Sir, I am 13 years old now, and I started watching your lectures from when I was 11. As an Indian ,I am very thankful to you and your lectures sir! Thank you!
Hii, sir I am Rishabh from India. I am in class 11 but still I watch you because the you explain is outstanding. ❤❤ YOU WILL BE ONE OF THE GREATEST TEACHER OF ALL TIME. ❤
I hope you understand that your work as a teacher has truly inspired people into studying physics, and emphasize how wonderful it truly is. I studied your lectures in depth when I first started, and every single one still amaze me. Now I'm in highschool and studying quantum mechanics and nuclear physics, without you I would have reached this point in my knowledge. I truly thank you for your impact on my life, and other aspiring scientists.
Hi Professor, I was a bit confused by the concepts of Electric fields, dipoles, and forces. However, the 8.02x lectures have strengthened my core understanding of these concepts and induced my love for physics to a greater extent. Thank you, Professor, for helping millions of students by creating such wonderful lectures that one truly longs for! Wishing you excellent health and happiness!
Good afternoon sir This side parth from india I am very lucky to be an era where such great teachers like you exist who make such tough concepts very easy I start you all your 94 lectures And now my conceptual knowledge and thinking ability towards tough concepts has increased thank you very much sir! I hope i will meet you one day because now all i can say is thank you ❤
Hi sir. I just wanted to tell you that your lectures have been extremely helpful to me. In our schools, teachers do not show interest in teaching physics, i mean by using real life examples. You really gave me the passion for physics and i thank you very much. YOU ARE TRULY THE GREATEST
(a) By total energy conservation, K_i + U_i = K_f + U_f The spacecraft could escape the planet's gravity when it is at infinite distance i.e., U_f = 0 K_i + U_i ≥ 0 1/2 *m(v_0)² - GMm/R ≥ 0 Therefore, (Vo)min = (2GM/R)^(1/2) (b) torque t = dL/dt, since there is no external force t = 0 hence L = constant L at R = m(Vo)R sin(30) L at 15R = m(V)15R We get, Vo/V = 30 (c) By conservation of total energy, 1/2 * m(Vo)² - GmM/R = 1/2 *m(V)² - GmM/15R On solving, We find that V = ((28/15)(GmM/899R))^(1/2) Therefore TE = (59/899)GmM/R or 0.065(GmM/R) (d) Since TE is the same, TE at farthest distance is also (59/899)* GmM/R
If it wasnt for your lectures and bi-weekly/monthly physics problem i wouldnt have made it into the physics program at KSU. Thank you for everything. Thank you for being honest.
Solutions: A: V0 = sqrt(2*G*M/R) B: V/v0 = sqrt(188790)/13485, approx 3.222% C: At launch, Etotal = -59/899*G*M*m/R, approx -0.0656*G*M*m/R D: At apsis, Etotal = -59/899*G*M*m/R, approx -0.0656*G*M*m/R Negative sign indicates its path is bounded, and will have an elliptical orbit if allowed to continue as a freely moving body. Supporting reasoning: In part A, the rocket drifts away indefinitely, which is the escape velocity. Energy changes are independent of launch angle, since gravity is conservative. Initial GPE = -G*M*m/R Final GPE = 0, for infinitely far away KE needed to achieve final GPE: 1/2*m*V0^2 = G*M*m/R Everybody brought rocket mass to the party: 1/2*V0^2 = G*M/R Solve for V0: V0 = sqrt(2*G*M/R) In part B, r = 15*R at its apsis point. To find its speed at apsis, use conservation of angular momentum and conservation of energy to lock-down the two unknowns (initial velocity vi and apsis velocity V). Initial & apsis angular momentum Launch: L = vi*R*sin(30 deg) Apsis: L = V*15*R Equate and solve for va: V = vi/30 Also apply conservation of energy, for an additional constraint. 1/2*m*vi^2 - G*M*m/R = 1/2*m*V^2 - G*M*m/(15*R) Everybody brought rocket mass to the party: 1/2*vi^2 - G*M/R = 1/2*V^2 - G*M/(15*R) Combine equations and solve for V V = 2*sqrt(7/13485)*sqrt(G*M/R) Compare V to V0: V/V0 = sqrt(188790)/13485 = 3.222% Parts C and D are equal, by conservation of energy. Calculate total energy at apsis as answer to both parts. Etotal = 1/2*m*V^2 - G*M*m/(15*R) Etotal = 14/13485*G*M*m/R - G*M*m/(15*R) Etotal = -59/899*G*M*m/R My initial thought to solve this problem, was to construct an ellipse with a focus at the origin, a circle centered at the origin for the planet, tangent and radial lines at launch, and the angle. Then, adjust the parameters of the ellipse, and find curvature at the apsis, such that the centripetal acceleration equals acceleration of gravity at the apsis. The results are consistent with the above result, but a lot more challenging to solve in closed form.
Context-1 a) It's basically the escape velocity required to get out of its gravitation which √2(GM/R). Any higher velocity than this would provide a non zero kinetic energy at infinity. Context-2: b) Since no other external force exists and force is along radius vector. Torque=dL/dt is conserved along the radius vector. Hence Vosin30R=Vsin15R. V is Vo/30. c)T.E is conserved assuming no energy loss outside the system KE+PE is conserved. Applying conservation of total energy and find V or Vo in standard terms. Total energy=-885/(15*899)GMm/r. d) Same as above
1) Energy is conserved, therefore by conservation of energy We get the escape velocity v = √(GM/R) 2) Taking center of planet as the axis, only force on the planet (gravitational force) passes through the axis always. So angular momentum is conserved. mVoR sin 30° = mV× 15R So Vo/V = 30 3) Total Energy = 1/2 m Vo^2 - GMm/R 4) Energy is conserved so, it is still the same as 3rd answer
(a) sqrt(2GM/R) {By conservation of energy} This velocity is the escape velocity for the planet and does not depend upon the angle of projection (b) By conservation of angular momentum, mv0Rsin30 = mv(15R) {at max distance velocity is perp to radius vector} v0/v = 30 (c) Initial energy of the spacecraft is 0.5mv0^2 - GMm/R = 450mv^2 - GMm/R {v0 = 30v} Final energy of the spacecraft (At r=15R) is 0.5mv^2 - GMm/(15R) Equating both by conservation of energy: 450mv^2 - GMm/R = 0.5mv^2 - GMm/(15R) So 449.5 mv^2 = 14GMm/(15R) So mv^2 = 14GM / (6742.5 R) Now initial energy = 450mv^2 - GMm/R = 442.5/6742.5 (-GMm/R) = 0.06563 (-GMm/R) (d) Total energy when spacecraft is at farthest point is the same, i.e., 0.06563 (-GMm/R) PS: We can see that the orbit will be bounded since total energy is negative.
Hi sir hope this text finds you fine. Sir I'm from India and recently our NEET examination was conducted, despite I was able to solve enough paper to get selection I would probably miss that. I don't know how i filled my booklet number wrong and then i messed up with the question answer sequence. And now I'm not able to overcome from this trauma. Whenever feel alone or sit in corner, some fear overwhelmes me. Need some words from your side, perhaps they might help.
one of my close Indian friends also failed the NEET examination. Yet he ended up at some of Indians best colleges (but not IT). I gave a talk at his college in 2014. After his degree there he was accepted at a Univeristy in Sweden where he got a degree in Physics and he will soon get his Dr degree at a university in Switzerland.
Sir I'm reading in class 9th and I have a question that....when I was seeing a Hypergolic fuel Nitrogen tetroxide and I came to a equation of this showing ΔH^θ what is it actually?....Enthalpy?
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 😢
Thank you so much sir , I wish if I could have a 1:1 Video Call session with you for guidance . That was also my dream since 8th Grade . If you could accept it it would be a pleasure and dream come true moment for me . If you could think of it then please 🙏 🙂
@@Physicslover0193 30% of my 1.7 million subscribers want to know my email address, my home phone and they want to make video calls with me. *All are off limit. Email + address top secret*
A) the second speed: (2GM/R)^(1/2) B)imagine there is no speed when reaching the 15R; all the way, gravity equal to GMm/r^2, where r is the distance from the center; the work done equal to integral (GMm/r^2)dr from R to 15R which is GMm(1/R-1/15R); the initial speed is {2GM/(1/R-1/15R)}^0.5 C) D) the work done by the gravity above
Hello sir ! I'm from India.... Sir I had a doubt in class 10 physics which I did asked to my teachers but wasn't satisfied with the answers... " Sir does convex mirror form a real image ? If yes , then how ?" My teachers said that yes , if the object is virtual , but how is that practically possible? Sir pls reply as I have been confused over this question since a long period of time....🙂
Concave mirrors can produce both real and virtual images depending on the distance from the mirror to the object and the curvature of the mirror, while convex mirrors produce only virtual images.
Hello sir im from india currently pursuing Bachelor's in CS i was a formal JEE aspirant at present preparing for Btech sems and other examinations i will be honoured to get a reply from you ❤😇
Sir, I read your book 'For the love of physics' finished it yesterday and I just want to say thank you for writing that wonderful book:) and I am so sorry for your loss from WW2.... PS: Can you give me some of the best physics book for Indian Institute of science I just got in 11th grade(India) thank you
Physics is difficult but i tried to understand but i also a NEET aspirants then physics seems so difficult please professor give advice to understand how physics become easy.
Hi sir I am from India I want to research in physics suggest me some topic I have not any one who help me in that work so sir please suggest me some topics in which I will research
a) G * m * M / R = 1/2 * m * v0² or v0 = √(2 * G * M / R) b) G * m * M / (15*R) = 1/2 * m * v² or v =√(2/15 * G * M / R) and v0 / v = √15 c) and d) Energy conservation: Eafter launch = Efar = - G * m * M / (15 * R) + 1/2 * m * v² = 0
@@lecturesbywalterlewin.they9259 oh my god the great professor walter lewin replied to my comment.Sir i topped the class 12 examination CBSE in physics and chemistry which subject should i major in?
(a) The minimum speed for escape velocity occurs when the spacecraft can reach infinity with zero speed. Thus both kinetic and potentional energy becomes zero. => ½mv₀² - GMm/R = 0 Answer (a): v₀ = √[2GM/R] (b) Conservation of angular momentum: mv₀R sin α = mvd sin 90° α = 30° => sin α = ½, sin 90° = 1 d = the distance from the center of Earth to apogee. d = 15R => ½ v₀R = vd => v₀/v = d/(½R) = 15R/(½R) = 2·15 Answer (b): v₀/v = 30 (c) The total energy of the spacecraft immediately after launch: Answer (c): Eᵢ = Kᵢ + Uᵢ = ½mv₀² - GMm/R (d) The total energy of the spacecraft when it is farthest away from the planet: E₀ = K₀ + U₀ = ½mv² - GMm/d Answer (d): E₀ = K₀ + U₀ = ½mv² - GMm/(15R)
Addition. Since both angular momentum and total energy are conserved we can set the equations in (c) and (d) equal and find out the actual velocities: v₀/v = 30 and ½mv² - GMm/d = ½mv² - GMm/(15R) gives v ≈ 360 m/s and v₀ ≈ 10810 m/s.
Hi Professor Lewin, I'm Saptarshi from India, currently in the 10th standard. I wrote a paper on Relativity for my school magazine which I wanna share with you. Could you pls give me your official contact details?
