Doctor, I have asked you about this before, and did not understand until this video. I can not thank you enough. 21 minutes of your time, has cleared up all my misunderstandings. You have shined light into spaces I did not even know I had. Thank you.
Just got back from Hopkinsville, Kentucky where my wife, my 11 year old son and myself experienced 2 minutes and 40 seconds of totality. I even saw the diamond ring. Great show.
Sir, can you please describe about who initiated the poles in MAGNETISM, how the south pole confirms that it should be SOUTH/ NORTH and why not WEST/ EAST?..
I'm not Walter but I liked your question so I'm going to take a stab at it. North/South are directions named long ago by tradition. The magnets in compasses line up with the global magnetic field and the end of the compass magnet that points to the geographic North lends its name to the naming of the North magnetic pole. However, since opposite magnetic poles attract, the magnetism at the North geographic pole actually corresponds to the South magnetic pole. Is that clear as mud? Further complicating matters, the magnetic poles are not fixed at the geographic poles. From year to year, the magnetic poles wander and before GPS, sailors local to my part of the world had to purchase new Eldridge maps almost every year, in order to navigate. Also, the magnetic poles reverse suddenly about every 10,000 years (we are due!) When that happens it is thought the magnetic poles shift in steps from North geographic to South geographic so that there will be periods of time soon when the magnetic poles do point East/West. Thanks to Big Al for GPS being invented before the shift happens but hey, shift happens!
First off, if you happen to read this...THANK YOU!!.. 2nd, sadly I'm going to miss the eclipse event as I am in Thailand at the moment. But again thanks for the exercises, I will work on them and hopefully travel to a place where a total eclipse will happen between now and 2030. Looking forward to such magnificent experience and being friends for ever!
Here's a summary of the problem. You're asked: What's the radius over the surface of the Earth where a partial (or total) eclipse will be seen? How many total eclipses will there be before 2030? (you're asked to look it up on Google) Which of those will be within 1000 km from where you live? (also look it up on Google) This is the data you're provided: Distance Earth-Sun: 152*10⁶ km (max) , 146*10⁶ km (min) Distance Earth-Moon: 406.7*10³km (max) , 356*10³km (min) Sun's radius: 696*10³km Moon's radius: 1737km I aprecciate the time and effort of Walter in making all of these problems very much, no doubt about that, but a 20 minute video for a simple question makes me want him to get to the point once and for all. Maybe someone else will find this summary useful.
I will have a total solar eclipse extremely close to me in about 7 years, it will practically pass right over me :) Thank you for the wonderful video on eclipses!
Thank you for your great lectures and videos on youtube!! I'm a physics student from Germany and you really gave me a lot of inspiration and motivation to get through my exams ;) especially your series about electricity and magnetism is fantastic! :D
it would be a circular shadow of 80km diameter for total eclipse and 6884.97km for partial. Love your videos professor, i have been solving every single problem you post but havent been posting the answer. i will from now on :)
HI Professor. I have little confidence in my answer, but this is how I approached the problem. First I took a line from the centre of the Sun to the centre of the Moon. I then took a line parallel to this from the edge of the Sun and another from the same edge of the Sun to the opposite edge of the Moon. This makes a right angle triangle with adjacent side 152m Km and opposite side 0,7m Km (half the diameter of the Sun - I ignored the additional 1/2 diameter of the Moon). The angle is therefore 0.264 degrees. A similar triangle from the Moon's edge to the Earth would have an opposite side of 1639 Km. Doubling this and adding the diameter of the Moon, 3474 Km, gives me a penumbra diameter of approximately 6752 Km.
Dear Dr. Lewin, I was in Gallatin, TN very near the center of the Path of Totality. I was still searching for an acceptable observation point when the Solar Eclipse Timer running on my phone sounded the alert at the time of First Contact (12:59 PM EDT) so I knew I didn't have much time to find it. Fortunately we spotted a nice open park like area where others had gathered and got my cameras set up with about a half hour to spare during which we observed the partial eclipse with our eclipse glasses. Second Contact at 2:27:25 EDT - the timer alert went off as predicted, as the Solar disk became increasingly obscured (but still blinding) until suddenly, a dazzling burst of white light formed the Diamond Ring. Then, slowly dimming until it became the Corona, persisting over the next 2 minutes 39 seconds. Unfortunately the sky only revealed Venus; I could not see the other planets or stars as I had hoped. The horizon appeared just as expected, the twilight being all around us. Third Contact at 2:30:04 EDT - the burst of white light yet again and another Diamond Ring, becoming increasingly bright until one could bear no more without the glasses. It was one of the most memorable moments of my life, as I'm sure it was for my companions and fellow observers that day.
Annular solar eclipse on 26 December 2019 within the range of 2000 Km here in South India particularly in Coimbatore.... I will go for it....Thank you for sharing and wish you a very healthy life sir. Namaste
Will you be going to the lane of totality? I will be! I am worried about traffic though, on account of nearly half of the state traveling to see it. Hopefully it would be too bad.
A good way to explain with a demonstration; use a video camera to present the eye, a hanging ping pong ball to present the moon, and a hanging beach ball to present the sun ( and having fun changing the distances)
@Walter Lewin, Dear Sir, its really nice of you to come up with these videos to share your knowledge. Sir i have a request: Can you please make a video on Hermholtz coils and Mu Metal Can's Concepts and calculations?
Can you calculate an approximate PROBABILITY of a total solar eclipse occurring by chance throughout the universe in other solar systems? Assuming there are approx 10 to the 21 stars in the universe and assume that each star has an average of 10 planets orbiting each star and that each planet has at least 1 moon orbiting it. One must calculate that that moon must be in the perfect position to properly occlude the star to perfectly occlude the disk of the star. (Our situation is that the moon is 400 times smaller than sun but that its also exactly 400 times the distance to perfectly cover the sun.) That kind of coincidental geometry is amazing
I got about about 6800 km (as the diameter of the region). I did have to look up the distance between the earth and the sun and the radius of the sun though. Also, unfortunately, the next total solar eclipse in my country will be in 2034. Looking forward to more problems in the future!
Professor, thanks for the video! One thing that I have not been able to explain adequately is the west-to-east motion of the umbra. Experts point out that the moon orbits west-to-east and that the tangential velocity of the moon in its orbit is greater than the tangential velocity of an observer at the equator. This faster orbital velocity allows the moon to move easterly faster than the surface of the earth, causing the umbra to travel west-to-east. However, the ANGULAR velocity of the spinning earth is some 27x faster than the angular velocity of the moon. This, I believe, explains why we see the moon rising in the east and setting in the west. Multi-exposure images of the full eclipse clearly show the moon and the sun drifting westward from the perspective of an earth observer. Clearly, the umbra moved west-to-east, but I've not been able to explain how this is possible given the much greater angular velocity of the earth vs. the moon. Would you care to comment?
On Aug 21 during the solar eclipse, the Moon moved onto the sun in the direction from West to East. That's what the Moon does EVERY day, moving 15 degrees per day relative to the sun. Thus the West coast of the US would see the eclipse first (due to parallax).
Daym! You're an insanely intelligent person! Really hope to meet you in person someday! Oughta be the best day of my life, After I figure out where you live. 😅
Using: Sun-earth distance 152e6 km Moon-earth distance 356e3 km Radius of the sun 695700 km Radius of the moon 1737 km Radius of the earth 6371 km The "diameter" of the penumbra (measured over the curvature of the earth) is about 7054 km
Someone asked, if the eclipse could be rescheduled to weekend, because children go to school on monday :-) There has been one total eclipse in my lifetime right where I live, and it was cloudy. So it just got dark for a while, which was interesting, but quite a bit less than I hoped to see.
I look through Glass, at the turtles under the water, They look really close, But when I look on top of water, at the Turtles head peaking out he is, about 3-4 times the perceived distance. How can I be convinced that a similar refraction of light is not happening during the eclipse?
Calculated (using similar triangles)and sun furthest distance, moon closest as per your figures, and worked out partial will extend 3375 km either side of centre of totality. As for total eclipses until 2030, I see from the web 8 such occurances, the last one on 25 nov 2030 will be within 1000 km of where I live however cannot drive there as is is over the sea!! Will definitely be in the partial zone. Can you guess where I live?...clue - it is an island south of equator and also a continent!
Good morning professor. In 2 August 2027 a solar eclipse will be visible in Italy, only in the little island of Lampedusa will be total. Time to book my holiday ;)
Dear Prof., The diameter of the penumbra is 6750 km. This post is more precious to me just because you are looking at it. ❤️Love Physics❤️Love You Prof.❤️
And, on the upcoming Total Solar Eclipse question, there are 8 Occurrences i believe before 2030. I live in India and unfortunately, am not within 1000km of path of totality for any of those. But the one in 2027 is the closest, though the totality falls on Indian Ocean, I would be in Penumbra region.
Bonus question for anyone who is truly enthusiastic: Calculate approximately the time that the partial eclipse lasts, and the total eclipse lasts. (Answers: partial ~ 120 minutes, total ~ 120 seconds) Hint: Think about the relative motion of a point on the Earth's surface to the moon, and hence its shadow.
Hello sir... I am a new viewer of your lectures... I have just recently started following your lectures and have seen 7 to 8 lectures... I have a question that we have a phenomenon of diamond ring effect... I am a bit confused because moon blocks the sun almond completely just an outer ring of sunlight is formed as you said it annular eclipse... Is there any use of diffraction of sun rays in diamond ring effect... As for the answer of your 1st question i got 3480 km on the surface of will be the diameter of penumbra means penumbra will be extended in an area of 10927.2 squared km...
my results are 1.2816*10^6 km or 1,2816 thousands km of penumbra. I would like to know how to calculate umbra too but i dont have any idea. I considered the distance of the moon from the sun at 152*10^6 km and the distance between the moon and earth at 356*10^3 km
correction: i have made an unwarranted assumption on the distance to the moon, i change the answer Lpenumbra=6691km, taking the radius of the earth to be 6371km
Professor, why does the umbra move from west to east, rather than the normal east-to-west 'motion'? Especially since the 'center' of the eclipse moves way faster than the point on the earth below the moon?
sir if everything in this universe is moving. I mean if every motion is relative and we can not measure the real speed so how the speed of light is same in every frame of reference
With max Sun and min Moon distances the intersection point is 378 000 km from Moon, so the penumbra cone size at earth distance is 6742 km. The size on Earth's surface depends on how it hits the Earth, but if it's right in the center it's 7094 km. There wil be total eclipses in 2019, 20, 21, 24, 26, 27, 28 and 30. None of them are within 1000 km from me.
I witnessed the total eclipse 1999. Traveled to Luxembourg. It was amazing. The rings and the corona were sparkling. But the blue light the world was shrouded in was ...mind boggling. The whole world went still. All birds went quiet. I'm most amazed by the astronomically small chance that from earth the sun and the moon appear nearly equal in size. I mean the variation in the universe is seemingly endless but somehow we earthlings have a sun AND a moon that appear the size of a pea at arms length.... Now I don't believe in coincidences like that..... at least I question that coincidence. But so far I found no theory to explain this correlation to have a scientific reason. So it seems to be strictly coincidental. Is it?
>>>>I'm most amazed by the astronomically small chance that from earth the sun and the moon appear nearly equal in size.>>> Yes that is remarkable. If the lunar orbits were just a little larger total solar eclipses would NEVER be possible. There is ABSOLUTELY No scientific reason for this. *ABSOLUTELY NOT!*
Hi Prof. Lewin, I have a question to ask. Is time a vector quantity or scalar quantity? There is a concept called 'Arrow of time', so can I interpret time as a vector quantity? Yours sincerely Shaoren Ong
"In physics101, scalar quantities are defined to be ones which have magnitude only, and no direction, where "direction" in this context means a direction in *three dimensional* space. Time clearly has no such direction. ... But Time can be a vector, but what it means depends on the context."
Lectures by Walter Lewin. They will make you ♥ Physics. is that bcuz the moon block the light just a small surface area on Earth so that we still able to see the light outside the moon's shadow?
Hi! because of similar triangles and proportionality and after doing some math I arrived : penumbra= 6750 Km with the distance to the sun of 152 .10^3 km and distance to the moon of 356.10^3km. Finally the distance from the earth to the intersección lines is 733515 km. . Thanks for your attention.
Hellow, Prof. lewin I am little confused with the concept of relative motion. I don't understand how can we say if frame of reference is inertial or not... should I look if a force is being applied on me( if I the observer ofcourse) My another confusion is related to magnetism... Suppose I have two non conducting balls having positive charges on them, I go in a cab and put them in my lap... the cab starts moving now, for me balls are applying only the repulsive electrostatic force on each other... but for my friend on the ground... the charges are moving with the cab and there should be a magnetic force of attraction between them???
+Lectures by Walter Lewin. They will make you ♥ Physics. Thanks for the reply. But If the velocity is heigh enough... My friend should them coming together and I would see them going apart... Can both be true?
Using distance of the Moon at the minimum and distance of the sun at the maximum, and the part of the Earth is flat, the radius of the penumbra is 1745km. If take the curve of the Earth into account, about 1767km. Although when I try to measure the eclipse zone on google map, the radius seems to be closer to 4000km. Of course the fact that the centres of Earth, Moon and the Sun are not perfectly aligned is not taken into account here, so when the shadow is off center it will be larger due to the curveness of the Earth. Can't think of any other factors for the moment.
Just to make sure, we are given the distance from the center of the sun to the surface of Earth, the distance from the surface of the moon to the surface of Earth, the radius of the sun, and the radius of the moon? And you want us to find the area of the shadow on the Earth, or just the radius of the shadow?
Diameter of penumbra of the moon on earth is 6751.00128 km considering max distance from the sun 152 million km and min distance of the moon 356000 km. Additionally, the distance from Earth to the intersection angle between the sun and the moon is 736979.556 km. There will 8 total eclipses till 2030. The total solar eclipses dates are on 2 July 2019, 14 Dec 2020, 4 Dec 2021, 8 April 2024, 12 August 2026, 2 August 2027 ( this will be in the country where I live), 22 July 2028, 25 November 2030.
I first tried using similar triangles with the chord of the penumbra, the diameter of the moon, and the diameter of the sun. But using values for the current eclipse, I got too small a value (7.29e3 km instead of nearly 10e3 km). (For reference, I got 7.10e3 using the 152e6 & 356e3 values.) Then I realized my triangles were wrong: you can't see the south pole of the sun from the north pole of the moon! This will require a different strategy...
The sun, moon and Earth are spheres. If they line up *precisely* (for a given distance to the sun and the moon) the diameter of the Moon's penumbra on Earth is determined regardless of where the totality is. Notice the word *"precisely"* .
I was assuming the centers are co-linear. But I was drawing lines from the bottom of the Sun to Earth, touching the top of the Moon, which meant the line passed through the sun and the moon. And I used the distance from the Earth to the Moon as the distance from the shadow to the Moon, figuring it was "close enough". I tried again, starting from thinking of the penumbra as the near intersection of the Earth and the double cone tangent to the Sun and the Moon (with apex between them.) Looking over the intermediate values, the dominant difference from before was the approximation for the distance between the shadow and the moon. Overall, my answer shrank by around 1%. I used a couple of inverse sines, which doesn't feel like the kind of solution you are hinting is out there. And, of course, I expected a bigger answer, not a smaller one! I'll just have to wait for your solution on Saturday :) Thank you for the problem!
I found 6 of them, sir 2019,2020,2021,2023,2024,2026. I live in India and Indonesian eclipse happens to be the shortest distance from my place(2023 eclipse)
Finding the Shadows of an eclipse. I lined up all objects so that their centers all fell on a common line, and then used all of the measurements given in the problem to be the coordinates used to find the size of the two shadows. I used a Cartesian Plain using the y-axis to define the line that the centers of the Earth, Moon and Sun are on. So that the points I used as: Sun (-696,000 , 152,000,000) & (696,000 , 152,000,000) Moon (-1737 , 356,000) & (1737 , 356,000) Earth’s surface at the center of the shadows (0,0) Penumbra Using the point slope formula, and choosing the right side of the Sun and the Left side of the moon, I get the points (696,000 , 152,000,000) & (-1737 , 356,000) Slope = (152000000 - 356000)/(696000, -(-1737)) = 217.3369049 y-y1=m(x-x1) for (696,000 , 152,000,000) y-152000000=217.3369049(x-696000) y-152000000=217.3369049x-151266485.8 y=217.3369049x-151266485.8+152000000 y=217.3369049x+733518.2038 And at the Earths surface, Where y=0 0=217.3369049x+733518.2038 -733518.2038=217.3369049x x=3375.008954 km. This is the radius of the Penumbra Umbra Choosing the points that are on the right side of the Sun and Moon Points - (696,000 , 152,000,000) & (1737 , 356,000) Slope = (152,000,000-356,000)/( 696,000-1737) = 218.4244299 y-y1=m(x-x1) for (696,000 , 152,000,000) y-152,000,000=218.4244299(x-696,000) y-152,000,000=218.4244299x-152030363 y-152,000,000=218.4244299x-152030363+152000000 y=218.4244299x-30363 Again finding the size of the shadow at point y=0 0=218.4244299x-30363 30363=218.4244299x x=139.002812 This is the radius of the Umbra Sig Figs only 3, so Radius of Umbra is 139 km Radius of Penumbra is 3380 km There are seven eclipses from now until 2030 and the closest for me will be around San Antonio, Texas in 2024.
Penumbra=2*((Dem/Dsm)*(Rs+Rm)+Rm)=6750km Dem = distance between Earth and Moon = 356,000km Dsm = distance between sun and moon = 152*10^6 - 356,000 km Rs = radius of Sun = 696,000 km Rm = radius of moon = 1737 km
about 6750 km absolute distance. Earth surface distance ( circumferential distance ) about 7461 km. Funny i have the same answer as Aaron Hangcock below. My sheet has several tan and atan functions in there.... Aaron has obviously the more elegant way of calculating this :-)
question : guess this is the newtonian answer. distances are maybe such that relativity comes into play ....is there a non newtonian answer ? Is it different ?
Sr does NO at all cone into play here. The speed of these objects are a minute fraction of c. The speed of the Earth in orbit is 30 km/sec which is 0.0001 c
I want NO black board and NO white board. At MIT I had 9 huge black boards and *3 cameras.* 2 of these cameras could move. Thus i would rarely every block the vision of any camera as I was writing on the black board. My lecture hall could hold 600 students. At home my auditorium is about 2x3 m. No room for even 1 person to watch me. If I worked on a blackboard or white board with 1 fixed camera I would block my own writing for 95% of the time.
I got you.......I just figured you were so used to nice set-up's at MIT that you just never thought to get a small one for your home office/studio......I don't think you can do your "dot thing" on one of those anyway !
show her the blue sky, the clouds, the sun, the Moon, the planets, your glasses, refrigerator, radio, TV, iPhone, internet, cars, airplanes, rockets, trains, street cars, calculators, telephones, day and night, sun rise, sun set . . . . . tell her that all that is Physics. Our whole world is Physics. Without Physics no life could exist.
tan a = (Rs + Rm) / (Ds - Dm) = ( 696000 + 1737 ) / ( 152000000 - 356000 ) = 697737 / 151644000 = 4.601/1000 Re = Rm + Dm * tan a = 1737 + 356 *4.601 = 3375km diameter is (supposed to be) 6750km R are radius. D are distances m is Moon, s is Sun, e is eclypse. Earth is assumed flat (it's not), diameter of Earth is considered too small. With Earth diameter : Re = Rm + (Dm - Rt) * tan a = 1737 + (356-6.366) *4.601 = 3345.6km. Diameter = 6691km Rt radius of Earth 6366km
i'm going to go with 6676 km (final answer), which is slightly less than two moon diameters. this is the diameter of the shadow that arrives at the plane that is tangent to the earth at the sun-moon-earth line. i used walter's distances for sun-earth and moon-earth (152000*10^3 km and 356*10^3 km.) to arrive at this 6676 number i translated a moon back to the sun along one of the tangent lines to where it was tangent with the sun. then i translated this back to the point where the moon center touched the earth-tangent at the sun-earth line. this created a simplifying geometry with two congruent triangles where it was easy to calculate the angles. part 2: i count 7 total eclipses between tuesday and 2030. only one of which i will not have to update my passport for: 4/8/24 extra credit: what is the diameter of the shadow that arrives at the plane that intersects the earth center and is perpendicular to the sun-moon-earth line? I get 6734 km. Which is only 0.8% larger. This is the maximum width of the penumbra shadow. Your mileage may vary, though, because the entire penumbra shadow may not be intercepted by the Earth. extra extra credit: watch this animation which shows the penumbra sweeping across the earth: ua-cam.com/video/dJyiWQcVmiM/v-deo.html
many people have 6750 which is not correct. This puzzles me. What have you assumed about the distance to the sun and the distance to the Moon? I will post my solution on Saturday.
Please send me how you arrived at the 6750 (I want to see your calculations). If I use your 336 thousand km to the Moon (with is not possible), I do not find 6750 km. Since several people sent me 6750 km I want to know how they got that number.
I also find 6750 with Similar Triangles. If we call O the point where the 2 lines defining the partial eclips cross each other and x the distance from the Earth to O. Let's call R_S, R_M the radius of Sun and Moon ; d_S, d_M the distance from Earth to Sun respectively Moon. We want to know 2*S (S for Shadow) Similar Triangles gives us : R_S/(d_S-x) = R_M/(x-d_M) = S /x. The fisrt equality gives us x= (R_S*d_M+R_M*d_S)/(R_S+R_M) and also x-d_M=R_M/(R_S+R_M)*(d_S-d_M) The second equalty gives us : S=(R_M*x)/(x-d_M) S= (R_S*d_M+R_M*d_S)/(d_S-d_M) S= 3375 2S = 6750 What's wrong ? Rgds.
I LOVE PHYSICS! AND MY ANSWER IS 6750.012881 KM. PENUMBRA LENGTH ON EARTH IS 6750 kms. And that is 50% of Earth Dia!... :) ... so... very sad for people in antartica and arctic..:)
quite a few people have 6754. I wonder why. What assumptions did you make about the distance to the Sun and to the Moon? I took 152 million km for the distance to the sun and 356x10*3 km for the distance to the moon. I will post my solution this Saturday.
Doctor, I have asked you about this before, and did not understand until this video. I can not thank you enough. 21 minutes of your time, has cleared up all my misunderstandings. You have shined light into spaces I did not even know I had. Thank you.
😊
Just got back from Hopkinsville, Kentucky where my wife, my 11 year old son and myself experienced 2 minutes and 40 seconds of totality. I even saw the diamond ring. Great show.
:)
Simply the greatest living man in history. Thank you for your sharing your amazing knowledge 👍
:)
Thank you, Dr. Lewin!
Sharing knowledge is the greatest pleasure of LIFE, you mean it sir. "Really you're the KING of the PHYSICS"..
:)
Sir, can you please describe about who initiated the poles in MAGNETISM, how the south pole confirms that it should be SOUTH/ NORTH and why not WEST/ EAST?..
I'm not Walter but I liked your question so I'm going to take a stab at it.
North/South are directions named long ago by tradition. The magnets in compasses line up with the global magnetic field and the end of the compass magnet that points to the geographic North lends its name to the naming of the North magnetic pole. However, since opposite magnetic poles attract, the magnetism at the North geographic pole actually corresponds to the South magnetic pole. Is that clear as mud?
Further complicating matters, the magnetic poles are not fixed at the geographic poles. From year to year, the magnetic poles wander and before GPS, sailors local to my part of the world had to purchase new Eldridge maps almost every year, in order to navigate. Also, the magnetic poles reverse suddenly about every 10,000 years (we are due!) When that happens it is thought the magnetic poles shift in steps from North geographic to South geographic so that there will be periods of time soon when the magnetic poles do point East/West. Thanks to Big Al for GPS being invented before the shift happens but hey, shift happens!
Thanks for the video Mr. Levin, it is so useful!
:)
First off, if you happen to read this...THANK YOU!!..
2nd, sadly I'm going to miss the eclipse event as I am in Thailand at the moment.
But again thanks for the exercises, I will work on them and hopefully travel to a place where a total eclipse will happen between now and 2030.
Looking forward to such magnificent experience and being friends for ever!
:)
Physics was boring until I bump into your lectures on youtube. Thanks sir for making physics understandable for all category of students. :)
*You are such a big inspiration to Me !!! I really hope to meet you !!!*
:)
Here's a summary of the problem. You're asked:
What's the radius over the surface of the Earth where a partial (or total) eclipse will be seen?
How many total eclipses will there be before 2030? (you're asked to look it up on Google)
Which of those will be within 1000 km from where you live? (also look it up on Google)
This is the data you're provided:
Distance Earth-Sun: 152*10⁶ km (max) , 146*10⁶ km (min)
Distance Earth-Moon: 406.7*10³km (max) , 356*10³km (min)
Sun's radius: 696*10³km
Moon's radius: 1737km
I aprecciate the time and effort of Walter in making all of these problems very much, no doubt about that, but a 20 minute video for a simple question makes me want him to get to the point once and for all. Maybe someone else will find this summary useful.
I will have a total solar eclipse extremely close to me in about 7 years, it will practically pass right over me :) Thank you for the wonderful video on eclipses!
great
Thank you for your great lectures and videos on youtube!! I'm a physics student from Germany and you really gave me a lot of inspiration and motivation to get through my exams ;) especially your series about electricity and magnetism is fantastic! :D
great
it would be a circular shadow of 80km diameter for total eclipse and 6884.97km for partial. Love your videos professor, i have been solving every single problem you post but havent been posting the answer. i will from now on :)
Thank you,thank you very much for those information ❤❤
:)
you are really awesome sir you are truely excellent
:)
i have been waiting for that eclipse for years!!!......
:)
Darn it! Too bad it won't be visible where I live.
how about in the next 20-40 years?
I just love physics....all because of you
:)
Fortune willing, I will enjoy viewing the totality Aug 21, 2017. I've got my glasses and I live 3 miles from the lane! WOW! Peace to you, Walter :-)
lucky U!
Lectures by Walter Lewin. They will make you ♥ Physics.
It was AMAZING! Thanks for posting your pics of crescent shadows, fun
Also look for the shadow bands...wish you best of luck.
hard to see
HI Professor. I have little confidence in my answer, but this is how I approached the problem. First I took a line from the centre of the Sun to the centre of the Moon. I then took a line parallel to this from the edge of the Sun and another from the same edge of the Sun to the opposite edge of the Moon. This makes a right angle triangle with adjacent side 152m Km and opposite side 0,7m Km (half the diameter of the Sun - I ignored the additional 1/2 diameter of the Moon). The angle is therefore 0.264 degrees. A similar triangle from the Moon's edge to the Earth would have an opposite side of 1639 Km. Doubling this and adding the diameter of the Moon, 3474 Km, gives me a penumbra diameter of approximately 6752 Km.
Dear Dr. Lewin, I was in Gallatin, TN very near the center of the Path of Totality. I was still searching for an acceptable observation point when the Solar Eclipse Timer running on my phone sounded the alert at the time of First Contact (12:59 PM EDT) so I knew I didn't have much time to find it. Fortunately we spotted a nice open park like area where others had gathered and got my cameras set up with about a half hour to spare during which we observed the partial eclipse with our eclipse glasses.
Second Contact at 2:27:25 EDT - the timer alert went off as predicted, as the Solar disk became increasingly obscured (but still blinding) until suddenly, a dazzling burst of white light formed the Diamond Ring. Then, slowly dimming until it became the Corona, persisting over the next 2 minutes 39 seconds. Unfortunately the sky only revealed Venus; I could not see the other planets or stars as I had hoped. The horizon appeared just as expected, the twilight being all around us.
Third Contact at 2:30:04 EDT - the burst of white light yet again and another Diamond Ring, becoming increasingly bright until one could bear no more without the glasses. It was one of the most memorable moments of my life, as I'm sure it was for my companions and fellow observers that day.
nice
Annular solar eclipse on 26 December 2019 within the range of 2000 Km here in South India particularly in Coimbatore.... I will go for it....Thank you for sharing and wish you a very healthy life sir. Namaste
go for it!
Awesome
:)
Will you be going to the lane of totality? I will be! I am worried about traffic though, on account of nearly half of the state traveling to see it. Hopefully it would be too bad.
:)
great sir
:)
I am from europe and i saw a partial eclipse(somewhere like 90%) 5am in the morning . It was awesome
:)
I had planned on seeing this over 5 years ago. I'm missing the first day of college just to see it.
are you in the lane of totality?
A good way to explain with a demonstration; use a video camera to present the eye, a hanging ping pong ball to present the moon, and a hanging beach ball to present the sun ( and having fun changing the distances)
:)
@Walter Lewin, Dear Sir, its really nice of you to come up with these videos to share your knowledge. Sir i have a request: Can you please make a video on Hermholtz coils and Mu Metal Can's Concepts and calculations?
use google
Can you calculate an approximate PROBABILITY of a total solar eclipse occurring by chance throughout the universe in other solar systems? Assuming there are approx 10 to the 21 stars in the universe and assume that each star has an average of 10 planets orbiting each star and that each planet has at least 1 moon orbiting it. One must calculate that that moon must be in the perfect position to properly occlude the star to perfectly occlude the disk of the star. (Our situation is that the moon is 400 times smaller than sun but that its also exactly 400 times the distance to perfectly cover the sun.) That kind of coincidental geometry is amazing
Well, even the zone of totality can be as long as 1650 kilometers if it's edge touches the pole.
I am very fortunate I live in Nashville Tennessee which is in the path of totality
ENJOY!!!!!!! I hope there are no clouds during your 2 minutes of totality
yeppp sir thank u sir for uploading these videos and one thing sir ur the second generation Einstein Sir
:)
I got about about 6800 km (as the diameter of the region). I did have to look up the distance between the earth and the sun and the radius of the sun though.
Also, unfortunately, the next total solar eclipse in my country will be in 2034.
Looking forward to more problems in the future!
Professor, thanks for the video! One thing that I have not been able to explain adequately is the west-to-east motion of the umbra.
Experts point out that the moon orbits west-to-east and that the tangential velocity of the moon in its orbit is greater than the tangential velocity of an observer at the equator. This faster orbital velocity allows the moon to move easterly faster than the surface of the earth, causing the umbra to travel west-to-east.
However, the ANGULAR velocity of the spinning earth is some 27x faster than the angular velocity of the moon. This, I believe, explains why we see the moon rising in the east and setting in the west. Multi-exposure images of the full eclipse clearly show the moon and the sun drifting westward from the perspective of an earth observer.
Clearly, the umbra moved west-to-east, but I've not been able to explain how this is possible given the much greater angular velocity of the earth vs. the moon. Would you care to comment?
On Aug 21 during the solar eclipse, the Moon moved onto the sun in the direction from West to East. That's what the Moon does EVERY day, moving 15 degrees per day relative to the sun. Thus the West coast of the US would see the eclipse first (due to parallax).
Daym!
You're an insanely intelligent person! Really hope to meet you in person someday! Oughta be the best day of my life, After I figure out where you live. 😅
:)
Immona stalk you everywhere because just the info of you living in Cambridge doesn't cut it. 😂
😊
6,687km - but i had to look up the radius of the earth!!
watch my solution on Sat
Your an absolute legend :D
:)
:D
To hec with his physics....I want his shirt!
Using:
Sun-earth distance 152e6 km
Moon-earth distance 356e3 km
Radius of the sun 695700 km
Radius of the moon 1737 km
Radius of the earth 6371 km
The "diameter" of the penumbra (measured over the curvature of the earth) is about 7054 km
Someone asked, if the eclipse could be rescheduled to weekend, because children go to school on monday :-) There has been one total eclipse in my lifetime right where I live, and it was cloudy. So it just got dark for a while, which was interesting, but quite a bit less than I hoped to see.
I am sure it can be rescheduled.
I look through Glass, at the turtles under the water, They look really close, But when I look on top of water, at the Turtles head peaking out he is, about 3-4 times the perceived distance. How can I be convinced that a similar refraction of light is not happening during the eclipse?
Of course there is refraction. The refraction index of 1 atm air 1.00029
Calculated (using similar triangles)and sun furthest distance, moon closest as per your figures, and worked out partial will extend 3375 km either side of centre of totality. As for total eclipses until 2030, I see from the web 8 such occurances, the last one on 25 nov 2030 will be within 1000 km of where I live however cannot drive there as is is over the sea!! Will definitely be in the partial zone. Can you guess where I live?...clue - it is an island south of equator and also a continent!
Good morning professor. In 2 August 2027 a solar eclipse will be visible in Italy, only in the little island of Lampedusa will be total. Time to book my holiday ;)
>> 2 August 2027>>>
yes I know - it will be a total solar eclipse
Dear Prof., The diameter of the penumbra is 6750 km.
This post is more precious to me just because you are looking at it. ❤️Love Physics❤️Love You Prof.❤️
And, on the upcoming Total Solar Eclipse question, there are 8 Occurrences i believe before 2030. I live in India and unfortunately, am not within 1000km of path of totality for any of those. But the one in 2027 is the closest, though the totality falls on Indian Ocean, I would be in Penumbra region.
❤️
Bonus question for anyone who is truly enthusiastic:
Calculate approximately the time that the partial eclipse lasts, and the total eclipse lasts.
(Answers: partial ~ 120 minutes, total ~ 120 seconds)
Hint: Think about the relative motion of a point on the Earth's surface to the moon, and hence its shadow.
:)
Hello sir... I am a new viewer of your lectures... I have just recently started following your lectures and have seen 7 to 8 lectures... I have a question that we have a phenomenon of diamond ring effect... I am a bit confused because moon blocks the sun almond completely just an outer ring of sunlight is formed as you said it annular eclipse... Is there any use of diffraction of sun rays in diamond ring effect... As for the answer of your 1st question i got 3480 km on the surface of will be the diameter of penumbra means penumbra will be extended in an area of 10927.2 squared km...
diamond ring - use google
my results are 1.2816*10^6 km or 1,2816 thousands km of penumbra. I would like to know how to calculate umbra too but i dont have any idea. I considered the distance of the moon from the sun at 152*10^6 km and the distance between the moon and earth at 356*10^3 km
incorrect
NASA has come out with guidelines of using a welding lens...
READ the details
correction: i have made an unwarranted assumption on the distance to the moon, i change the answer Lpenumbra=6691km, taking the radius of the earth to be 6371km
I would like it if you provided a pdf with the problem written down because sometimes I don't have time to watch a 20 minute video Thanks
😎
Your profile picture, I thought there was a hair strand on my screen lol
Professor, why does the umbra move from west to east, rather than the normal east-to-west 'motion'? Especially since the 'center' of the eclipse moves way faster than the point on the earth below the moon?
bcoz the moon relative to the sun moves from West to East. Tomorrow at this time the moon will be about 15 degrees east of the sun.
Thank you!
sir if everything in this universe is moving. I mean if every motion is relative and we can not measure the real speed so how the speed of light is same in every frame of reference
That''s the power of Einstein's SR. You are used to think Newtonian. SR is NOT Newtonian.
With max Sun and min Moon distances the intersection point is 378 000 km from Moon, so the penumbra cone size at earth distance is 6742 km. The size on Earth's surface depends on how it hits the Earth, but if it's right in the center it's 7094 km. There wil be total eclipses in 2019, 20, 21, 24, 26, 27, 28 and 30. None of them are within 1000 km from me.
superrrrr 😊
I googled "angular eclipse" and I got some remarkable pictures of some JavaScript code XD
:)
hi sir ..electron is revolving arround the nucleus why it is not radiate energy
coulomb's law is not valid for moving chagre why?
That's why we need QM
I witnessed the total eclipse 1999. Traveled to Luxembourg. It was amazing. The rings and the corona were sparkling. But the blue light the world was shrouded in was ...mind boggling. The whole world went still. All birds went quiet.
I'm most amazed by the astronomically small chance that from earth the sun and the moon appear nearly equal in size. I mean the variation in the universe is seemingly endless but somehow we earthlings have a sun AND a moon that appear the size of a pea at arms length.... Now I don't believe in coincidences like that..... at least I question that coincidence. But so far I found no theory to explain this correlation to have a scientific reason. So it seems to be strictly coincidental.
Is it?
>>>>I'm most amazed by the astronomically small chance that from earth the sun and the moon appear nearly equal in size.>>>
Yes that is remarkable. If the lunar orbits were just a little larger total solar eclipses would NEVER be possible. There is ABSOLUTELY No scientific reason for this. *ABSOLUTELY NOT!*
Hi Prof. Lewin, I have a question to ask.
Is time a vector quantity or scalar quantity?
There is a concept called 'Arrow of time', so can I interpret time as a vector quantity?
Yours sincerely
Shaoren Ong
"In physics101, scalar quantities are defined to be ones which have magnitude only, and no direction, where "direction" in this context means a direction in *three dimensional* space. Time clearly has no such direction. ... But Time can be a vector, but what it means depends on the context."
Lectures by Walter Lewin. They will make you ♥ Physics. Thank you, Sir
can we observe those stars behind the sun during total Eclipse ? I hope I can see it once in my life
yes you can see bright planets and bright stars. However, your 360 degree horizon is NOT dark! You should be able to tell why.
Dr Lewin, he is referring to the sun behind the eclipse, he wants to see the stars behind through gravitational lensing method
Lectures by Walter Lewin. They will make you ♥ Physics. is that bcuz the moon block the light just a small surface area on Earth so that we still able to see the light outside the moon's shadow?
Sir please make a video in the topic 22 degree halo.I watched this thing only one time in 2016.Whats reason behind it???????????
use google or watch my 8.02 lecture where I may cover the 22 degree halo.
Hi! because of similar triangles and proportionality and after doing some math I arrived : penumbra= 6750 Km with the distance to the sun of 152 .10^3 km and distance to the moon of 356.10^3km. Finally the distance from the earth to the intersección lines is 733515 km. . Thanks for your attention.
correct
Hellow, Prof. lewin
I am little confused with the concept of relative motion.
I don't understand how can we say if frame of reference is inertial or not... should I look if a force is being applied on me( if I the observer ofcourse)
My another confusion is related to magnetism...
Suppose I have two non conducting balls having positive charges on them, I go in a cab and put them in my lap... the cab starts moving now, for me balls are applying only the repulsive electrostatic force on each other... but for my friend on the ground... the charges are moving with the cab and there should be a magnetic force of attraction between them???
1. use google
2. yes B-fields are the result of Lorentz contraction. Your friend will see B-fields, you will not.
+Lectures by Walter Lewin. They will make you ♥ Physics. Thanks for the reply.
But If the velocity is heigh enough... My friend should them coming together and I would see them going apart...
Can both be true?
you should take a course in SR
+Lectures by Walter Lewin. They will make you ♥ Physics. I have just passed my 12th grade, can I do it?
Any good youtube channel that you know of? :)
SR is not simple. Search the web! There nay be talks for beginners.
Lewin sir... The level you teach here. Is it sufficient to get into MIT??
How to get into MIT?
ua-cam.com/video/hRp3ND-fBNw/v-deo.html
Using distance of the Moon at the minimum and distance of the sun at the maximum, and the part of the Earth is flat, the radius of the penumbra is 1745km. If take the curve of the Earth into account, about 1767km. Although when I try to measure the eclipse zone on google map, the radius seems to be closer to 4000km. Of course the fact that the centres of Earth, Moon and the Sun are not perfectly aligned is not taken into account here, so when the shadow is off center it will be larger due to the curveness of the Earth. Can't think of any other factors for the moment.
incorrect
Just to make sure, we are given the distance from the center of the sun to the surface of Earth, the distance from the surface of the moon to the surface of Earth, the radius of the sun, and the radius of the moon? And you want us to find the area of the shadow on the Earth, or just the radius of the shadow?
:)
sir ,why does mercury does not go in elliptical path.......does volcan really exist
question unclear
gravitational time dilation explains all of the anomalous precession of Mercury's perihelion without need for additional planets.
Diameter of penumbra of the moon on earth is 6751.00128 km considering max distance from the sun 152 million km and min distance of the moon 356000 km. Additionally, the distance from Earth to the intersection angle between the sun and the moon is 736979.556 km. There will 8 total eclipses till 2030.
The total solar eclipses dates are on 2 July 2019, 14 Dec 2020, 4 Dec 2021, 8 April 2024, 12 August 2026, 2 August 2027 ( this will be in the country where I live), 22 July 2028, 25 November 2030.
:)
sir If I give a infinite force to a body of infinite mass and infinite friction will the body move??
THINKKKKKKKKKK
I first tried using similar triangles with the chord of the penumbra, the diameter of the moon, and the diameter of the sun. But using values for the current eclipse, I got too small a value (7.29e3 km instead of nearly 10e3 km). (For reference, I got 7.10e3 using the 152e6 & 356e3 values.) Then I realized my triangles were wrong: you can't see the south pole of the sun from the north pole of the moon!
This will require a different strategy...
The sun, moon and Earth are spheres. If they line up *precisely* (for a given distance to the sun and the moon) the diameter of the Moon's penumbra on Earth is determined regardless of where the totality is. Notice the word *"precisely"* .
I was assuming the centers are co-linear. But I was drawing lines from the bottom of the Sun to Earth, touching the top of the Moon, which meant the line passed through the sun and the moon. And I used the distance from the Earth to the Moon as the distance from the shadow to the Moon, figuring it was "close enough".
I tried again, starting from thinking of the penumbra as the near intersection of the Earth and the double cone tangent to the Sun and the Moon (with apex between them.) Looking over the intermediate values, the dominant difference from before was the approximation for the distance between the shadow and the moon. Overall, my answer shrank by around 1%. I used a couple of inverse sines, which doesn't feel like the kind of solution you are hinting is out there. And, of course, I expected a bigger answer, not a smaller one!
I'll just have to wait for your solution on Saturday :) Thank you for the problem!
It's Ok - Your value is too high but not ridiculously high - watch my solution on Saturday
6946Km..........using .559 deg for close moon, radius of moon, and distances of sun and moon to Earth.
:)
i always will to have teacher like you.... because i think actually physics is funny, interesting, shocking, true....
:)
Where's problem #12?
my dog ate it
d_{penumbra}=3328km
using d_{sun}=152*10^6km
and d_{moon}=356*10^3km
:)
I found 6 of them, sir 2019,2020,2021,2023,2024,2026. I live in India and Indonesian eclipse happens to be the shortest distance from my place(2023 eclipse)
there are 8
Lectures by Walter Lewin. They will make you ♥ Physics. Sir, do you have any videos on circuits, resistors, and stuff?
sir we should consider earth to be approximately flat?
that's good enough
+Lectures by Walter Lewin. They will make you ♥ Physics.
thank you for your answer! and sorry for the delayed answer. You are always an inspiration
Finding the Shadows of an eclipse.
I lined up all objects so that their centers all fell on a common line, and then used all of the measurements given in the problem to be the coordinates used to find the size of the two shadows. I used a Cartesian Plain using the y-axis to define the line that the centers of the Earth, Moon and Sun are on. So that the points I used as:
Sun (-696,000 , 152,000,000) & (696,000 , 152,000,000)
Moon (-1737 , 356,000) & (1737 , 356,000)
Earth’s surface at the center of the shadows (0,0)
Penumbra
Using the point slope formula, and choosing the right side of the Sun and the Left side of the moon, I get the points (696,000 , 152,000,000) & (-1737 , 356,000)
Slope = (152000000 - 356000)/(696000, -(-1737)) = 217.3369049
y-y1=m(x-x1) for (696,000 , 152,000,000)
y-152000000=217.3369049(x-696000)
y-152000000=217.3369049x-151266485.8
y=217.3369049x-151266485.8+152000000
y=217.3369049x+733518.2038
And at the Earths surface, Where y=0
0=217.3369049x+733518.2038
-733518.2038=217.3369049x
x=3375.008954 km. This is the radius of the Penumbra
Umbra
Choosing the points that are on the right side of the Sun and Moon
Points - (696,000 , 152,000,000) & (1737 , 356,000)
Slope = (152,000,000-356,000)/( 696,000-1737) = 218.4244299
y-y1=m(x-x1) for (696,000 , 152,000,000)
y-152,000,000=218.4244299(x-696,000)
y-152,000,000=218.4244299x-152030363
y-152,000,000=218.4244299x-152030363+152000000
y=218.4244299x-30363
Again finding the size of the shadow at point y=0
0=218.4244299x-30363
30363=218.4244299x
x=139.002812 This is the radius of the Umbra
Sig Figs only 3, so
Radius of Umbra is 139 km
Radius of Penumbra is 3380 km
There are seven eclipses from now until 2030 and the closest for me will be around San Antonio, Texas in 2024.
:)
Where I'm at it is 99.1% of totality, so I will drive to the umbra lane.
super make sure you know where to go. The lane of totality is very narrow.
I got roughly 6800 Km
:)
There will be a total solar eclipse in 2027 and will be partially visible from greece. I hope I'll be at southern spain by that time to watch it!
:)
Penumbra=2*((Dem/Dsm)*(Rs+Rm)+Rm)=6750km
Dem = distance between Earth and Moon = 356,000km
Dsm = distance between sun and moon = 152*10^6 - 356,000 km
Rs = radius of Sun = 696,000 km
Rm = radius of moon = 1737 km
:)
about 6750 km absolute distance. Earth surface distance ( circumferential distance ) about 7461 km. Funny i have the same answer as Aaron Hangcock below. My sheet has several tan and atan functions in there.... Aaron has obviously the more elegant way of calculating this :-)
right ... tan() and atan() cancel out ... i dont need the actual angle
:) I was only asking for the diameter of the penumbra (cross-section of Earth). Perhaps that was not clear. That's what you have: 6750
question : guess this is the newtonian answer. distances are maybe such that relativity comes into play ....is there a non newtonian answer ? Is it different ?
Sr does NO at all cone into play here. The speed of these objects are a minute fraction of c. The speed of the Earth in orbit is 30 km/sec which is 0.0001 c
My solution for minimal moon distance and maximal sun distance is 6750 km.
:)
Does that mean that i am wrong or right ? Because if i am wrong i want to try again.
BTW: love your lectures.
I never comment on right or wrong. On Sat Aug 26 I will post my solution.
Oh ok. Could you please put the right answer or a timestamp in the description so that i could try it again without knowing the right derivation?
will somebody please donate a whiteboard for Walter's Home ?.........if he has a place to put it mind you
I want NO black board and NO white board. At MIT I had 9 huge black boards and *3 cameras.* 2 of these cameras could move. Thus i would rarely every block the vision of any camera as I was writing on the black board.
My lecture hall could hold 600 students.
At home my auditorium is about 2x3 m. No room for even 1 person to watch me. If I worked on a blackboard or white board with 1 fixed camera I would block my own writing for 95% of the time.
I got you.......I just figured you were so used to nice set-up's at MIT that you just never thought to get a small one for your home office/studio......I don't think you can do your "dot thing" on one of those anyway !
I will do my dotted lines once on a small blackboard just to surprise my viewers. In any case dotted lines are not possible on YOUR white boards.
we will invent a white board pen that does dotted lines, just for you!
sir what do you think how can u answer a child the question what is physics??????????????
show her the blue sky, the clouds, the sun, the Moon, the planets, your glasses, refrigerator, radio, TV, iPhone, internet, cars, airplanes, rockets, trains, street cars, calculators, telephones, day and night, sun rise, sun set . . . . . tell her that all that is Physics. Our whole world is Physics. Without Physics no life could exist.
Lectures by Walter Lewin. They will make you ♥ Physics. the best answer I ever got.😂
Is it 6750.01 km ????
superrrrr 😊
tan a = (Rs + Rm) / (Ds - Dm) = ( 696000 + 1737 ) / ( 152000000 - 356000 )
= 697737 / 151644000 = 4.601/1000
Re = Rm + Dm * tan a = 1737 + 356 *4.601 = 3375km
diameter is (supposed to be) 6750km
R are radius. D are distances
m is Moon, s is Sun, e is eclypse.
Earth is assumed flat (it's not), diameter of Earth is considered too small.
With Earth diameter :
Re = Rm + (Dm - Rt) * tan a = 1737 + (356-6.366) *4.601 = 3345.6km. Diameter = 6691km
Rt radius of Earth 6366km
superrrrr 😊
superrrrr 😊
I got 6750 km approx., sir.
😊
i'm going to go with 6676 km (final answer), which is slightly less than two moon diameters. this is the diameter of the shadow that arrives at the plane that is tangent to the earth at the sun-moon-earth line. i used walter's distances for sun-earth and moon-earth (152000*10^3 km and 356*10^3 km.) to arrive at this 6676 number i translated a moon back to the sun along one of the tangent lines to where it was tangent with the sun. then i translated this back to the point where the moon center touched the earth-tangent at the sun-earth line. this created a simplifying geometry with two congruent triangles where it was easy to calculate the angles.
part 2: i count 7 total eclipses between tuesday and 2030. only one of which i will not have to update my passport for: 4/8/24
extra credit: what is the diameter of the shadow that arrives at the plane that intersects the earth center and is perpendicular to the sun-moon-earth line? I get 6734 km. Which is only 0.8% larger. This is the maximum width of the penumbra shadow. Your mileage may vary, though, because the entire penumbra shadow may not be intercepted by the Earth.
extra extra credit: watch this animation which shows the penumbra sweeping across the earth: ua-cam.com/video/dJyiWQcVmiM/v-deo.html
I will also be within 1000 kn of the totality on April 8, 2024
excellent. let's organize a future eclipse party! let's bring lots of croissants!
Hi professor, is the answer 6750km diameter?
many people have 6750 which is not correct. This puzzles me. What have you assumed about the distance to the sun and the distance to the Moon? I will post my solution on Saturday.
Lectures by Walter Lewin. They will make you ♥ Physics. I took 152 million to sun and 336000 to moon.
the closest distance to the moon is 356 thousand km.
Please send me how you arrived at the 6750 (I want to see your calculations). If I use your 336 thousand km to the Moon (with is not possible), I do not find 6750 km. Since several people sent me 6750 km I want to know how they got that number.
I also find 6750 with Similar Triangles.
If we call O the point where the 2 lines defining the partial eclips cross each other and x the distance from the Earth to O.
Let's call R_S, R_M the radius of Sun and Moon ; d_S, d_M the distance from Earth to Sun respectively Moon.
We want to know 2*S (S for Shadow)
Similar Triangles gives us :
R_S/(d_S-x) = R_M/(x-d_M) = S /x.
The fisrt equality gives us
x= (R_S*d_M+R_M*d_S)/(R_S+R_M)
and also
x-d_M=R_M/(R_S+R_M)*(d_S-d_M)
The second equalty gives us :
S=(R_M*x)/(x-d_M)
S= (R_S*d_M+R_M*d_S)/(d_S-d_M)
S= 3375
2S = 6750
What's wrong ?
Rgds.
My answer is about 7100 km
:)
I won't see anything here in Croatia :I
:)
My answer is 6750 km.
superrrrr 😊
I LOVE PHYSICS! AND MY ANSWER IS 6750.012881 KM. PENUMBRA LENGTH ON EARTH IS 6750 kms. And that is 50% of Earth Dia!... :) ... so... very sad for people in antartica and arctic..:)
superrrrrr 😊
:D
Hello Professor, my answer is 6754Km
quite a few people have 6754. I wonder why. What assumptions did you make about the distance to the Sun and to the Moon? I took 152 million km for the distance to the sun and 356x10*3 km for the distance to the moon. I will post my solution this Saturday.
First