of JEE and i am in 9th standard and i want to study as much as science as possible for these competitive entrance exam. What do you recommend me to learn first or complete basics before I am able to solve tough questions such as these and for your knowledge we in India nowadays are starting to prepare from class 7 in which we have NSEJS and NTSE type Olympiad Examinations which have questions of Kinematics at the level of JEE Advanced and much difficult students are waking up at 3 in morning to study and they study 10+ hours even tho they arent in 11 or 12th Grade. So please help me out as im not late but also not to early to start.
Thanks, 🙏🏼
Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, SolutionsY & Lecture Notes".
8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
Love you@@lecturesbywalterlewin.they9259
Ok sir I'll start watching ❤️
a) What is the density of the shell (SI units 3 digit
precision)?
M = mass of the shell = 7.0×10²² kg
R = outside radius = 1500 km
d = thickness of the shell = 30 km
V = (4/3)π[R³ - (R-d)³]
=> V ≈ 8.31379×10¹⁷ m³
ρ = M/V ≈ 7.0×10²²/ 8.31379×10¹⁷
Answer a): ρ ≈ 8.42×10⁴ kg/m³
b) Compare this with the density of water.
Density of water is about 1000 kg /m³
Answer b): Relative density is about 84
Inside the shell at a distance of 250 km from its
center is a mass of 3500 kg.
c) What is the gravitational force on that mass
(SI units 3 digit precision)?
Answer c): Gravitational force is zero due to symmetry.
a) 4953.99 kg/m³
Sir you looked handsome in your old photographs.
Sir I have a question but not related to this scenario...That is can the graph which shows maxwell boltsnman's distribution intersect at zero. Kinetic energy 0 molecules are 0, but kinetic energy is zero when in 0Kelvin. So There can be particles with 0 kinetic energy in 0Kelvin. So some people draw graph intersect at zero some shouldn't...What's the correct one sir ? Can you please explain this sir by a reply...😭
use google
@@lecturesbywalterlewin.they9259 Ok.Thank you sir...
@@GayathriLokuge
The Maxwell-Boltzmann distribution describes the distribution of speeds among the particles in a sample of gas at a given temperature. The distribution is often represented graphically, with particle speed on the x-axis and relative number of particles on the y-axis. google *MB-distribution - shows nice plots*
@@lecturesbywalterlewin.they9259 Thank You very much sir... I understood the concept... I am very thankful to your support. 🙏😇
Sir i have done some research on bursting pulsars furthur from your studies can you please give a reply
many aricles have been publised about bursting pulsars. I suggest you read them first. If you then think you can still make a major contributing - submit your article to a refereed journal.
As you are apparently an expert, can you not only find the value but actually prove part (c) without just referring to Newton's proof?
Hello sir, I am 'Akash' I am preparing for the Jee exam as I am watching your lecture 8.02 and Solving the Question that you provided But it is difficult to solve the Question.
So, I need your help and guidance please sir help me out ...
Hello Sir: Could you pls tell how would the total time of flight change if we consider air resistance in projectiles? If wind blows horizontally, there would be some air resistance which would be tangential to the projectile and its vertical component would point downwards and add to gravity due to which time of ascent would decrease, and likewise in the other half the vertical component would be opposite to gravity so time of descent would increase, but what about the total time of flight?
You have to work out the forces on the aircraft first. You need the characteristic lengths, drag coefficient, the mass of the plane, the velocity v, the altitudes throughout the flight, weather, pressure, wind ...etc.
If you have these relevant details, you can work out the drag force (usually proportional to velocity squared), then put it all together in a differential equation and solve for v.
If you have v, you can easily
find the flight time t.
🌬️ ✈️
@@The_Green_Man_OAP thanks, so to sum up - the total time of flight will depend upon a lot of factors and will keep changing as these factors change, right?
Anyone from india ❤
Sir I'm reading in class 9th now and I want to know about Bayesian Probability.....please tell me....can't understand.....Google...
en.wikipedia.org/wiki/Bayesian_probability#:~:text=Bayesian%20probability%20(%2F%CB%88be%C9%AA,quantification%20of%20a%20personal%20belief.
"A visual guide to Bayesian thinking"
- Julia Galef : "I use pictures to illustrate the mechanics of "Bayes' rule," a mathematical theorem about how to update your beliefs as you encounter new evidence. Then I tell three stories from my life that show how I use Bayes' rule to improve my thinking."
"Bayes theorem, the geometry of changing beliefs" -3Blue1Brown
Sir pls tell me how to learn physics concepts easily for inter
eat yogurt every day but *never on Fridays.* That also worked well for Einstein and for me.
Read books, watch videos like Prof. Lewin's lectures, interact with others who are interested in the sciences, use AI, use search engines, use libraries and most importantly pay attention to your teachers or tutors if you are lucky enough to have any.
@@lecturesbywalterlewin.they9259
First of all Thank you for your response sir..
I will practice it until my neet exam sir ,after my result I tell you how it had worked for me sir😃...
Just wondering if you have a second channel if not there is another fake channel with same name picture and two videos on as well
yes there are 2 Lewin channels none of them are by me as I already died 4 years ago. The other 2 are run by my twin brothers; 1 was 3 yr younger than I was and one 2 yr older.
@@lecturesbywalterlewin.they9259 sorry I was a bit confused I didn't know you had any brothers have a great day or night as it is in the uk
@@lecturesbywalterlewin.they9259 wait what??? You died 4 years ago
Somebody has been watching "Moonfall" again...
😂 🌝⤵️
I miss your solution of part C) of 202
@@lecturesbywalterlewin.they9259
Well, I know it's zero but I want to prove it my way. It'll take me a while to do that.
I need to find the center of mass for a hollow spherical cap which I call ”object A" and then for the rest, "object B". I already have the masses for these hypothetical objects.
Then I will just work out the difference between the fields. Hopefully it'll end up being zero. 😊
- It's tricky figuring out the cog for B.
I need a way of combining/averaging the cog positions of the ring and the hollow hemisphere in B.
It does sound like the specs of a hypothetical hollow moon.
@@carultch Feel free to carry on with what I started for part (c), if you think it's going somewhere. Maybe we'll come up with something interesting...❔I want more than just "zero" as an answer.
@@lecturesbywalterlewin.they9259
Icymi, I did write something else but I decided to change my strategy.
It was to do with a comparing a contiguous model to a discrete model of the mass distribution in the shell.
But a discrete model would help explain why g is roughly proportional to π² at the surface...
I believe it comes from π²≈6(1+1/2²+1/3²+..+1/N²), for large N and the way the meter was defined by the French. If this is true, then GM is something to do with the internal geometry of large masses.
-I worked it out in detail a while back, so maybe I'd have to dig that up.
I think that different results would be obtained for each model, but of course the discrete model would limit to contiguous model as the number particles increases and their size relative to the shell decreases.
Hii sir
[INCORRECT❌] Density = 6.391 * 10^12 kg/m³; Relative density = 6.391*10^9.... I thought the last one was going to be just as easy but no 🤣 . A 30-min video later , turns out it is 0 . Which only somewhat makes sense , I need time to take that in , I did hear this in one of those what if you dig a hole through earth videos . I really need to learn more math. ALR I'll do that .
Why is the gravitational force 0? Is it because there is no mass located at that point?
@@synchronium24All the perpendicular forces cancel out via symmetry wrt the axis formed by connecting the 250km point to the centre of the sphere. The parallel components add up as if there were 2 opposing masses either side of the 3500kg mass, call them A and B. Say A was the closer mass. It's closer but consists of less mass than B which is a further distance out. The masses of A & B and their distances from the 3500kg mass conspire so that the net force of A & B on the test mass is ZERO.
@@The_Green_Man_OAP thank you so much for this explanation 🙏
how old are u sir ?
stone age
First comment😂
a) 8,42 10^4 kg/m3
b) a bit more than 84 times more dense than water - there is no such material.
c) zero according to en.wikipedia.org/wiki/Shell_theorem