Hi I saw your vids before my exam today and I know you may not have that many views, but my friends and i just found your vids and youre honestly one of the best if not the best that we've seen. i hope you know you're appreciated and behalf of us all thank you for all the effort you put into make these, it helped us a lot 😄👌
This is the best physics content I’ve seen on TH-cam. Other videos either have good information at the cost of being mind numbing or super simple examples. This video was very entertaining and actually had useful applications of equations.
lol mine is even worst, he told me to present and TEACH this topic BY MYSELF IN The Class eventhough I know nothing and doesn't have any clue at all with this topic. After reading some pages in my textbook, I think I need youtube for serious help...... What is even worse is that i learn physics in another language in high school and now I learn it in english in higher education and I'm still not confortable using it :/ alot of terms(in english) are still a mess in my head .
Fun fact: 7:42 Under Newtonian physics, this is without a doubt true. But in modern physics, we know that gravity propagates at the speed of light and therefore any object that is moving faster than the speed of light relative to us through the expansion of the universe is untraceable by gravity.
I just discovered flyping physics today OMG y didn't i know u guys all along . I've always thought other famous tutors which i wont mention here were the only good ones ,it turns out u guys are amoung the best .I mean i can literally Understand and remember coz u show real life demonstrations and also are funny😍😆
In the satellite problem, I don't understand why you set Gmsme/r^2 = vc^2/r. I understand why you would use vc^2/r. Does mass in F=ma equal to the gravitational equation?
+WerewolfXD13 1) As far as I can tell "Gmsme/r^2 = vc^2/r" never appears in this problem. 2) what is "c" in "vc^2/r" 3) "Does mass in F=ma equal to the gravitational equation?" Most likely the answer is "no", however, I have no idea what you mean by the "gravitational equation".
Mg is not entirely equivalent to GMm/r^2 as you stand on a spinning planet for there is also a centripetal force thus making the gravitational force the addition of centripetal force and mg according to parallelogram law
If a planet is close enough to a sphere that you wouldn't immediately think it is anything other than a sphere from a photo of it, you will find that the centrifugal effect you describe is insignificant. Take our own planet for instance. The lowest value of g we have on Earth's surface is 9.76 N/kg, as the apparent gravitational field strength. This occurs on a mountain in Ecuador, which as the country's name implies, is very close to the equator. The true gravity at this location is 9.79 N/kg, as you would calculate with G*M/r^2. The reduction in gravity of the centrifugal effect turns out to be 0.0334 N/kg. That's 0.3% difference between true gravity and effective apparent gravity. Any planet that is close enough to a sphere that you wouldn't notice the difference, will have a centrifugal effect that is a very small fraction of the true gravity. If you were on an oblong world like Haumea, whose ellipsoidal shape is immediately obvious to us, the difference true gravity and apparent gravity would be a lot more significant.
@@dougefresh97 It is a factually accurate statement to say that jet fuel can't melt steel beams. However, jet fuel CAN bring steel beams to a temperature that is hot enough that their Young's modulus will decrease, causing them to bend like a wet noodle, and be useless as a beam, despite still being solid.
+lanre Mabogunje Thanks. I really wish I had the time to tutor, however, I think my spending time making these videos is a way I can reach a wider audience and have a larger positive effect on the world.
sooo gravitational potential energy is always less than or equal to 0 because of infinity? sorry i guessed i missed something but it doesnt seem like this video answered that question very well.
It is arbitrary where we define gravitational potential energy to equal zero. This is true for electric potential energy as well. There is a convention to define it to equal zero, when it is infinitely far away from the objects with which it interacts. It makes the math more convenient by doing so. If you generalize the potential energy so that you define it to equal zero somewhere else, you would add a +C term to it. Instead of using GPE = -G*M*m/r, the formula would become GPE = - G*M*m/r + C, where C is an arbitrary constant. Due to the fact that most applications for which you'll ever need this formula will end up being one value of GPE subtracted from another, the +C constant will cancel, as if you didn't even need it in the first place. This is why we define GPE to equal zero between masses M and m, when r=infinity. You see a +C term in Calculus all the time when you integrate, and give your answer for the indefinite integral (also known as anti-derivative). When finding the definite integral between two bounds of integration, the +C terms end up cancelling out of the equation. However, there are applications of integration where you do care to keep the +C term around, and will eventually assign a value other than zero to it. One application is Euler's beam theory, where you use integration to solve for the equation of the elastic curve for a deflected beam under the load.
The reason why GPE is consistently negative when using this formula, is that an object will always receive positive work from gravity due to the object in question, as it falls from infinitely far away. Bring the objects closer together, and the potential energy decreases. The speed would increase if you didn't have another force involved that would stop them from picking up speed. Because positive work was done by gravity, GPE has to decrease. Decreasing from zero means that it will be negative.
Question. Has anyone thought of replacing "F" for applied Force with "E" for applied Energy? Eg vs Fg? Would that work? If we were to know how strong or weak the “gravity” is, then we would know how much energy is being applied from within a planet or black hole. Cause, without energy there is no gravity, magnetism, fields, waves, motion, etc. After all, Force does not exist "physically" in the same way that an object with mass does. "Force", as we know it, turns out to be nothing more than an expression to express an idea like one would use the word "Love" to express one's feelings. Meaning, Force or Net Force does not push, pull or work for that matter.
+Ammar Ali I am making some guesses, however, here is what I think you are asking. "In the Orbiting Satellite Example problem, why don't you include the force of gravity from the moon on the satellite?" Assuming that is what you are asking. Do me a favor, look up the distance between the earth and the moon, and then the altitude of geo-stationary orbit and then calculate the ratio between the force of gravity caused by the Earth on the satellite, versus the force of gravity caused by the moon on the satellite. Let me know what you get. (Hint: The Earth's Force of Gravity with be _astronomically_ larger than the Moon's Force of Gravity.)
@@FlippingPhysics Great explanation. To add onto this point, another reason the moon's gravity matters a lot less than the Earth's gravity for satellites in Earth orbit, is that Earth also accelerates toward the moon due to the moon's gravity on Earth. A satellite also has to follow this reference frame, since earth is not stationary. Most of the moon's gravity on the satellite will be "used up" in causing the satellite's path center to follow the reference frame.
ur teacher speaks all the concepts in 10 mins and doesn't explain anything, once the students get their tests the teacher sees they are very weak in these concepts, ur teacher is just flexing just to show off his knowledge about this
whom else cramming for this afternoon. i've hit new lows my friends
RCHSDrama yeah unfortunately that's me
2 years late but I am cramming as well.
@@juiceontheloose123 How did you do?
@@bharatkrishnamurthy2652 i got a 4 on the ap physics 1 exam!
@@juiceontheloose123 nice!
The way you explain things (and your 3 students!) makes it click. Thank you so much for your help!!
You are welcome!
Hi I saw your vids before my exam today and I know you may not have that many views, but my friends and i just found your vids and youre honestly one of the best if not the best that we've seen. i hope you know you're appreciated and behalf of us all thank you for all the effort you put into make these, it helped us a lot 😄👌
This is the best physics content I’ve seen on TH-cam. Other videos either have good information at the cost of being mind numbing or super simple examples. This video was very entertaining and actually had useful applications of equations.
This man needs more recognition!!! Life saver! Thank you :)
AP Physics 1 test is tomorrow, and my teacher didn’t teach this at all. Thanks to this video I have a chance of getting it right
I would suggest you look over some of my more detailed videos on this topic. www.flippingphysics.com/algebra.html#ug
anyone else binging these videos for AP exam next week? Super helpful
Saweet. Perhaps you could "like" while you binge?
Good luck tomorrow!
2:37, Everybody Brought Mass (of the object) to the Party! [Insert relaxed funky synth music] Everybody Brought Mass! (Mass! Mass! (fade echo))
The seventh in my set of AP Physics 1 review videos. Enjoy! #PhysicsED #flipclass #APPhysics1
Flipping Physics brightened my day that a video was posted!
😊
I NEEDED THIS SO BAD. MY TEACHER DIDN'T EXPLAIN THIS AS GOOD. COME THRU TO MY SCHOOL
BRING BO TOO
Sorry. Bo is too busy staring at the wall to come to your school.
lol mine is even worst, he told me to present and TEACH this topic BY MYSELF IN The Class eventhough I know nothing and doesn't have any clue at all with this topic.
After reading some pages in my textbook, I think I need youtube for serious help......
What is even worse is that i learn physics in another language in high school and now I learn it in english in higher education and I'm still not confortable using it :/ alot of terms(in english) are still a mess in my head .
i have my exam in an hour; this is an amazing review, you're brilliant
this was LIFESAVING, 2 minutes in and already half my questions were answered, it's crazy how helpful you are
so glad to help!
Binge watching this the day before my exam - very helpful
Glad I can help you learn and best of luck to you on your exam tomorrow!
quick review for tomorrow! thank you so much!
You are welcome. Good luck tomorrow!!
Good thing I found these, lol. Had most of the math questions down due to repetition but concept questions had me messed up. Thanks!
You are welcome. Good luck today!
Am I the only one that expected "Everybody Brought Mass To the Party" when Mr. P cancelled out the masses?
l
thank you so much! Great review and a lot of effort put into the video!!!!
+Alice Balashova You are welcome. Yep. Lots of effort. Hopefully lots of learning too.
Getting ready for next week....
I like your video sir,I enjoy learning with you too,love from Malaysia!
Thanks and welcome
Fun fact: 7:42 Under Newtonian physics, this is without a doubt true. But in modern physics, we know that gravity propagates at the speed of light and therefore any object that is moving faster than the speed of light relative to us through the expansion of the universe is untraceable by gravity.
I just discovered flyping physics today OMG y didn't i know u guys all along . I've always thought other famous tutors which i wont mention here were the only good ones ,it turns out u guys are amoung the best .I mean i can literally Understand and remember coz u show real life demonstrations and also are funny😍😆
Thank you for your lovely words!
Thanks, These Vids are awesome and you clearly put a lot of effort into these videos. Will you make Physics 2 videos that I can use next year aswell?
Not in my current plan. Still have to finish AP Physics 1 videos.
this dude is slowing saving my physics grade
Hopefully it's not that slow...
Awesome stuff.
Thanks.
Thanks for the quick review
Hey, great video as always, but i realized there were some topics not covered. Could you possibly explain density and how its used in PE?
Thank you so much sir! I owe a lot to you!! And you're the BEST!!! And why is Bo not taking any notes?
There is always 1 student whom I cannot convince to take notes.
Haha!
In the satellite problem, I don't understand why you set Gmsme/r^2 = vc^2/r.
I understand why you would use vc^2/r. Does mass in F=ma equal to the gravitational equation?
+WerewolfXD13
1) As far as I can tell "Gmsme/r^2 = vc^2/r" never appears in this problem.
2) what is "c" in "vc^2/r"
3) "Does mass in F=ma equal to the gravitational equation?" Most likely the answer is "no", however, I have no idea what you mean by the "gravitational equation".
Thank you so much Sir! Your videos are very helpful
You are welcome!
What's the difference between Fg and Ug. Besides one has r on the denominator and the other has r squared.
Fg is the force being applied to the objects and Ug is the potential energy that the objects have.
Hey love you videos but with the equation gm1m2/infinity has to equal 0 what happens if its infinity/infinity wouldn't that equal 1
Mg is not entirely equivalent to GMm/r^2 as you stand on a spinning planet for there is also a centripetal force thus making the gravitational force the addition of centripetal force and mg according to parallelogram law
If a planet is close enough to a sphere that you wouldn't immediately think it is anything other than a sphere from a photo of it, you will find that the centrifugal effect you describe is insignificant. Take our own planet for instance. The lowest value of g we have on Earth's surface is 9.76 N/kg, as the apparent gravitational field strength. This occurs on a mountain in Ecuador, which as the country's name implies, is very close to the equator. The true gravity at this location is 9.79 N/kg, as you would calculate with G*M/r^2. The reduction in gravity of the centrifugal effect turns out to be 0.0334 N/kg. That's 0.3% difference between true gravity and effective apparent gravity.
Any planet that is close enough to a sphere that you wouldn't notice the difference, will have a centrifugal effect that is a very small fraction of the true gravity. If you were on an oblong world like Haumea, whose ellipsoidal shape is immediately obvious to us, the difference true gravity and apparent gravity would be a lot more significant.
Why do the 3 guys in the deal look like the teacher O_o
4 of them are the same person =-=
Nicolas Lopez I'm joking too
SIMPLY AWESOME :D
Jet fuel can't melt steel beams
depends on the steel my man, mild steel melts at about 2700 degree f
@@dougefresh97 It is a factually accurate statement to say that jet fuel can't melt steel beams. However, jet fuel CAN bring steel beams to a temperature that is hot enough that their Young's modulus will decrease, causing them to bend like a wet noodle, and be useless as a beam, despite still being solid.
Do you provide tutoring??? If so what are your rates?
+lanre Mabogunje Sorry, nope. I can barely keep up with teaching in a classroom and making these videos.
Either way we definitely applause your efforts and will stay tuned.
+lanre Mabogunje Thanks. I really wish I had the time to tutor, however, I think my spending time making these videos is a way I can reach a wider audience and have a larger positive effect on the world.
Didnt explain KE on the gravitational field
sooo gravitational potential energy is always less than or equal to 0 because of infinity? sorry i guessed i missed something but it doesnt seem like this video answered that question very well.
It is a review video and therefore would not answer that question very well. This one will: www.flippingphysics.com/universal-law-gravitation.html
It is arbitrary where we define gravitational potential energy to equal zero. This is true for electric potential energy as well. There is a convention to define it to equal zero, when it is infinitely far away from the objects with which it interacts. It makes the math more convenient by doing so.
If you generalize the potential energy so that you define it to equal zero somewhere else, you would add a +C term to it. Instead of using GPE = -G*M*m/r, the formula would become GPE = - G*M*m/r + C, where C is an arbitrary constant. Due to the fact that most applications for which you'll ever need this formula will end up being one value of GPE subtracted from another, the +C constant will cancel, as if you didn't even need it in the first place. This is why we define GPE to equal zero between masses M and m, when r=infinity.
You see a +C term in Calculus all the time when you integrate, and give your answer for the indefinite integral (also known as anti-derivative). When finding the definite integral between two bounds of integration, the +C terms end up cancelling out of the equation. However, there are applications of integration where you do care to keep the +C term around, and will eventually assign a value other than zero to it. One application is Euler's beam theory, where you use integration to solve for the equation of the elastic curve for a deflected beam under the load.
The reason why GPE is consistently negative when using this formula, is that an object will always receive positive work from gravity due to the object in question, as it falls from infinitely far away. Bring the objects closer together, and the potential energy decreases. The speed would increase if you didn't have another force involved that would stop them from picking up speed. Because positive work was done by gravity, GPE has to decrease. Decreasing from zero means that it will be negative.
I have a question
What is the difference between
1) Gravity
2) gravitation
3) Gravitational force
Please help me 🙏🙏
They are all the same thing.
@@FlippingPhysics thank you sir ☺️☺️
Good luck tomorrow guys
Good luck today all
Fliiiiiipppppiiinnnnnggggg phyyyyyyysics 🎶
Thanks 🖤
Question. Has anyone thought of replacing
"F" for applied Force with "E" for applied Energy? Eg vs Fg?
Would that work? If we were to know how strong or weak the “gravity” is, then
we would know how much energy is being applied from within a planet or black
hole. Cause, without energy there is no gravity, magnetism, fields, waves,
motion, etc. After all, Force does not exist "physically" in the same
way that an object with mass does. "Force", as we know it, turns out
to be nothing more than an expression to express an idea like one would use the
word "Love" to express one's feelings. Meaning, Force or Net Force
does not push, pull or work for that matter.
AM I THE ONLY ONE HERE IN 2024?!?!
I LOVE U MAN
nah
sir what about the moon and the earth
R is not the radius
+Bouncing Rhino Speak the truth!
I mean that gravitation between the moon and earth
+Ammar Ali I am making some guesses, however, here is what I think you are asking. "In the Orbiting Satellite Example problem, why don't you include the force of gravity from the moon on the satellite?" Assuming that is what you are asking. Do me a favor, look up the distance between the earth and the moon, and then the altitude of geo-stationary orbit and then calculate the ratio between the force of gravity caused by the Earth on the satellite, versus the force of gravity caused by the moon on the satellite. Let me know what you get. (Hint: The Earth's Force of Gravity with be _astronomically_ larger than the Moon's Force of Gravity.)
@@FlippingPhysics Great explanation. To add onto this point, another reason the moon's gravity matters a lot less than the Earth's gravity for satellites in Earth orbit, is that Earth also accelerates toward the moon due to the moon's gravity on Earth. A satellite also has to follow this reference frame, since earth is not stationary. Most of the moon's gravity on the satellite will be "used up" in causing the satellite's path center to follow the reference frame.
hey y’all
You Crammin?
Dude nice socks!
+Bouncing Rhino Thanks, dyed them myself.
11-asp guys hi
hi
hi
when ur teacher is more cringe than the views
ur teacher speaks all the concepts in 10 mins and doesn't explain anything, once the students
get their tests the teacher sees they are very weak in these concepts, ur teacher is just
flexing just to show off his knowledge about this
Dude, this is a review video. If you want more detailed explanations of each of these concepts, see here: www.flippingphysics.com/algebra.html#ug