Problem 199 - Newtonian Mechanics 

Walter Lewin Sir even when I eat I watch your lectures because your teaching is so interesting. As your channel says you made me love physics . It's amazing that you still connected with your students by a youtube channel. your memories are so awesome.

Hello Walter Lewin. You can record your screen using software. It will improve the quality of the video and the video file will be on your laptop or computer

Professor! I am so happy to see you in a good health. 12 years ago I enjoyed your Mechanics and Electricity and Magnetism courses, despite the fact I am a pure mathematician. Thank you for inspiration! Thank you for emotions! Thank you for being an example!

I really love you lecture even though I don’t have opportunity to enroll at MIT but your online teaching so amazing . Good luck professor

I cant get tired of Professor's lectures.absolutely interesting, here in Uganda

Sir, I have a question that is pinning me If there was nothing before Big bang and the Big bang happened it means something was there that triggered Big bang?

next week i have my interviwe fot physics special degree.i want to thank you for you and ocw for providing such great lecthures.they heleped me a lot for comming this far

Good problem. Thanks 🙏

What Set of Books would you suggest for a Physics Beginner to deep dive..?

a) 9,81 m/s² b) 0 m/s² c) -6.25 * 10⁷ J d) -9.38 * 10⁷ J

very wise teacher ♥

Sir im was zero in physics but now im able to understand the beauty of physics my aim is to become like you i have to struggle a lot to understand a concept but how to improve my thinking skills because I'm an ordinary person who was who is who will work hard even if things are going wrong ❤❤❤❤❤

You are great sir

Hi professor Lewin, In physics, do we really gloss over why a magnetic field always points north to be clockwise around a current carrying wire? Do we know anything more about how or why it points clockwise as opposed to counterclockwise given the same direction of current, other than “that’s just how the universe works and here's the right hand corkscrew rule to remember it.” Why is the field around the current a clockwise helix, not a counterclockwise helix? Thank you!

You are a great physicist. Is the book available online?

You are literally the best sir! wishing you good health and happiness. Your videos make physics seem way more interesting than it already is! ❤

Great Walter ,you are the God of physics

Sir Walter I adore physics and i want to study it but many ((all)) people tells me that there is no future in it , especially Astrophysics and theoretical physics (which i love the most)... So i really want your opinion BTW the answer : 1st Q: 9.810 ms^-2 2nd Q: 0 ms^-2 3d Q: -6.253 ×10^7J Last Q : 0 The explanation of the last answer: the PE undefined as we cannot divide by zero. However, the concept of gravitational potential energy assumes that the value of ( r ) is the distance from the object to the point where the gravitational potential energy would be zero (which is at an infinite distance from the Earth). Since we are considering the center of the Earth, the mass of the Earth above the object would not exert a gravitational pull, and thus the PE would be zero. Therefore, the potential energy of a 1 kg mass at the center of the Earth would be 0 J (joules) with three-digit precision. This is because the mass is being pulled equally in all directions by Earth’s gravity, resulting in no net force and thus no work done on the mass. Hence, its potential energy is zero.

Good day to you Mr Lewin, a) At the surface, a = g = GM / R**2 = 9.81 m/s**2 b) At the center, net force is zero so a = 0 c) P.E. of 1 kg = - GM / R = - 6.253 * 10**7 J d) At r < R, Fgravity becomes due to Mr where Mr = M * r**3 / R**3 (constant density assumption) so Fgravity = G * M * r / R**3 Finally, P.E. of 1 kg at the center = P.E. at the surface + integral evaluated from R to r = 0 of (Fgravity dr) which equals - 3GM / 2R = - 9.3796 * 10**7 J

Sir, I am preparing for JEE and SAT exams. Please give me some tips. You made me love physics.

Sir I am reading in class 9 now and few days ago I told my mother to buy me a book for Quantum Mechanics....and she bought me the book yesterday...the book is very good....B.H. Bransden and C.J. Joachain...are the writers of the book..

I have a question. As Mr. Musk and Spacex are going to launch a great many rockets and that they will be going to Mars. Will this change the mass of the Earth and Mars to disrupt the orbits?

Done ✅👍🏼

(a) The gravitational acceleration at the surface of Earth: F = GMm/R² = mg => g = GM/R² ≈ 9.81 m/s², where G = the gravitational constant = 6.674⋅10ᐨ¹¹ M = the mass of Earth = 5.972⋅10²⁴ kg R = the radius of Earth = 6.374⋅10⁶ m Answer (a): 9.81 m/s² (b) The gravitational acceleration at the center of Earth: a = 0 since equal forces from all directions. (Net force = 0 => a = F_net/m = 0). Answer (b): a = 0 m/s² (c) Potential energy at the surface of Earth: PE = U(R) = ∫ (GMm/r²) dr {from ∞ to R} = -GMm/R, where m is a mass of 1 kg. => -GMm/R ≈ -6.25 ·10⁷ J We let PE at infinity, U(∞) = 0. Answer (c): PE ≈ -6.25 ·10⁷ J (d) PE at the center of Earth: U(0) = U(R) + {lim r-> 0} ∫ (GM’m/r²) dr {from R to r} where M’ = M·(r³/R³). M’ is the mass of a concentric spheric volume inside Earth with the same mass density and radius r. => U(0) = -GMm/R + ∫ (GM·(r³/R³)m/r²) dr = -GMm/R + (GMm/R³) ∫ r dr = -GMm/R + (GMm/R³)·[r²/2] {from R to r} = -GMm/R + (GMm/R³)·[r²/2 - R²/2] Let r->0 => U(0) = -GMm/R - ½ GMm/R = -(3/2) GMm/R m = 1 kg Answer (d): PE = -(3/2) GM/R ≈ -9.38 ·10⁷ J

whre is problem 198

Sir what is the basic requirements to start wit your 8.03 lectures .... From where i can learn that Thanks

a) 9.8102 SI Units b) 0 c) - 6.253* 10^7 SI Units d) 0

sir if i want to get scholarship in mit then in which month scholarship portal opens

Hi,sir walter I am a 15 years old iraqi student who loves physics,i am in 9th grade and i am wondering if i can take the lectures that you had posted,i know the 12th's grade calculus and some other maths

Wow you are still alive!

1)GM/R INWARDS 2)0 3)IF IT WAS POTENTIAL -GM/R HERE IT IS PE -GMm/R^2 4)AGAIN IF IT WAS POTENTIAL -GM/R(LIKE ELECTROSTATICS FOR THIS CASE ONLY) BUT PE IS STILL -GMm/R^2? not particularly sure of that prof please mention in the reply is it right or wrong and i'll do the calculations accordingly ps:M is the mass of object and m is the mass of Earth

Good eve professor ... Somehow i am able to get pdf of both books you used for your 8.03 mit lectures ...... So in case you want them i can send you Thanks sir

Sir i am in class 10,i have completed most of the chapter of physics of class 10 So,Should i try to study physics of class 11 as i want to apper for jee

Sir how do we send you the solution

Good Afternoon respected Lewin sir I am from India following your lectures for my Neet Journey as for me its the best way to learn physics from you ... But sir lecture and problems pdf are not downloading... Also on google these pdfs are unavailable... I request you to please do something regarding this ... I hope soon you will fix this thing At last : Best wishes from my side ❤

Prof walter lewin sir (a) gravitational acceleration at the surface of the earth =GM/R²= 6.674*10^-11* 5.972*10^24/(6374*10^3)^2 = 9.810 SI Units (b)gravitational acceleration at the centre of the earth = 0 SI units (c)PE of 1 kg mass at the surface of the earth = GM/R=(6.674*10^-11)*(5.972*10^24)/(6374*10^3) = 62530793.850 SI units (d)PE of 1 kg mass at the centre of the earth = 3/2 GM/R=(3/2)*(6.674*10^-11)*(5.972*10^24)/(6374*10^3) = 93796190.775 SI units

Too easy…..but a nice highschool problem:-)

Lewin sir I have accepted your request ... Now you can check out books And pls let me know ... Thanks a lot sir

Gravitational acceleration at the surface of Earth : 9.81 m/s^2 @ the centre g = 0 PE of 1KG at surface -6.25 * 10^7 J @ the centre 0

Sir love u 🥺🥺🥺

Answers: A: g_surface = 9.81 N/kg B: g_center = 0 N/kg C: GPE_surface = -62.5 MJ for a 1 kg object D: GPE_ center = -93.8 MJ for a 1 kg object Note: when accounting for geologists' PREM model of Earth's interior, instead of uniform density, I get -112 MJ/kg for GPE at Earth's center. Supporting Calculations: Start with Newton's law of gravity, F = G*M*m/r^2. Since F=m*g, divide by little m to find g. The r in question is capital R, the Earths' radius. All data on the right is given, so we can plug it in directly: g_surface = G*M/R^2 In Earth's interior, the shell theorem tells us for spherical symmetry, that only the mass contained within each radial shell, contributes to the gravitational field, and gravity due to outer shells adds up to zero. At the Earth's center, zero mass is in the shell, thus zero gravity at the Earth's exact center. g_center = 0 N/kg Integrate F, from r to infinity, to get potential energy at the surface GPE_surface = -G*M*m/r For potential energy in the Earth's interior, determine how g varies, integrate it, and then subtract from the GPE at its surface. GPE_center = GPE_surface - deltaGPE To find g(r), start by finding M(r), the mass contained within a sphere of radius r: M(r) = 4/3*pi*rho*r^3 g(r) = G*M(r)/r^2 To find density rho, we construct this formula at the surface, and solve for it: M = 4/3*pi*rho*R^3 rho = 3/(4*pi)*M/R^3 Combine: M(r) = M*r^3/R^3 Thus: g(r) = G*M/R^3 * r Integrate g(r) from 0 to R, to find the change in GPE per unit mass: deltaGPE = m*integral G*M/R^3 * r dr deltaGPE = G*M*m/R^3 * integral r dr = G*M*m/R^3 * [1/2*r^2] eval from r = 0 to R deltaGPE = 1/2*G*M*m/R^2 Subtract from the surface value: GPE_center = -3/2*G*M*m/R^2

🎉

Hello sir, i tried the question and im submitting the answer here also i cant write the equations in the comment box so sorry for that A) gravitational acceleration at the surface of the earth = 9.810 m/s² B) gravitational acceleration at the centre of the earth = 0 C) PE of 1kg mass at the surface of the earth = -62530793.850 joules D) potential energy of 1kg mass at the centre of the earth = ∞ I used the equations that you gave in your 8.01 lectures I may have done some mistakes.

a) g=GM/R² , R=6374㎞, M=5.972*10²⁴kg. Magnitude of g at the surface: (6.674*10⁻¹¹)(5.972*10²⁴)/(6374*10³)² → (6.674)(5.972*10¹³⁻¹²)/(6.374)² ≈ 6.7×60÷6.4² ≈ 9.81 ㎨=g (same either way). If you fell through a hole, your inertial acceleration would be 9.81 ㎨ downwards ↓, with the g-field. But, if you were supported by the ground, on average it would be zero, and you'd actually be bouncing up and down a tiny amount, roughly in SHM. Say the bounce amplitude was δh, then the acceleration “a' ” on the upwards bounce would be proportional to δh, via Hooke’s law, a’≈kδh/m, however the only acceleration in common here with gravity would be initially, before any ground penetration took place. However, if the object of mass m is just free falling or bouncing back up & inertial acceleration “a” is roughly constant, then δh≈ u(δt)±½a(δt)² u≈0=> a≈2δh/(δt)² , which should be the same as 2h/t² for m falling from a higher height h, i.e. |a|= g, as long as no other fields are nearby (eg. normal reaction force is μ*mg, not mg exactly => different acceleration on close contact). b) All the mass around the center of the spherical Earth of uniform density would contribute g-fields to an object in the center in such a way as to completely cancel out, so g=0 ㎨ here b/c of symmetry. c) GPE= U = -GMm/R = -R•mg → -(6.674)(5.972*10)(1)/(6.374) ≈ -9.81(6.374*10⁶) (1) ≈ -62.5 million Nm (or Joules). You get this from integrating g↓•dr↓from "infinity" where the potential is zero to the surface r=R, i.e. 1/r² → -1/r +0. d) You'd think it would be 0 Joules for same reasons as (b), HOWEVER... A solution of Poisson's equation, 𝝯²Φ=4πGρ where g= -𝝯Φ, gives: U=mΦ= -(GMm/2R)(3-(r/R)²) for the gpe, and at r=0 this is: U= -³/₂GMm/2R=³/₂mg → 1.5*answer (c) ≈ -93.75 million Joules This is b/c the g-field is linear inside, so you can confirm the Poisson solution by adding the potential on the surface r=R to the difference in potential between r=R & r=0. The gravitational potential Φ is the area under the g(r) vs r position curve and there are two functions for g(r), outside ~1/r² and inside ~r. To get the magnitude of change in Φ(inside), just use: ½*height*base= ½g(surface)·R At the surface, U(r=R)= -mgR and the gpe difference inside using above is ∆U=-½mgR, So the value at r=0 is sum of the areas: U(r=0)= U(r=R)+∆U = -³/₂mgR

Respected Lewin sir I request you if you can send me pdf of books you used for 8.01 ,8.02 and 8.03 It will be great for me .. Also i am thankfull to you from bottom of my heart . Also i want to tell you that i came to know you after alakh sir used your name in one of his video At last Thanks sir

a) g = 9.81 m/s^2 b) infinity c) U = - 6.27 x 10^-13 Nm d) -infinity

Hello Sir, help how to get pdf Ohanian volume 1.Thanks in advance ,much love From Tanzania EAC

A) 9.810 ms^-2 B) 0 ms^-2 obviously C) -6.253 x 10^7 J D) 0 Kindly bump up the difficulty

Respected sir I want to become a scientist.i am currently preparing for NEET. HOW can I become a scientist please tell me.

Good Morning sir from India I want to start with you lectures which are in three parts I want to know what is prerequisite to start with for 8.01,8.02,8.03 .. And also from where i can learn that .. At last pls reply to this Thanks sir

Sir i feel like i am failure neet dropper 2024 my 3rd attempt,lost every hopeee..

Sir if you're reading this please reply im in a very big problem

helllo professor ❤❤❤

a) 9.810 m/s/s b) 0 c) -6.253e7 J d) -9.380e7 J

I like sir's hello,hello,hello

Dear Sir, As we approach problem number 200 After we looked at problem 198 on removing AC noise And we look to problem 199 on Newton (simple, though one might argue even the AC noise is simple) What comes to my mind is thoughta on time and life. So I wish you good health and to pass your time with moderate pleasure. I am too unwell to solve the problem (mentally under, but Recovering slowly!) But I will say I saw some 8.01 lectures and this does not seem to be a difficult challenge, some basic algebra should be all that is needed here.

a) 9.810m/s^2 b) 0 c) -6.253*10^7 Joules d) -9.379*10^7 Joules

Sir pls start teaching again plssss

Respected sir As you told to contact publisher of Ohanian book .. I did the same twice But this is what they replied Reply no.1) Walter Lewin doesn't work for wwnorton its an april fool prank Reply no. 2) we dont send pdf of books Even after you convinced them they are doing this .. I am also sorry as i am bothering you but only coz of you i got this spark for physics .. Thats why asking you all this Is there anything you can do sir Thanks thanks a lot for your time and love

Tala gach mar kuni jayai..

W walter

Saleem sir st 50 years

Gravitational acceleration zero at the centre of earth in my NEET physics.

hello sir

Professor, do you believe in God?