a) Period (T) and orbital speed (v) Kepler's 3rd law tells us that: T^2/r^3=const and we can calculate that constant with info of Thethys, I get const=0.1391. So to obtain the periods (2nd column of the following table) we just do (for Dione for example): T^2/3.77^3= 0.1391 and solve for T T=sqrt(0.1391*3.77^3)=2.73 days. Idem for the rest of the Moons: r(10^5 km) Period (days) v (km/s) 2,95 1,89 11,35 3,77 2,73 10,04 5,27 4,51 8,49 12,22 15,93 5,58 35,6 79,23 3,27 and the orbital speed v (last column) is calculated in the following manner: v=(2πr)/T b) The constant in Kepler's law is 4π^2/(GM) so we can solve for M M= 4π^2/(G*const)= 5.69x10^26 kg (making use of const in S.I. units of course)
a) R(*10^5 km) T (days) V (km/s) Tethys 2.95 1.89 11.35 Dione 3.77 2.73 10.04 Rhea 5.27 4.51 8.49 Titan 12.22 15.93 5.58 Iapetus 35.6 79.23 3.27 b) the mass of Saturn is 5.66*10^26 kg.
(a) Periods and orbital speeds. Keplers 3:rd Law says: T² ∝ R³ where T is the orbital period and R is the length of the semi-major axis of its orbit. (Here we assume circular orbits). The distances for all moons from Saturn is given, but the period is only given for the nearest moon. The orbital speed can be found by: v = 2πR/T For the first moon: v₁ = 2πR₁/T₁ We can write Keplers Law as: T = k·R³ᐟ² where k is a constant. For the first moon: k = T₁/ R₁³ᐟ² Next, for the other moons: vᵢ = 2πRᵢ/Tᵢ and Tᵢ = k·Rᵢ³ᐟ² or Tᵢ = (T₁/ R₁³ᐟ²)·Rᵢ³ᐟ² for i = 2,3,4 and 5 (b) The mass of Saturn. The gravitational acceleration: a = GM/r² and the centripetal acceleration: a = v²/r Setting them equal gives: M = v²r/G G = 6.674·10ᐨ¹¹ Mass of Saturn can be calculated as: M = v₁²R₁/G Answer (a): I put the formulas in an Excel sheet and got the following numbers: Moon Distance (km) Period (days) Orbital speed (km/s) --------------------------------------------------------------------------------------------------- Tethys 2.950E+05 1.89 11.35 Dione 3.770E+05 2.73 10.04 Rhea 5.270E+05 4.51 8.49 Titan 1.222E+06 15.93 5.58 Japetus 3.560E+06 79.23 3.27 Answer (b): M = v₁²R₁/G => mass of Saturn was calculated to about 5.7·10²⁶ kg. (5.695·10²⁶)
Remember guys... Change is defined as our differences between our final and initial positions. Avg is your total change in x divided by your total change in time. And if you are moving circular then just think of your change in your angle and how that relates to your change on the movement around the circumference of the circle. You should know how to convert radian to degrees. Radians to whatever unit of measure you require. Just take things slow. Use Feynman's problem solving methods, make tables, underline key words. Its not hard if you just try. You will love it and it feels so good to be able to describe the movements of huge celestial bodies. You know the distances You know the period of the first moon. You know how to find the circumference of a circle. You know how to find avg velocity. You know newton's law of gravitation. And it couldnt hurt to also know keplers 2nd law. And with that you can get everything you need to answer this. You can do it!
Mass of Saturn can be calulated with, M = 4π²r³/GT² Using the information from the 1st row. After finding the mass of Saturn, we can find the period & orbital velocity of the moons using, T=√(4π²r³/GM) & v=√(GM/r) Also, Thank you sir Walter Lewin. I have watched most of your lectures of 8.01 & 8.02 and it has really helped me get a better understanding of the concepts❤️
Maybe. It's about time I do another one. Busy. Question : With all this moons, is there not a risc of colliding and would we notice such an event on earth ? Or maybe heavily influenced ?
Thank you so much for sharing this type of question with elaborate description. All time, from your precious lectures video I learn many things. Thank you sir
Hello Hello Hello Professor ! I am very grateful to you for changing the way I view the world and understand physics. I will be giving the JEE and would be thankful for your well wishes
Hello Mr Lewin, a) From Tethys distance 'r' and period 'T' we obtain the constant of proportionality 'k' of Kepler's third law and use it to obtain the periods of the other moons: T = sqrt(k*r^3) The orbital speed, v = 2pi*r/T 1) v = 11.35 km/s 2) T = 2.73 days, v = 10.04 3) T = 4.51, v = 8.5 4) T = 15.93, v = 5.58 5) T = 79.19, v = 3.27 b) Mass of Saturn = (r*v^2)/G = 5.7*10^26 kg
Sir, I am sorry its out of topic, i just cannot put my mind to study fluid mechanics without clearing this doubt. If you can please answer this We derive formula for Archemedes principle by using the formula P = hpg., Where p is density and h is the height of fluid from the surface we are measuring therefore F=hpgA If we take a cubical container with dimension d, The buoyant force must be the subtracted value of force acting on the top and bottom of the container which is F=(h+d)pg-hpg F=dpgA. But I have a doubt. We derive p=hpg by taking the force mg acting on it by the whole fluid above it, ie in a cubical container at a depth h, F=mg P=mg/A =mgh/Ah =mgh/V =hpg But if we take a cube, The top side of the container can be calculated by the above formula and it will be hpg But then on the bottom side we cant take mg of fluid above it only, as the bottom surface is in a depth of h+d,till height h we can account it for the weight of the fluid, but for height d , there is no fluid infact in height d there is the solid part of the cube, then how can we write (h+d)pg
a) To find the orbital speed of the moons of Saturn, we can use the formula for orbital speed, which is derived from Kepler's third law. The formula for the orbital speed (\(v\)) of an object in a circular orbit is given by: \[ v = \frac{2\pi r}{T} \] where: - \( r \) is the orbital radius (distance from the center of Saturn), - \( T \) is the orbital period. Given the distances and periods, we'll first convert all the given distances to kilometers and periods to seconds to ensure consistent units. Then we'll use the formula to calculate the orbital speeds. Here are the given values: - Tethys: \( r = 2.95 \times 10^5 \) km, \( T = 1.89 \) days - Dione: \( r = 3.77 \times 10^5 \) km - Rhea: \( r = 5.27 \times 10^5 \) km - Titan: \( r = 12.22 \times 10^5 \) km - Iapetus: \( r = 35.60 \times 10^5 \) km Let's convert the periods of the other moons into days, assuming they follow Kepler's third law: \[ T_Dione = Tethys \times \left(\frac{Dione_{distance}}{Tethys_{distance}} ight)^{3/2} \] \[ T_Rhea = Tethys \times \left(\frac{Rhea_{distance}}{Tethys_{distance}} ight)^{3/2} \] \[ T_Titan = Tethys \times \left(\frac{Titan_{distance}}{Tethys_{distance}} ight)^{3/2} \] \[ T_Iapetus = Tethys \times \left(\frac{Iapetus_{distance}}{Tethys_{distance}} ight)^{3/2} \] After calculating the periods, we can then use the orbital speed formula. First, let's convert the given period of Tethys to seconds: \[ T_{Tethys} = 1.89 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} \] Now, calculate the periods for the other moons using the ratios: \[ T_{Dione} = 1.89 \times \left(\frac{3.77 \times 10^5}{2.95 \times 10^5} ight)^{3/2} \text{ days} \] \[ T_{Rhea} = 1.89 \times \left(\frac{5.27 \times 10^5}{2.95 \times 10^5} ight)^{3/2} \text{ days} \] \[ T_{Titan} = 1.89 \times \left(\frac{12.22 \times 10^5}{2.95 \times 10^5} ight)^{3/2} \text{ days} \] \[ T_{Iapetus} = 1.89 \times \left(\frac{35.60 \times 10^5}{2.95 \times 10^5} ight)^{3/2} \text{ days} \] Finally, calculate the orbital speeds: \[ v = \frac{2\pi r}{T} \] Let's perform these calculations step by step. Here are the calculated orbital speeds of the moons of Saturn: - **Tethys**: \( 11350.80 \) m/s - **Dione**: \( 10040.76 \) m/s - **Rhea**: \( 8492.43 \) m/s - **Titan**: \( 5577.01 \) m/s - **Iapetus**: \( 3267.48 \) m/s These speeds show how the orbital velocity decreases with increasing distance from Saturn, consistent with Kepler's laws of planetary motion. b) To find the mass of Saturn, we can use Kepler's third law, which relates the orbital period and radius of an orbiting moon to the mass of the planet. The formula is given by: \[ T^2 = \frac{4\pi^2 r^3}{G M} \] where: - \( T \) is the orbital period, - \( r \) is the orbital radius, - \( G \) is the gravitational constant (\(6.67430 \times 10^{-11} \text{ m}^3 \text{kg}^{-1} \text{s}^{-2}\)), - \( M \) is the mass of Saturn. Rearranging the formula to solve for \( M \): \[ M = \frac{4\pi^2 r^3}{G T^2} \] We'll use the data for Tethys to find the mass of Saturn. Given: - \( r_{Tethys} = 2.95 \times 10^5 \) km \( = 2.95 \times 10^8 \) m - \( T_{Tethys} = 1.89 \) days \( = 1.89 \times 24 \times 3600 \) seconds Plugging these values into the formula, we get: \[ M = \frac{4\pi^2 (2.95 \times 10^8 \text{ m})^3}{(6.67430 \times 10^{-11} \text{ m}^3 \text{kg}^{-1} \text{s}^{-2}) (1.89 \times 24 \times 3600 \text{ s})^2} \] Let's calculate this step by step. The calculated mass of Saturn is approximately \(5.69 \times 10^{26}\) kilograms.
I get: Mass of Saturn 3.897x10^25 slugs (5.687 x10^26 KG) Velocity of Tethys 37, 240 FPS Dione: 2.73 days / 32,942 FPS Rhea: 4.51 Days / 27,862 FPS Titan: 15.93 days / 18,297 FPS Upetus: 79.2 days / 10,720 FPS I used the radius of Tethys as 10^5 KM. The exponent was a little hard to see.
My try: (a) Orbital Period of Tethys, T₀ = 1.89 days Its distance from Saturn, R₀ = 2.95×10⁵ km According to Kepler's third law, T² ∝ R³ So, T²/T₀² = R³/R₀³ Time period, T = T₀(R/R₀)^(3/2) … [Eq. 1] And orbital velocity, v = (2πR)/T … [Eq. 2] For Tethys, Orbital velocity, v = (2πR₀)/T₀ = (2×3.1416×2.95×10⁵)/(1.89×24×60×60) = 11.35 km/s Using Eq. 1 & 2, we can calculate the time period T, and the orbital velocity, v of the mentioned planets. For Dione, T = 1.89×(3.77/2.95)^(3/2) days = 2.73 days v = (2×3.1416×3.77×10⁵)/(2.73×24×60×60) = 10.04 km/s For Rhea, T = 1.89×(5.27/2.95)^(3/2) days = 4.51 days v = (2×3.1416×5.27×10⁵)/(4.51×24×60×60) = 8.5 km/s For Titan, T = 1.89×(12.22/2.95)^(3/2) days = 15.93 days v = (2×3.1416×12.22×10⁵)/(15.93×24×60×60) = 5.58 km/s For Iapetus, T = 1.89×(35.6/2.95)^(3/2) days = 79.23 days v = (2×3.1416×35.6×10⁵)/(79.23×24×60×60) = 3.27 km/s (b) Considering the mass of Saturn, M And mass of the moon, m Gravitational attraction force = (GMm)/R² Required centripetal force = (mv²)/R [Assuming the orbits to be circular] Thus, (GMm)/R² = (mv²)/R Simplifying we obtain, M = (4π²R³)/(T²G) Using values for Tethys, M = (4(3.1416)²(2.95×10⁸)³)/(1.89×24×60×60)²(6.67×10⁻¹¹) = 5.7×10²⁶ kg. (Answer) Priyo
a) The speeds of all the moons are 11351 m/s Tdione = 2.42 days Trhea = 3.38 days Ttitan = 7.83 days Tlapelus = 22.01 days b) The mass of Saturn is 5.683×10^26 Kg
@@lecturesbywalterlewin.they9259In case RU-vid is even blocking you from seeing my burn link, I also posted it in my own channel description. The answers are there.
M(Saturn) = 5.69e+26 kg Tethys: T = 1.89d, v = 11.35 km/s Dione: T = 2.73d, v = 10.04 km/s Rhea: T = 4.51d, v = 8.50 km/s Titan: T = 15.93d, v = 5.58 km/s Lapetus: T = 79.23d, v = 3.27 km/s
Moon r (x 10^5 Km) T (days) v (Km/sec) Tethys 2.95 1.89 11.35 Dione 3.77 2.73 10.04 Rhea 5.27 4.59 8.49 Titan 12.22 15.93 5.58 Iapetus 35.6 79.22 3.27 Mass of Saturn = 5.7 x 10^26 Kg
I know this is very inappropriate sir but we students are trying to bring every sort of person concerned with us to know about what's happening to neet students in INDIA. NEET- National Eligibility cum entrance test is conducted by National testing agency of India every year for entry to MBBS colleges. The exam is of 3 hours 20 minutes with 200 questions and 720 marks. Over 25 lakhs student appeared in neet 2024 Neet 2024 was held on 5 May 2024, but allegations of paper leak started by 3 May 2024. Several Candidates said that they received telegram messages and phone calls luring them to buy question paper beforehand. However no previous investigation was set up into the matter. On the day of examination too, there were several incidents of cheating and paper leak in patna and Rajasthan. There were also issues with late paper distribution and lack of time for some candidates. After the examination, on 6 may police caught 7 people over issues of paper leak Students were outraged and demanded RENEET. An plea was filed in patna high Court and further in Supreme Court of India. However, on 18 May, the Supreme Court refused allegations over paper leak, and asked NTA to proceed further with the process. Amid chaotic Indian lok sabha elections whose vote counting was being done on 4 june 2024, NTA released the NEET 2024 results. The result showed several flaws including 718 and 719 marks which were practically impossible as per the marking criteria of neet exam. (You have to attempt only 180 questions out of 200.Each question has 4 marks and -1 negative marking for every wrong answer) Later, nta gave a clarification for giving 718 and 719 marks as grace marks for those who complaint about getting less time. This was done under a judgment of hon'ble apex Court dated 13.06.2018. From the student point of view the failure of NTA in holding fair examination for neet students lies below. 1. Why was a judgment which was dated as of 13. 06. 2018 was suddenly and for the first time used for normalization of students without informing the other concerned candidates ? 2. How many marks have been awarded as grace marks and to how many students? What is the criteria of grace marks? 3. This year 2200 students secured more than 700 marks in a national level exam, clearly, this points towards failure of NTA as a testing authority where 25 lakh students are concerned. 4. Why was the result declared on such a busy day where all the national bodies including media were absorbed in the election results. 5.Under this rising pressure of competition on students, why hasn't the NTA tried to change its testing methods. Also, the students comment that this inflation is a result of paper leak which was silenced by nta in the previous half only. 6. The cutoff for admission to mbbs college has seen a historic rise from 610 in 2023 to 655 in 2024. Senior mbbs students who had passed neet in previous years say that nta has failed to be eligible for testing medical aspirants. Under this massive rise in cutoff, students are suffering from mental health issues, which NTA has failed to curb. 7. A list of top 100 students of the neet 2024 as published by NTA reveals that nearly 89 students were getting 720/720 out of which nearly 8 were from one common exam centre. This year there were 571 exam centers. The probability of 8 students bagging full marks at one common centre is nearly impossible. This clearly point towards a malpractice. 8. Every year the number of suicides by aspirants in INDIA (mostly kota: the coaching capital), keeps on increasing. Many of us suffer from depression, anxiety and have family issues due to being unable to fir under the qualifying cutoff. A large number of candidates say this was their 3rd or 4th attempt. With NTA not addressing the major problems of neet candidates, our future lay in dark. The condition of the student is pitiful, not something I would like to write in a few words. I would say, they have been preparing for this exam since 3-4 years, same books over and over again, have paid 2-3 lakhs to coaching institutes, have cut themselves off from society for the sake of the exam , missing family celebrations, vacations, festivals, and any companionship with their friends. Locked in a room, made to study for hours, giving mock tests, revising and feeling isolated is a story of millions of such students. There's so much to write about them, but I would say each of them is an individual warrior, to having survived such a gruesome journey only to find out that they can not get a medical college because their rank isn't enough this year.
@@lecturesbywalterlewin.they9259 your biography sir, I wonder why would you wish me to see that? After watching it I realised I had hope, life is surprising enough! I never thought one day I would be writing a comment to you about my/our grievances, but I did and you did reply to me. Sometimes I wished my life would be perfect, a route I devised, and I was ready to keep it running at any expense of hardwork. I have reflected on this ridiculous idea, it's straining me morally and mentally, tied to one pole and i ran around covering the largest circle I could. I have lost myself in it, mourned for what I have become, a perfect rat in a race, a creature who has come so far in his plan that he accepts self destruction over destruction of his world. The change outside is not apparent, but inside it is grave, I lost my creativity, my imagination, and my sympathy for myself. For more than 17 years, I wanted to be an artist, or a writer, both I self trained myself, enjoying my own classes. But, there's a debt a lot of us are born with, our parents' sacrifice. I am the eldest child in my family, a girl at that too. My parents have sacrificed their entire earnings to pay my Education fee, I cannot escape now! My love should move aside, take my dreams along with it and go away. Sometimes, some other things, bring these thoughts back.... but I think I can see your love for physics, forgive me if I am wrong. In your video I put myself at that same position where our minds would start, when it had something to tell everyone, an idea, a creation, a simple discovery which it knows that everyone will appreciate, it delights the mind. It's the driving force we evolve on our arts, without audience the artist can exist but not evolve. When he does't have audience, he turns into himself, and finds an audience there, this is mentally exhausting, he continues to evolve, the process is his food. But, other than that he needs real food to exist too. That was the thorn I was poked with again and again by family and society, the thorn of reality until my fabric of stubbornness shattered and I took over a new one, that of a perfect path, a path to be a doctor and caring for my parents till the end. I have said too much sir 😅, I apologize for this story telling, but among the list of people that will inspire me, your class has joined. I still train myself for one or other reason, and your class will remind me to be a teacher like you and an audience like yours. Thankyou Walter Lewin sir! Here I leave an experience I shall recall many a times now.
m * v² / r = G * m * M / r² or M = v² * r / G and T = 2 * pi * r / v or v = 2 * pi * r / T With G = 6.674 * 10^-11 m³/(kg*s²) i get: v Tethys = 1.135 * 10^4 m/s and M = MassSaturn = 5.69*10^26 kg For the other Values i can use the two equations: v = √(G * M / r) and T = 2 * pi * r / v Dione: v = 1.00 * 10^4 m/s and T = 2.73 d Rhea: v = 8589 m/s and T = 4.51 d Titan: v = 5575 m/s and T = 15,94 d Lapetus: v = 3266 m/s and T = 79,27 d
@@lecturesbywalterlewin.they9259 sorry sir but I didn't get your point is it correct? This address is not found.... Actually My name is Fawad Ahmed Abbasi and I am from Pakistan. I am BS physics semester 8th student in Pakistan. I am also a podcastor I want a podcast with you. It will be a great hounur not only for me but for Pakistan to host you. Kindly accept my request I will be really thankful to you. Best regards you big fan.
@@FAWADPODCAST 30% of my 1.7 million subscribers want to do a potcast with me, 80% want to know my email address and my phone. *All of this is top secret* - I do not video chat no potcasts with anyone.
@@lecturesbywalterlewin.they9259 it's ok dear professor no problem stay blessed and have good health....we all want this because we love you I know it's difficult for you to manage this.
