Sir i from india , and aspirating for an engineering exam JEE , i belong to general category and in our country their is a huge problem of reservation... This creates trouble for students from general category, thus i request you to post lectures regarding 11th class physics classes i entirely trust ur teachings sir., we just need ur blessings
(a) At a distance 'r' from center, let's say the speed is 'v' We obtain the expression for v by conservation of total energy. (K_i + U_i = K_f + U_f) (Initially, object starts from rest so K_i = 0) U_i = -GMm/R_2, U_f = -GMm/R_1 By further simplifying, we get v = (2GM(1/r - 1/R_2))^(1/2) Also, here v = -dr/dt Now we need to integrate, dr/(1/r - 1/R_2)^(1/2) from R_2 to R_1 on LHS and -(2GM)^1/2 dt from 0 to t on RHS. Put r = R_2*(sin u)^2 and integrate We get R_2(u - 1/2* sin(2u)) = -((2GM/R_2)^1/2)*t Finally, t = (π/2)*( (R_2)^3/2)/(2GM)^1/2 - ((R_2)^3/2)/(2GM)^1/2 * arcsin((R_1/R_2)^1/2) + (R_2)*(R_1/2GM)^1/2 * (1-(R_1/R_2))^1/2 (b) By putting all the given values, time taken from R_2 to R_1 = 0.396 hours.
Hello dear Mr. Walter Levin. Is it possible to use Kepler's second law to solve this problem? I was unable to integrate the equation I received. I hope you answer
Using a spreadsheet and 5 minute increments, I'm getting approximately 87 hours, but I really have to double check it and try to do the complex math to check if the spreadsheet has mistakes or is anywhere close.
Respected walter lewin sir❤ Answer for a) 2/3(2GM)^-1/2×[(R1)^3/2 - (R2)^3/2] b) 36.46 hours Please sir respond to my answers whether they are correct or not! Thank you sir
Today also a very well good morning to a great physicist of this century Mr Walter Lewin sir. please... You tell me from where I start the physics because I also want to become like you . I am in 11th grade and I am from India and I want to give something this world
Starting from conservation of energy between starting point R2 and a generic distance r: variation in kinetic energy = variation in potential energy, we get: (1/2)v^2 = k•(1/r-1/R2) k=MG v=dr/dt = - sqrt(2k(1/r-1/R2)) minus sign is because the direction of motion is towards decreasing distance r Rearranging: -dr/sqrt(2k(1/r-1/R2)) = dt Integral between R2 and R1 of left hand term equals the time delta. After looking up the primitive (sorry for that) of 1/sqrt(1/x-1/c), which is: (x-c)/sqrt(1/x-1/c)-c^1.5 • arctan(sqrt(c/x-1)) and taking the sign into account, we get: T=((1/sqrt(2MG))•((R2-R1)/sqrt(1/R1-1/R2)+R2^1.5 • arctan(sqrt(R2/R1-1)) Plugging in the values of R1 and R2, I get: T=313426 s = 87h 3m
Using the free fall theorem, we can find the time taken from R2 to R1 is a.) t = sqrt(2R²(R2-R1)/GM) And if R2=50R and R1=5R, using the equation above, we can find the exact time is b.) t = 7680 second or t = 2.1 hours I hope it's the correct answer Thanks for the problem, Sir. Can't wait for the explanation 😊
Hello sir ❤️🙏 I have a question to you, why inertia depends upon mass of an object rather than weight (amount of gravitational force act of an object), this is why because mass is a fixed quantity but inertia is not, for example: we put same mass having object at different planets here mass is same but inertia is different i.e due to gravitational force of planets. So sir can we say inertia depends upon weight (amount of gravitational force act of an object)?? Sir if you see so please clear my doubt please❤️🙏
Sir could you please explain what is the potential of an electron moving in a magnetic field and the interaction energy associated with it.I was asked this question in an interview
a) Time t is an integral from R2 to R1 and... where the gravitational force is getting larger each second the probe is getting closer to the surface of the Earth, due to: F = GMm/R² and... where the acceleration a equals F/m ➔ a = GM/R² and... where velocity v equals current velocity u plus the acceleration a × 1 second and... where the total distance traveled should be increased by (old) velocity u × 1 second plus ½ times the acceleration a times 1 second squared and... until the total distance traveled is equal to, or larger than R2 - R1 While (R2 - R1) > D D = D + (v * 1s) + (½ * a * 1s²) ➔ D = D + ((u + (a * 1s)) * 1s) + (½ * a * 1s²) ➔ D = D + ((u + GM/(D + R1)²) * 1s) + (½ * GM/(D + R1)² * 1s²) ➔ (final:) Iteration: D = D + (u + GM/(D + R1)²) + (½ * GM/(D + R1)²) Compute New: u = u + GM/(D + R1)² * 1s (u=u+a*1s) Time = Time + 1 second Print Time b) # Initialize variables u = 0 D = 0 T = 0 Re = 6400000 Me = 6 * 10**24 G = 6.67 * 10**(-11) # Calculate constants C = G * Me R2 = 50 * Re R1 = 5 * Re Rc = R2 - R1 # Run loop while Rc > D: D = D + (u + C/(D + R1)**2) + (0.5 * C/(D + R1)**2) u = u + C/(D + R1)**2 T = T + 1 # Print result print(T);(T/3600) 72340 20.094444444444445 Thus 20 hours.
Interesting computational approach. With 1 second steps I would expect a close approximation, but part b) is well off the analytic answer. I can't see an obvious error but it must be in there somewhere.
@@KeithandBridget The iteration should be v(t+dt) = v(t) + a(t)*dt and s(t+dt) = s(t) + v(t+dt) * dt. The formular v = a * t and s = 0.5 * a * t² is valid, if a = const. That is not the case.
"s" should be around a linear function of t to the power of 2/3. I used "r" and I found it also helpful to simplify things by changing the variable to x=r/6.4*10⁶. For this variable, I found x≈ 0.138(134896.4 -t)^(2/3).
(a) 1. Integrate acceleration, a, due to gravity. a = dv/dt = (dv/dr)·(dr/dt) where dr/dt = velocity = v => a = v·(dv/dr) = -GM/r² => v dv = -(GM/r²) dr ∫ v dv = -∫(GM/r²) dr {from R₂ to r} where R₂ = 50R and r is a point below R₂. => v²/2 = -GM∫dr/r² = -GM [-1/r] {from R₂ to r} => v² = 2GM(1/r - 1/R₂) v = dr/dt = -√(2GM)·√(1/r - 1/R₂) We take the negative root since motion is downwards.
After some additional steps: dt = -√[R₂/(2GM)]·√[r/(R₂ - r)] dr 2. Integrate one more time. Substitution: u = √[r/(R₂ - r)] => u² = r/(R₂ - r) => r = R₂ - R₂/(1 + u²) dr/du = 2R₂u/(1 + u²)² r = R₂ => u → ∞ , r = R₁ => u = √[R₁/(R₂ - R₁)] = =√[5R/(50R - 5R)] = 1/3 t = -√[R₂/(2GM)]·∫u·2R₂u/(1 + u²)² du {from ∞ to 1/3} t = -√[2R₂³/(GM)]∫u²/(1 + u²) du {from ∞ to 1/3} t = - √[2R₂³/(GM)]·{ ½[arctan(u) - u/(1+u²)] } {from ∞ to 1/3}
Answer (a): t = -√[R₂³/(2GM)]·[arctan(1/3) - 3/10 - π/2] …(1) (b) M = 6·10²⁴ kg R = 6400 km = 6.4·10⁶ m G = 6.67·10ᐨ¹¹ m³/(kg·s²) R₂ = 50R R₁ = 5R Answer (b): From equation (1) we get t ≈ 87 hours.
Hey very well good morning to you Walter Lenin sir I want to learn physics with you so I can give this world a new thought. My English is not so good I hope you will react . I am waiting
@@lecturesbywalterlewin.they9259 Hello again Professor Lewin. I was wondering if you can help me figure out why my method didn't work. Apparently the average force over space as calculated from the potential is different from the average force over time. This puzzles me greatly. Also, please do check my response to Keith in my solution. I found that using a approximate version of your method that makes the integral separable gives a similar answer to mine, but both are less than half of your answer, both in the 30s ... Also, there is an interesting relationship between my solution and yours... √(20/3)=√6.666 ... ≈2.58 ≈87/33.7
Sir plz 🙏🙏🙏 i am very inspire for you and very motivated sir But.... Physics is not understand for me sir.... My english are very weak but i am trying to do Best 👍💯to best Also i am preparing for neet In 2025......😢😢😢😢
My final solution. (a) The time "t" taken to fall from R2 to R1 can be obtained using the concepts of impulse and average force over the displacement R2-R1. Thus: t= Overall change in impulse/average force =(∆p)₂₁/⟨F⟩₂₁= m(v₁ - 0)/⟨F⟩₂₁= (v₁/GM)/⟨1/r²⟩₂₁ (r here is the position wrt Earth's CM) To calculate the average force over the distance travelled from R2 to R1, we need to consider the changing force of gravity along the path from R2 to R1. The force of gravity varies with distance, so we need to account for this variation in order to find the correct average force. We can determine the changing force of gravity at each point along the path from position R2 to R1 using the inverse square law. The force of gravity is inversely proportional to the square of the distance. We proceed as follows: 1. Calculate the work done by gravity along the path from R2 to R1. This work will be equal to the change in potential energy of the object as it falls. 2. Integrate the varying force of gravity with respect to distance over the path from R2 to R1. 3. Calculate the average force over the entire distance by dividing the total work done by the total distance travelled. 4. Use the total work done by gravity to find the final velocity of the object at position R1 using the work-energy principle. 5. Calculate the total transit time for the object to fall from R2 to R1 based on the distance travelled and the velocity at R1. By properly accounting for the varying force of gravity along the path of fall from radius R2 to R1 and integrating it with respect to distance, we can determine the accurate average force over the distance and calculate the total transit time for the fall. My answer for 1-5: Now I will change the variable from r wrt CM to r wrt the surface (position R), so that r becomes r+R. With limits from r2= R2 -R to r1=R1-R, the work done on mass m by Earth's gravity from R2 to R1 is: W₂₁= ₂₁∫F •dr = -(∆U)₂₁ = -m(∆V)₂₁ = -m[-GM/(r+R)]₂₁ Change in kinetic energy of mass m wrt Earth's CM = (∆K)₂₁= ∆ (½mv²)₂₁ = ½m(v₁² - v₂²) Here v₁ and v₂ are the final and initial velocities respectively. By the Work-Energy theorem, W₂₁ = (∆K)₂₁ Thus, as mass m is initially at rest v₂=0, and we have: W₂₁= GMm(1/(r₁+R) - 1/(r₂+R)) =GMm(1/R₁ - 1/R₂)=½mv₁² Using F=GMm/(r+R)² and letting R₂-R₁ = d₂₁ , we have: ₂₁∫F•dr = GMm(d₂₁/R₁R₂)= ½mv₁² ...(1) The average force over displacement d₂₁ is given by ⟨F⟩₂₁ = ₂₁∫F•dr / d₂₁ = -GMm/R₁R₂ Also, solving for v₁ in (1) and cancelling out m on both sides gives: v₁= √{2GM(d₂₁/R₁R₂)} Finally, we have t= √{(2GM(d₂₁/R₁R₂))}/GM/R₁R₂ =√{2d₂₁R₁R₂/GM} (b) is about (7680000/(4002)^.5)/3600~ 33.72 hours. With R1= 5R and R2=50R, R=64E5 [m], GM=(6.67*6)E(-11+24)=40.02E13, We have (in seconds): t=(64E5)√{2(64E5)(50-5)(5)(50)/40.02E13} =(64*8E5)√{(1E6)(9)(25)/40.02E12} =(64*8E5)(15)(1E(3-6))√{1/40.02} =768,000/√40.02 In hours this is: t=213⅓/√40.02~33.723 😉⏰🌚🚀⤵🍑🌎 Also, I dunno if anyone noticed but R1 is 32,000km away from Earth and this is about a Uranus equatorial radius away from the surface... ;)
@@KeithandBridget I checked out a few vids on UA-cam and it looks like my method is pretty standard, so it's puzzling me why it's apparently not working in this case. I tried it again, the same way you and WL did, but approximating the integral so that variables could be separated easily. I got a similar answer to my answer here... In the 30s. Approximately 36.29 hrs or 130,630.6 secs. If the variable r is changed to x= r/6.4*10⁶, then ẋ=dx/dt=10⁻⁴·√(6.106(50/x -1)) ≈2.471*10⁻⁴√(50/x -1). Simply square & rearrange to solve for x: => x=50/(1+16377333.77ẋ²) Approximate by getting rid of the 1: x=50/16377333.77ẋ² Now it is easily separated. Take the -ve root as dx is -ve: -√x · dx = 10⁻⁴·√(50(6.106))=√305.3 *10⁻⁴ dt Integrating both sides, with x=50 to x=5 over time interval "t" gives: (-⅔(5)³ᐟ²) - (-⅔(10)³ᐟ²)≈10⁻⁴·(√305.3)(t-0) 228.25≈ 10⁻⁴·(√305.3)t => t≈ 130,630.5969 seconds ≈ 36.29 hrs. or more precisely: t≈⅔√(1250/3053)(√1000 -1) seconds. This gives me doubts about the 87 hrs answer... Later on I will try again with a numerical integral, including the 1, and I'll get back to you.
I also got the same equation. But there is one thing , at position r2 the acceleration due to gravity (g) is GM/(r2^2). And at each point during the journey of that falling object the acceleration due to gravity is changing and by the time it reaches to distance r1 the g has increased to GM/r1^2. Therefore our equation is only true if we are considering it that g = GM/r^2. But in this problem it is not the case.
@@AbhijeetKumar-mq4vhhence finding the average acceleration using the two points 5R and 50R a1 = GM/(5R)^2 a2 = GM/(50R)^2 ā = (a1 + a2)/2 Skipping trivial steps ā = (101GM)/(5000R^2)
To solve this problem, we can use the concept of gravitational potential energy and conservation of mechanical energy. (a) The gravitational potential energy of an object at a distance \( r \) from the center of the Earth is given by: \[ U = -\frac{GMm}{r} \] Where: - \( G \) is the gravitational constant, - \( M \) is the mass of the Earth, - \( m \) is the mass of the object, and - \( r \) is the distance from the center of the Earth. The object falls freely from \( R_2 \) to \( R_1 \), so its total mechanical energy (\( E \)) remains constant: \[ E = K + U \] Where: - \( K \) is the kinetic energy, and - \( U \) is the gravitational potential energy. Since the object starts from rest at \( R_2 \), its initial kinetic energy (\( K_i \)) is zero. Therefore, the total mechanical energy at \( R_2 \) is equal to the gravitational potential energy: \[ E_{R_2} = U_{R_2} \] \[ 0 = -\frac{GMm}{R_2} \] At \( R_1 \), the object has kinetic energy (\( K_f \)) and gravitational potential energy (\( U_{R_1} \)). Therefore, the total mechanical energy at \( R_1 \) is: \[ E_{R_1} = K_f + U_{R_1} \] Using conservation of mechanical energy, we have: \[ E_{R_2} = E_{R_1} \] \[ -\frac{GMm}{R_2} = \frac{1}{2}mv^2 - \frac{GMm}{R_1} \] \[ -\frac{GM}{R_2} = \frac{1}{2}v^2 - \frac{GM}{R_1} \] Where: - \( v \) is the final velocity of the object. Since the object falls freely, its final velocity can be found using the equation for the velocity of an object undergoing constant acceleration: \[ v^2 = u^2 + 2a \Delta s \] Where: - \( u \) is the initial velocity (which is zero), - \( a \) is the acceleration due to gravity, and - \( \Delta s \) is the displacement. The acceleration due to gravity is given by \( a = \frac{GM}{r^2} \), and \( \Delta s = R_2 - R_1 \). Therefore, \[ v^2 = 0 + 2 \left(\frac{GM}{R_2^2} ight) (R_2 - R_1) \] \[ v^2 = 2 \frac{GM}{R_2} - 2 \frac{GM}{R_1} \] Substituting this expression for \( v^2 \) into the previous equation, we can solve for \( t \): \[ -\frac{GM}{R_2} = \frac{1}{2} \left(2 \frac{GM}{R_2} - 2 \frac{GM}{R_1} ight) - \frac{GM}{R_1} \] \[ -\frac{GM}{R_2} = \frac{GM}{R_2} - \frac{GM}{R_1} - \frac{GM}{R_1} \] \[ -\frac{GM}{R_2} = \frac{GM}{R_2} - \frac{2GM}{R_1} \] \[ \frac{GM}{R_2} = \frac{2GM}{R_1} \] \[ R_1 = 2R_2 \] But \( R_1 = 5R_1 \), so \( R_1 = 2R_2 \) is not possible. (b) Given: - \( R_2 = 50 \times R \) - \( R_1 = 5 \times R \) - \( M = 6 \times 10^{24} \) kg - \( R = 6400 \) km (which is \( 6400 \times 10^3 \) m) - \( G = 6.67 \times 10^{-11} \) N m²/kg² We'll use the derived equation for time \( t \) from part a: \[ t = \sqrt{\frac{2(R_2 - R_1)R_1^2}{GM}} \] Substituting the given values: \[ t = \sqrt{\frac{2((50 \times R) - (5 \times R))(5 \times R)^2}{GM}} \] \[ t = \sqrt{\frac{2(45 \times R)(25 \times R^2)}{GM}} \] \[ t = \sqrt{\frac{2250 \times R^3}{GM}} \] Now, let's substitute the values of \( M \), \( G \), and \( R \) and calculate \( t \): \[ t = \sqrt{\frac{2250 \times (6400 \times 10^3)^3}{(6 \times 10^{24}) \times (6.67 \times 10^{-11})}} \] \[ t ≈ \sqrt{\frac{2250 \times (26214400000000000000000)}{4.002 \times 10^{14}}} \] \[ t ≈ \sqrt{\frac{59016000000000000000000000}{4.002 \times 10^{14}}} \] \[ t ≈ \sqrt{14750749.375} \] \[ t ≈ 3840.03 \] Converting seconds to hours: \[ t ≈ \frac{3840.03}{3600} \] \[ t ≈ 1.07 \] Therefore, the time \( t \) it takes for the object to fall from \( R_2 = 50 \times R \) to \( R_1 = 5 \times R \) is approximately 1.07 hours (to 2 decimal places).
"Since the object falls freely, its final velocity can be found using the equation for the velocity of an object undergoing constant acceleration:" The acceleration is not constant.
Sir i from india , and aspirating for an engineering exam JEE , i belong to general category and in our country their is a huge problem of reservation... This creates trouble for students from general category, thus i request you to post lectures regarding 11th class physics classes i entirely trust ur teachings sir., we just need ur blessings
I can understand you bro
You can find mit lectures that enough
But that's not relevant @@UdaySharma-yf6dr
Bhai reservation ka rona band kr
Itna hi issue h toh ews bana le
Aur padhai kr
Main bhi gen open male hu
Sab lectures online free mil jaate h
@@jeeaspirant2232😂
Congratulations for the publication of 200th problem. Best wishes for the publication of the 300th one.
Greeting from Paris France. Thanks for your knowledge and enthusiasm in bringing to us. ❤
(a) At a distance 'r' from center, let's say the speed is 'v'
We obtain the expression for v by conservation of total energy. (K_i + U_i = K_f + U_f)
(Initially, object starts from rest so K_i = 0)
U_i = -GMm/R_2, U_f = -GMm/R_1
By further simplifying, we get
v = (2GM(1/r - 1/R_2))^(1/2)
Also, here v = -dr/dt
Now we need to integrate, dr/(1/r - 1/R_2)^(1/2) from R_2 to R_1 on LHS and -(2GM)^1/2 dt from 0 to t on RHS. Put r = R_2*(sin u)^2 and integrate
We get R_2(u - 1/2* sin(2u)) = -((2GM/R_2)^1/2)*t
Finally, t = (π/2)*( (R_2)^3/2)/(2GM)^1/2 - ((R_2)^3/2)/(2GM)^1/2 * arcsin((R_1/R_2)^1/2) + (R_2)*(R_1/2GM)^1/2 * (1-(R_1/R_2))^1/2
(b) By putting all the given values, time taken from R_2 to R_1 = 0.396 hours.
Right idea but you must have made a slip somewhere, wrong b).
Wow the 200th problem!
Sir please start an internship program for high school students so that we could learn more about physics.
Hello dear Mr. Walter Levin. Is it possible to use Kepler's second law to solve this problem? I was unable to integrate the equation I received. I hope you answer
Yes . I love you very much ❤❤
Very Very Beautiful Class sir 😇😇🙌
Is there a mark scheme for this? I’d love to see the answer
Sir , i respect from India up
Using a spreadsheet and 5 minute increments, I'm getting approximately 87 hours, but I really have to double check it and try to do the complex math to check if the spreadsheet has mistakes or is anywhere close.
A is square root of r1r2[r1-r2]/2gR^2 here i took GM=gR^2,sorry for that
b is 50/3 hrs which is16.67 hrs
love the way you teach
Love you walter lewin
Hello Mr Lewin
a) KE(at r) = GPE(initial) - GPE(at r)
We get an expression of v(r)
t = (5R to 50R) integral of [dr/v(r)]
b) t = 87 hours
I love you sir ❤❤
Respected walter lewin sir❤
Answer for
a) 2/3(2GM)^-1/2×[(R1)^3/2 - (R2)^3/2]
b) 36.46 hours
Please sir respond to my answers whether they are correct or not!
Thank you sir
Acceleration is not constant here.
@@KeithandBridgetdo you really think this equation will after taking constant acceleration?
Today also a very well good morning to a great physicist of this century Mr Walter Lewin sir. please... You tell me from where I start the physics because I also want to become like you . I am in 11th grade and I am from India and I want to give something this world
ask your teachers who know you.
Starting from conservation of energy between starting point R2 and a generic distance r:
variation in kinetic energy = variation in potential energy, we get:
(1/2)v^2 = k•(1/r-1/R2) k=MG
v=dr/dt = - sqrt(2k(1/r-1/R2))
minus sign is because the direction of motion is towards decreasing distance r
Rearranging:
-dr/sqrt(2k(1/r-1/R2)) = dt
Integral between R2 and R1 of left hand term equals the time delta.
After looking up the primitive (sorry for that) of 1/sqrt(1/x-1/c), which is:
(x-c)/sqrt(1/x-1/c)-c^1.5 • arctan(sqrt(c/x-1)) and taking the sign into account, we get:
T=((1/sqrt(2MG))•((R2-R1)/sqrt(1/R1-1/R2)+R2^1.5 • arctan(sqrt(R2/R1-1))
Plugging in the values of R1 and R2, I get:
T=313426 s = 87h 3m
Using the free fall theorem, we can find the time taken from R2 to R1 is
a.) t = sqrt(2R²(R2-R1)/GM)
And if R2=50R and R1=5R, using the equation above, we can find the exact time is
b.) t = 7680 second or t = 2.1 hours
I hope it's the correct answer
Thanks for the problem, Sir. Can't wait for the explanation 😊
Hope you are doing well, professor. Greetings from CO!
I will be attempting this problem soon!
Hello sir ❤️🙏
I have a question to you, why inertia depends upon mass of an object rather than weight (amount of gravitational force act of an object), this is why because mass is a fixed quantity but inertia is not, for example: we put same mass having object at different planets here mass is same but inertia is different i.e due to gravitational force of planets. So sir can we say inertia depends upon weight (amount of gravitational force act of an object)?? Sir if you see so please clear my doubt please❤️🙏
angular momentum =mvr kg.m^2/sec
t = sqrt(R2^3 / 2GM) * [arccos(sqrt(R1/R2)) + sqrt(R2*R1 - R1^2) / R2]; For the given values, t ~ 87 hours.
Respected professor, how may I send my solution to this problem to you?
send it here
❤
Sir could you please explain what is the potential of an electron moving in a magnetic field and the interaction energy associated with it.I was asked this question in an interview
I cover this in my 8.02 lectures. Plse watch them, that's what they are for.
@@lecturesbywalterlewin.they9259 thankyou sir..
a)
Time t is an integral from R2 to R1 and...
where the gravitational force is getting larger each second the probe is getting closer to the surface of the Earth, due to: F = GMm/R² and...
where the acceleration a equals F/m ➔ a = GM/R² and...
where velocity v equals current velocity u plus the acceleration a × 1 second and...
where the total distance traveled should be increased by (old) velocity u × 1 second plus ½ times the acceleration a times 1 second squared and...
until the total distance traveled is equal to, or larger than R2 - R1
While (R2 - R1) > D
D = D + (v * 1s) + (½ * a * 1s²) ➔
D = D + ((u + (a * 1s)) * 1s) + (½ * a * 1s²) ➔
D = D + ((u + GM/(D + R1)²) * 1s) + (½ * GM/(D + R1)² * 1s²) ➔
(final:)
Iteration: D = D + (u + GM/(D + R1)²) + (½ * GM/(D + R1)²)
Compute New: u = u + GM/(D + R1)² * 1s (u=u+a*1s)
Time = Time + 1 second
Print Time
b)
# Initialize variables
u = 0
D = 0
T = 0
Re = 6400000
Me = 6 * 10**24
G = 6.67 * 10**(-11)
# Calculate constants
C = G * Me
R2 = 50 * Re
R1 = 5 * Re
Rc = R2 - R1
# Run loop
while Rc > D:
D = D + (u + C/(D + R1)**2) + (0.5 * C/(D + R1)**2)
u = u + C/(D + R1)**2
T = T + 1
# Print result
print(T);(T/3600)
72340
20.094444444444445
Thus 20 hours.
Interesting computational approach. With 1 second steps I would expect a close approximation, but part b) is well off the analytic answer. I can't see an obvious error but it must be in there somewhere.
@@KeithandBridget The iteration should be v(t+dt) = v(t) + a(t)*dt and s(t+dt) = s(t) + v(t+dt) * dt. The formular v = a * t and s = 0.5 * a * t² is valid, if a = const. That is not the case.
@@michaelbruning9361 Agreed.
"s" should be around a linear function of t to the power of 2/3. I used "r" and I found it also helpful to simplify things by changing the
variable to x=r/6.4*10⁶. For this variable,
I found x≈ 0.138(134896.4 -t)^(2/3).
(a)
1. Integrate acceleration, a, due to gravity.
a = dv/dt = (dv/dr)·(dr/dt)
where dr/dt = velocity = v
=> a = v·(dv/dr) = -GM/r²
=> v dv = -(GM/r²) dr
∫ v dv = -∫(GM/r²) dr {from R₂ to r}
where R₂ = 50R and r is a point below R₂.
=> v²/2 = -GM∫dr/r² = -GM [-1/r] {from R₂ to r}
=> v² = 2GM(1/r - 1/R₂)
v = dr/dt = -√(2GM)·√(1/r - 1/R₂)
We take the negative root since motion is downwards.
After some additional steps:
dt = -√[R₂/(2GM)]·√[r/(R₂ - r)] dr
2. Integrate one more time.
Substitution: u = √[r/(R₂ - r)]
=> u² = r/(R₂ - r) => r = R₂ - R₂/(1 + u²)
dr/du = 2R₂u/(1 + u²)²
r = R₂ => u → ∞ , r = R₁ => u = √[R₁/(R₂ - R₁)] =
=√[5R/(50R - 5R)] = 1/3
t = -√[R₂/(2GM)]·∫u·2R₂u/(1 + u²)² du {from ∞ to 1/3}
t = -√[2R₂³/(GM)]∫u²/(1 + u²) du {from ∞ to 1/3}
t = - √[2R₂³/(GM)]·{ ½[arctan(u) - u/(1+u²)] } {from ∞ to 1/3}
Answer (a):
t = -√[R₂³/(2GM)]·[arctan(1/3) - 3/10 - π/2] …(1)
(b)
M = 6·10²⁴ kg
R = 6400 km = 6.4·10⁶ m
G = 6.67·10ᐨ¹¹ m³/(kg·s²)
R₂ = 50R
R₁ = 5R
Answer (b):
From equation (1) we get t ≈ 87 hours.
Hello sir ❤
😅
Professor , in problem Acceleration due to gravity is considered constant or it is varying?
Kid, learn basics first
Given the distance, it varies a lot.
Did we skip198?, it seemed like after 197 came 199th problem
Yes
Hey very well good morning to you Walter Lenin sir I want to learn physics with you so I can give this world a new thought. My English is not so good I hope you will react . I am waiting
eat yogurt every day but *never on Fridays* That also worked well for Einstein and for me.
Thank you sir for replying me . I think now it is first step towards success of this world by me with your advice. I am very grateful to you .
Hi sir i am preparing for neet please give me some advice for physics
eat yogurt every day but *never on Fridays* that also worked well for Einstein and for me
@@lecturesbywalterlewin.they9259
Hello again Professor Lewin. I was wondering if you can help me figure out why my method didn't work. Apparently the average force over space as calculated from the potential is different from the average force over time. This puzzles me greatly. Also, please do check my response to Keith in my solution.
I found that using a approximate version of your method that makes the integral separable gives a similar answer to mine, but both are less than half of your answer, both in the 30s ...
Also, there is an interesting relationship between my solution and yours...
√(20/3)=√6.666 ... ≈2.58 ≈87/33.7
Sir plz 🙏🙏🙏 i am very inspire for you and very motivated sir
But.... Physics is not understand for me sir....
My english are very weak but i am trying to do
Best 👍💯to best
Also i am preparing for neet
In 2025......😢😢😢😢
All the best
2. Try:
a) t=sqrt(R2/(2GM)) * [R2*arctan{sqrt((R2-R1)/R1)} - sqrt(R1*(R2-R1))]
b) 53.34 h
If my answer is correct, i can send the whole solution to a)
Sir i belong from nepal i🇳🇵🇳🇵
apod.nasa.gov/apod/ap240510.html
Answer is time t=√(R2-R1)/g
T = (R2^(3/2) ArcTan[Sqrt[R2/R1 - 1]] + Sqrt[R1 R2 (R2 - R1)]) / Sqrt[2 G M]
T = 87.06 h
My answer is approximetly 93,6 hours
Is this true ?
@@elbekelbek8205 No.
My final solution.
(a) The time "t" taken to fall from R2 to R1 can be obtained using the concepts of impulse
and average force over the displacement R2-R1.
Thus: t= Overall change in impulse/average force =(∆p)₂₁/⟨F⟩₂₁= m(v₁ - 0)/⟨F⟩₂₁= (v₁/GM)/⟨1/r²⟩₂₁
(r here is the position wrt Earth's CM)
To calculate the average force over the distance travelled from R2 to R1, we need to consider the changing force of gravity along the path from R2 to R1. The force of gravity varies with distance, so we need to account for this variation in order to find the correct average force.
We can determine the changing force of gravity at each point along the path from position R2 to R1 using the inverse square law. The force of gravity is inversely proportional to the square of the distance.
We proceed as follows:
1. Calculate the work done by gravity along the path from R2 to R1. This work will be equal to the change in potential energy of the object as it falls.
2. Integrate the varying force of gravity with respect to distance over the path from R2 to R1.
3. Calculate the average force over the entire distance by dividing the total work done by the total distance travelled.
4. Use the total work done by gravity to find the final velocity of the object at position R1 using the work-energy principle.
5. Calculate the total transit time for the object to fall from R2 to R1 based on the distance travelled and the velocity at R1.
By properly accounting for the varying force of gravity along the path of fall from radius R2 to R1 and integrating it with respect to distance, we can determine the accurate average force over the distance and calculate the total transit time for the fall.
My answer for 1-5:
Now I will change the variable from r wrt CM to r wrt the surface (position R), so that r becomes r+R.
With limits from r2= R2 -R to r1=R1-R, the work done on mass m by Earth's gravity from R2 to R1 is:
W₂₁= ₂₁∫F •dr = -(∆U)₂₁ = -m(∆V)₂₁ = -m[-GM/(r+R)]₂₁
Change in kinetic energy of mass m wrt Earth's CM = (∆K)₂₁= ∆ (½mv²)₂₁ = ½m(v₁² - v₂²)
Here v₁ and v₂ are the final and initial velocities respectively.
By the Work-Energy theorem, W₂₁ = (∆K)₂₁
Thus, as mass m is initially at rest v₂=0, and we have:
W₂₁= GMm(1/(r₁+R) - 1/(r₂+R)) =GMm(1/R₁ - 1/R₂)=½mv₁²
Using F=GMm/(r+R)² and letting R₂-R₁ = d₂₁ , we have:
₂₁∫F•dr = GMm(d₂₁/R₁R₂)= ½mv₁² ...(1)
The average force over displacement d₂₁ is given by ⟨F⟩₂₁ = ₂₁∫F•dr / d₂₁ = -GMm/R₁R₂
Also, solving for v₁ in (1) and cancelling out m on both sides gives:
v₁= √{2GM(d₂₁/R₁R₂)}
Finally, we have t= √{(2GM(d₂₁/R₁R₂))}/GM/R₁R₂ =√{2d₂₁R₁R₂/GM}
(b) is about (7680000/(4002)^.5)/3600~ 33.72 hours.
With R1= 5R and R2=50R, R=64E5 [m], GM=(6.67*6)E(-11+24)=40.02E13,
We have (in seconds):
t=(64E5)√{2(64E5)(50-5)(5)(50)/40.02E13}
=(64*8E5)√{(1E6)(9)(25)/40.02E12}
=(64*8E5)(15)(1E(3-6))√{1/40.02}
=768,000/√40.02
In hours this is:
t=213⅓/√40.02~33.723
😉⏰🌚🚀⤵🍑🌎
Also, I dunno if anyone noticed but R1 is 32,000km away from Earth and this is about a Uranus equatorial radius away from the surface... ;)
b) not correct. I think your averaging is over simplifying things.
@@KeithandBridget I made a math error. It's 33.7
@@KeithandBridget I checked out a few vids on UA-cam and it looks like my method is pretty standard, so it's puzzling me why it's apparently not working in this case.
I tried it again, the same way you and WL did, but approximating the integral so that variables could be separated easily.
I got a similar answer to my answer here...
In the 30s. Approximately 36.29 hrs or 130,630.6 secs.
If the variable r is changed to x= r/6.4*10⁶, then ẋ=dx/dt=10⁻⁴·√(6.106(50/x -1))
≈2.471*10⁻⁴√(50/x -1).
Simply square & rearrange to solve for x:
=> x=50/(1+16377333.77ẋ²)
Approximate by getting rid of the 1:
x=50/16377333.77ẋ²
Now it is easily separated.
Take the -ve root as dx is -ve:
-√x · dx = 10⁻⁴·√(50(6.106))=√305.3 *10⁻⁴ dt
Integrating both sides, with x=50 to x=5 over time interval "t" gives:
(-⅔(5)³ᐟ²) - (-⅔(10)³ᐟ²)≈10⁻⁴·(√305.3)(t-0)
228.25≈ 10⁻⁴·(√305.3)t
=> t≈ 130,630.5969 seconds ≈ 36.29 hrs.
or more precisely:
t≈⅔√(1250/3053)(√1000 -1) seconds.
This gives me doubts about the 87 hrs answer... Later on I will try again with a numerical integral, including the 1, and I'll get back to you.
@@KeithandBridget I found something curious...
√(20/3)=√6.666 ... ≈2.58 ≈87/33.7
❤❤first view🎉🎉
a)
F(r) = m a(r) = m GM/r^2;
a(r) = GM/r^2;
r(t) = INTEGRAL(a(r) t dt);
r^2 dr = GM t dt;
r^3(t) = 3/2 GM t^2;
then:
Δt^2 = 2 (R2-R1)^3 / (3GM);
b)
R2=50R; R1 = 5R;
substituting:
Δt = SQRT(2*45^3*R^3 / (3GM)) = 55 hours approx.
Sir I got answer 29.32 seconds for A). Is it correct???
No but at least you don't exceed the speed of light🙂
Helloo
I think the answer is t = 26.48 sec
Find average acceleration (ā) from the two points, 5R and 50R, with a = GM/R^2
Δx = 1/2at^2
Δx = ΔR = 45R
=> t = sqrt(90R/ā)
I also got the same equation. But there is one thing , at position r2 the acceleration due to gravity (g) is GM/(r2^2). And at each point during the journey of that falling object the acceleration due to gravity is changing and by the time it reaches to distance r1 the g has increased to GM/r1^2. Therefore our equation is only true if we are considering it that g = GM/r^2. But in this problem it is not the case.
@@AbhijeetKumar-mq4vhhence finding the average acceleration using the two points 5R and 50R
a1 = GM/(5R)^2
a2 = GM/(50R)^2
ā = (a1 + a2)/2
Skipping trivial steps
ā = (101GM)/(5000R^2)
Final solution
t = (300Rsqrt(5))/(sqrt(101GM))
@@StarWarsTherapy Average acceleration is not accurate enough.
Using a≈ 0.197 m/s² , I get t≈15 hrs using your method. Definitely an underestimate.
To solve this problem, we can use the concept of gravitational potential energy and conservation of mechanical energy.
(a)
The gravitational potential energy of an object at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{r} \]
Where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the object, and
- \( r \) is the distance from the center of the Earth.
The object falls freely from \( R_2 \) to \( R_1 \), so its total mechanical energy (\( E \)) remains constant:
\[ E = K + U \]
Where:
- \( K \) is the kinetic energy, and
- \( U \) is the gravitational potential energy.
Since the object starts from rest at \( R_2 \), its initial kinetic energy (\( K_i \)) is zero. Therefore, the total mechanical energy at \( R_2 \) is equal to the gravitational potential energy:
\[ E_{R_2} = U_{R_2} \]
\[ 0 = -\frac{GMm}{R_2} \]
At \( R_1 \), the object has kinetic energy (\( K_f \)) and gravitational potential energy (\( U_{R_1} \)). Therefore, the total mechanical energy at \( R_1 \) is:
\[ E_{R_1} = K_f + U_{R_1} \]
Using conservation of mechanical energy, we have:
\[ E_{R_2} = E_{R_1} \]
\[ -\frac{GMm}{R_2} = \frac{1}{2}mv^2 - \frac{GMm}{R_1} \]
\[ -\frac{GM}{R_2} = \frac{1}{2}v^2 - \frac{GM}{R_1} \]
Where:
- \( v \) is the final velocity of the object.
Since the object falls freely, its final velocity can be found using the equation for the velocity of an object undergoing constant acceleration:
\[ v^2 = u^2 + 2a \Delta s \]
Where:
- \( u \) is the initial velocity (which is zero),
- \( a \) is the acceleration due to gravity, and
- \( \Delta s \) is the displacement.
The acceleration due to gravity is given by \( a = \frac{GM}{r^2} \), and \( \Delta s = R_2 - R_1 \).
Therefore,
\[ v^2 = 0 + 2 \left(\frac{GM}{R_2^2}
ight) (R_2 - R_1) \]
\[ v^2 = 2 \frac{GM}{R_2} - 2 \frac{GM}{R_1} \]
Substituting this expression for \( v^2 \) into the previous equation, we can solve for \( t \):
\[ -\frac{GM}{R_2} = \frac{1}{2} \left(2 \frac{GM}{R_2} - 2 \frac{GM}{R_1}
ight) - \frac{GM}{R_1} \]
\[ -\frac{GM}{R_2} = \frac{GM}{R_2} - \frac{GM}{R_1} - \frac{GM}{R_1} \]
\[ -\frac{GM}{R_2} = \frac{GM}{R_2} - \frac{2GM}{R_1} \]
\[ \frac{GM}{R_2} = \frac{2GM}{R_1} \]
\[ R_1 = 2R_2 \]
But \( R_1 = 5R_1 \), so \( R_1 = 2R_2 \) is not possible.
(b)
Given:
- \( R_2 = 50 \times R \)
- \( R_1 = 5 \times R \)
- \( M = 6 \times 10^{24} \) kg
- \( R = 6400 \) km (which is \( 6400 \times 10^3 \) m)
- \( G = 6.67 \times 10^{-11} \) N m²/kg²
We'll use the derived equation for time \( t \) from part a:
\[ t = \sqrt{\frac{2(R_2 - R_1)R_1^2}{GM}} \]
Substituting the given values:
\[ t = \sqrt{\frac{2((50 \times R) - (5 \times R))(5 \times R)^2}{GM}} \]
\[ t = \sqrt{\frac{2(45 \times R)(25 \times R^2)}{GM}} \]
\[ t = \sqrt{\frac{2250 \times R^3}{GM}} \]
Now, let's substitute the values of \( M \), \( G \), and \( R \) and calculate \( t \):
\[ t = \sqrt{\frac{2250 \times (6400 \times 10^3)^3}{(6 \times 10^{24}) \times (6.67 \times 10^{-11})}} \]
\[ t ≈ \sqrt{\frac{2250 \times (26214400000000000000000)}{4.002 \times 10^{14}}} \]
\[ t ≈ \sqrt{\frac{59016000000000000000000000}{4.002 \times 10^{14}}} \]
\[ t ≈ \sqrt{14750749.375} \]
\[ t ≈ 3840.03 \]
Converting seconds to hours:
\[ t ≈ \frac{3840.03}{3600} \]
\[ t ≈ 1.07 \]
Therefore, the time \( t \) it takes for the object to fall from \( R_2 = 50 \times R \) to \( R_1 = 5 \times R \) is approximately 1.07 hours (to 2 decimal places).
incorrect
@@lecturesbywalterlewin.they9259 why sir
@@lecturesbywalterlewin.they9259 Ok sir thank you for your quick response.
"Since the object falls freely, its final velocity can be found using the equation for the velocity of an object undergoing constant acceleration:" The acceleration is not constant.
T=rs(2(r2-r1)/g) - 33 years
a) t= 1/sqrt{ 2GM } *[ R_2^{3/2} .arccos sqrt{R_1 /R_2} + R_1 .R_2 .sqrt {1/R_1 - 1/R_2}],
b) 87.063h.
Answer for
a) 2/3(2GM)^-1/2×[(R1)^3/2 - (R2)^3/2]
b) 36.46 hours
Acceleration is not constant here.