Solution Problem 201 - Kepler Orbits

Conservation of Angular Momentum and Energy

[@World_Complex_The]

I love your videos sir they are actually priceless❤

[@African.Football]

Love you from Kenya professor,you made me love physics

[@turan772]

Yes me too

[@duckyoutube6318]

You are not alone in that sentiment!

[@gaming_with_bhadola]

Love you sir , love from hills of uttarakhand ❤️❤️❤️

[@garfielddexter6224]

Thank you for very interesting video🙂 very impressive 😃👍

[@turan772]

Hello Mr. Genius. Why can't you eat yogurt on Friday? He said it happens every other day. Please answer the question when you see it. Thanks in advance

[@AyanHaider_5mdcat]

Thank u sir g I watched your videos interesting

[@lecturesbywalterlewin.they9259]

Most welcome

[@studytosuccess6501]

Sir , on full moon day, we can see full moon whole night but when it's crescent moon day we don't see it for longer period why?

[@carultch]

During a full moon, the moon is on the opposite side of the sun, and is visible to every part of the Earth that has nighttime. You can see a full moon at sunrise or sunset due to atmospheric refraction, but not during the middle of the day. During a crescent moon, the moon is on the same side of the Earth as the sun, at an acute angle from the line-of-sight to the sun. The vast majority of the land area that can see the moon at any given time, is experiencing daylight. During a waxing crescent, you can see it for a few hours after sunset. During a waning crescent, it rises a few hours before the sun.

[@qinga8]

Great and beautiful, I will do it again

[@lecturesbywalterlewin.they9259]

Have fun

[@Nikhil-gh7qr]

Sir please answer my query. The question goes as follows : A block of mass m is connected to another block of mass M by a massless spring of springs constant K initially the blocks are at rest and the spring is unstreatched when a constant force F starts acting on the block of mass M to pull, Find the max extension of the spring. If I try to solve it using centre of mass frame then the elongation comes out to be 2mF/(m+M) but when I solve it using inertial frame using equations, F-Kx = Ma ma = kx where x is max elongation then it is mF/(m+M) where the hell is 2 going?

[@lecturesbywalterlewin.they9259]

I do not solve problems - I teach Physics. This is an easy high school problem

[@carultch]

The non-intuitive part about this question, is the blocks do NOT have the same acceleration when the spring is stretched at the maximum. The center of mass will accelerate at F/(M + m), but the blocks will oscillate indefinitely (for the ideal case). In the steady state solution in reality, they settle to a common acceleration. But at a point before this, block m will lag behind. For the steady state case, the spring extension settles to x_ss = F*m/(k*(M + m)). For the transient behavior, it oscillates between 0 and twice this value. Assign capital A as the acceleration of M, and little a for m's acceleration. In the lab frame: F - k*x = M*A k*x = m*a The setup in the center of mass frame, has to account for a force term of -M*a_cm and -m*a_cm in each equation on the left side, due to an accelerating system. The accelerations on the RHS would be replaced with (A - a_cm) and (a - a_cm) respectively. We need a kinematic constraint to relate A to a. To do this, relate A & a to the 2nd derivative of x. The acceleration of the spring length is the difference of the two accelerations: x" = A - a Multiply the 2nd equation by M/m, so we can generate (A - a), and replace it with x" when combining equations. This gives us the differential equation: M*x" + k*(1 + M/m)*x = F The solution with all initial conditions of zero, for x(t) is: x(t) = F*m/(k*(M + m) * [1 - cos(ω*t)] where ω = k*(M + m)/(M*m) Its maximum value occurs when cos(ω*t) = -1. Which is therefore: x_max = 2*F*m/(k*(M + m))

[@hoffiliz3762]

Hello Professor Walter Lewin, I hope you are doing well. I have a physics question that is keeping me up at night. Could you help me elucidate it? If yes, here it goes: Consider an infinite parallel plate capacitor with equal and opposite charges on each plate, distributed uniformly. Suppose there is a tiny hole in each plate, such that a line connecting these two holes is perpendicular to both plates. Now imagine a positron (or just a positively charged particle) is placed into the capacitor from the side of the positive plate, with an initial velocity of zero. While inside the capacitor, it will be accelerated by a constant electric field proportional to the surface charge density of the plates, towards the hole in the opposite (negative) plate. When the positron exits the capacitor through the opposite hole, it will have kinetic energy which, when added to the energy lost to radiation due to acceleration, equals the total work done by the parallel plate capacitor. But energy cannot be created. So, the capacitor must lose something. To me, it seems that the thing it must lose is the charge stored on each plate. Otherwise, we could generate infinite energy through the same process I just described: placing charges through one hole infinitely and gaining kinetic (and radiation) energy. (Of course, this wouldn't be possible since infinite charge doesn't exist, but as the capacitors are infinite and the surface charge density is constant, we are already assuming this absurdity is possible in the first place). However, the capacitor is not "losing" the excess charge on each plate because there is no conductor connecting the plates to allow the charges to neutralize, and charge cannot be destroyed. So where does this energy needed to accelerate the positron come from? How does the capacitor lose energy when accelerating charges that pass through it? I am aware that the field inside the capacitor isn't exactly constant since the positron itself can alter the charge distribution of the plates. However, it would still accelerate and gain energy from the plates, right? Another possibility is that the distance between the plates must be smaller after the positron has exited the capacitor, so there is less electric field over the space at the end of the process. However, I am assuming that the plates are fixed in place. Thank you very much for your time and assistance. Greetings from Brazil, and thank you very much for your videos!

[@carultch]

This is just like the energy considerations of a diver jumping from a diving board into a pool. Consider a 5 meter dive, and an 80 kg diver. The diver starts with 4 kJ of GPE, by virtue of his initial position. On his way down, this becomes 4kJ of KE, and later 4kJ of thermal energy upon slowing to rest in the water. Other than the diver's position, the configuration of the masses that set up the GPE didn't change. The diver lost GPE by virtue of moving with positive work done by gravity. Likewise, the positron loses energy when it departs the positive plate and moves to the negative plate. It took work to bring the positron to its position on the positive plate, and that work becomes its kinetic energy as it moves toward the negative plate, and then away from it. The configuration of charges that set up the capacitor do not change, only the location of the positron. On net, the amount of EPE that the positron loses, is equal to the amount of work that had to be done, bringing it to the positive plate from infinitely far away.

[@mrxzod20]

Walter sir can u suggest the best book for physics fir jee examination India

[@lecturesbywalterlewin.they9259]

ask your teachers - they know you

[@pranav.mishra]

Sir please make a video on electric dipole As its not covered in your playlist It will be very helpful

[@lecturesbywalterlewin.they9259]

it is covered in my 8.02 lectures

[@pranav.mishra]

@@lecturesbywalterlewin.they9259 oh thank you sir I think I missed that part

[@DharaniDharan-rj6ps]

Sir, i love your teaching and approach to physics❤❤... Currently I am preparing for NEET medical entrance exam 2025 so I have a complete year in my hand. but still, here in my language there is no proper physics lectures in KZbin and also I can't afford paid courses😢......can I watch your lectures for physics and solve your example problems and test for my preparation??.... I have been in this dilemma for watching your lectures for my physics preparation.... 😕😕

[@lecturesbywalterlewin.they9259]

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.

[@DharaniDharan-rj6ps]

@@lecturesbywalterlewin.they9259 Thank you very much sir, for responding to my request💖💖..i will be grateful for your help✨😊

[@saaaalimmm]

walter sir like my comment