Sir, I am 13 years old now, and I started watching your lectures from when I was 11. As an Indian ,I am very thankful to you and your lectures sir! Thank you!
I hope you understand that your work as a teacher has truly inspired people into studying physics, and emphasize how wonderful it truly is. I studied your lectures in depth when I first started, and every single one still amaze me. Now I'm in highschool and studying quantum mechanics and nuclear physics, without you I would have reached this point in my knowledge. I truly thank you for your impact on my life, and other aspiring scientists.
Hii, sir I am Rishabh from India. I am in class 11 but still I watch you because the you explain is outstanding. ❤❤ YOU WILL BE ONE OF THE GREATEST TEACHER OF ALL TIME. ❤
Good afternoon sir This side parth from india I am very lucky to be an era where such great teachers like you exist who make such tough concepts very easy I start you all your 94 lectures And now my conceptual knowledge and thinking ability towards tough concepts has increased thank you very much sir! I hope i will meet you one day because now all i can say is thank you ❤
Hi sir. I just wanted to tell you that your lectures have been extremely helpful to me. In our schools, teachers do not show interest in teaching physics, i mean by using real life examples. You really gave me the passion for physics and i thank you very much. YOU ARE TRULY THE GREATEST
Hi Professor, I was a bit confused by the concepts of Electric fields, dipoles, and forces. However, the 8.02x lectures have strengthened my core understanding of these concepts and induced my love for physics to a greater extent. Thank you, Professor, for helping millions of students by creating such wonderful lectures that one truly longs for! Wishing you excellent health and happiness!
If it wasnt for your lectures and bi-weekly/monthly physics problem i wouldnt have made it into the physics program at KSU. Thank you for everything. Thank you for being honest.
(a) By total energy conservation, K_i + U_i = K_f + U_f The spacecraft could escape the planet's gravity when it is at infinite distance i.e., U_f = 0 K_i + U_i ≥ 0 1/2 *m(v_0)² - GMm/R ≥ 0 Therefore, (Vo)min = (2GM/R)^(1/2) (b) torque t = dL/dt, since there is no external force t = 0 hence L = constant L at R = m(Vo)R sin(30) L at 15R = m(V)15R We get, Vo/V = 30 (c) By conservation of total energy, 1/2 * m(Vo)² - GmM/R = 1/2 *m(V)² - GmM/15R On solving, We find that V = ((28/15)(GmM/899R))^(1/2) Therefore TE = (59/899)GmM/R or 0.065(GmM/R) (d) Since TE is the same, TE at farthest distance is also (59/899)* GmM/R
1) Energy is conserved, therefore by conservation of energy We get the escape velocity v = √(GM/R) 2) Taking center of planet as the axis, only force on the planet (gravitational force) passes through the axis always. So angular momentum is conserved. mVoR sin 30° = mV× 15R So Vo/V = 30 3) Total Energy = 1/2 m Vo^2 - GMm/R 4) Energy is conserved so, it is still the same as 3rd answer
(a) sqrt(2GM/R) {By conservation of energy} This velocity is the escape velocity for the planet and does not depend upon the angle of projection (b) By conservation of angular momentum, mv0Rsin30 = mv(15R) {at max distance velocity is perp to radius vector} v0/v = 30 (c) Initial energy of the spacecraft is 0.5mv0^2 - GMm/R = 450mv^2 - GMm/R {v0 = 30v} Final energy of the spacecraft (At r=15R) is 0.5mv^2 - GMm/(15R) Equating both by conservation of energy: 450mv^2 - GMm/R = 0.5mv^2 - GMm/(15R) So 449.5 mv^2 = 14GMm/(15R) So mv^2 = 14GM / (6742.5 R) Now initial energy = 450mv^2 - GMm/R = 442.5/6742.5 (-GMm/R) = 0.06563 (-GMm/R) (d) Total energy when spacecraft is at farthest point is the same, i.e., 0.06563 (-GMm/R) PS: We can see that the orbit will be bounded since total energy is negative.
Solutions: A: V0 = sqrt(2*G*M/R) B: V/v0 = sqrt(188790)/13485, approx 3.222% C: At launch, Etotal = -59/899*G*M*m/R, approx -0.0656*G*M*m/R D: At apsis, Etotal = -59/899*G*M*m/R, approx -0.0656*G*M*m/R Negative sign indicates its path is bounded, and will have an elliptical orbit if allowed to continue as a freely moving body. Supporting reasoning: In part A, the rocket drifts away indefinitely, which is the escape velocity. Energy changes are independent of launch angle, since gravity is conservative. Initial GPE = -G*M*m/R Final GPE = 0, for infinitely far away KE needed to achieve final GPE: 1/2*m*V0^2 = G*M*m/R Everybody brought rocket mass to the party: 1/2*V0^2 = G*M/R Solve for V0: V0 = sqrt(2*G*M/R) In part B, r = 15*R at its apsis point. To find its speed at apsis, use conservation of angular momentum and conservation of energy to lock-down the two unknowns (initial velocity vi and apsis velocity V). Initial & apsis angular momentum Launch: L = vi*R*sin(30 deg) Apsis: L = V*15*R Equate and solve for va: V = vi/30 Also apply conservation of energy, for an additional constraint. 1/2*m*vi^2 - G*M*m/R = 1/2*m*V^2 - G*M*m/(15*R) Everybody brought rocket mass to the party: 1/2*vi^2 - G*M/R = 1/2*V^2 - G*M/(15*R) Combine equations and solve for V V = 2*sqrt(7/13485)*sqrt(G*M/R) Compare V to V0: V/V0 = sqrt(188790)/13485 = 3.222% Parts C and D are equal, by conservation of energy. Calculate total energy at apsis as answer to both parts. Etotal = 1/2*m*V^2 - G*M*m/(15*R) Etotal = 14/13485*G*M*m/R - G*M*m/(15*R) Etotal = -59/899*G*M*m/R My initial thought to solve this problem, was to construct an ellipse with a focus at the origin, a circle centered at the origin for the planet, tangent and radial lines at launch, and the angle. Then, adjust the parameters of the ellipse, and find curvature at the apsis, such that the centripetal acceleration equals acceleration of gravity at the apsis. The results are consistent with the above result, but a lot more challenging to solve in closed form.
Context-1 a) It's basically the escape velocity required to get out of its gravitation which √2(GM/R). Any higher velocity than this would provide a non zero kinetic energy at infinity. Context-2: b) Since no other external force exists and force is along radius vector. Torque=dL/dt is conserved along the radius vector. Hence Vosin30R=Vsin15R. V is Vo/30. c)T.E is conserved assuming no energy loss outside the system KE+PE is conserved. Applying conservation of total energy and find V or Vo in standard terms. Total energy=-885/(15*899)GMm/r. d) Same as above
A) the second speed: (2GM/R)^(1/2) B)imagine there is no speed when reaching the 15R; all the way, gravity equal to GMm/r^2, where r is the distance from the center; the work done equal to integral (GMm/r^2)dr from R to 15R which is GMm(1/R-1/15R); the initial speed is {2GM/(1/R-1/15R)}^0.5 C) D) the work done by the gravity above
Sir here is your proud student . Sir I'm selected for International Scientific Physics Olympiad from Team India which is gonna held in MIPT Russia. Thank you so much sir for making me capable for this huge glory 😢
Thank you so much sir , I wish if I could have a 1:1 Video Call session with you for guidance . That was also my dream since 8th Grade . If you could accept it it would be a pleasure and dream come true moment for me . If you could think of it then please 🙏 🙂
@@Cyber_ghost7 30% of my 1.7 million subscribers want to know my email address, my home phone and they want to make video calls with me. *All are off limit. Email + address top secret*
Hello sir im from india currently pursuing Bachelor's in CS i was a formal JEE aspirant at present preparing for Btech sems and other examinations i will be honoured to get a reply from you ❤😇
Hi sir hope this text finds you fine. Sir I'm from India and recently our NEET examination was conducted, despite I was able to solve enough paper to get selection I would probably miss that. I don't know how i filled my booklet number wrong and then i messed up with the question answer sequence. And now I'm not able to overcome from this trauma. Whenever feel alone or sit in corner, some fear overwhelmes me. Need some words from your side, perhaps they might help.
one of my close Indian friends also failed the NEET examination. Yet he ended up at some of Indians best colleges (but not IT). I gave a talk at his college in 2014. After his degree there he was accepted at a Univeristy in Sweden where he got a degree in Physics and he will soon get his Dr degree at a university in Switzerland.
a) G * m * M / R = 1/2 * m * v0² or v0 = √(2 * G * M / R) b) G * m * M / (15*R) = 1/2 * m * v² or v =√(2/15 * G * M / R) and v0 / v = √15 c) and d) Energy conservation: Eafter launch = Efar = - G * m * M / (15 * R) + 1/2 * m * v² = 0
Physics is difficult but i tried to understand but i also a NEET aspirants then physics seems so difficult please professor give advice to understand how physics become easy.
Sir, I read your book 'For the love of physics' finished it yesterday and I just want to say thank you for writing that wonderful book:) and I am so sorry for your loss from WW2.... PS: Can you give me some of the best physics book for Indian Institute of science I just got in 11th grade(India) thank you
Hello sir ! I'm from India.... Sir I had a doubt in class 10 physics which I did asked to my teachers but wasn't satisfied with the answers... " Sir does convex mirror form a real image ? If yes , then how ?" My teachers said that yes , if the object is virtual , but how is that practically possible? Sir pls reply as I have been confused over this question since a long period of time....🙂
Concave mirrors can produce both real and virtual images depending on the distance from the mirror to the object and the curvature of the mirror, while convex mirrors produce only virtual images.
Sir I'm reading in class 9th and I have a question that....when I was seeing a Hypergolic fuel Nitrogen tetroxide and I came to a equation of this showing ΔH^θ what is it actually?....Enthalpy?
(a) The minimum speed for escape velocity occurs when the spacecraft can reach infinity with zero speed. Thus both kinetic and potentional energy becomes zero. => ½mv₀² - GMm/R = 0 Answer (a): v₀ = √[2GM/R] (b) Conservation of angular momentum: mv₀R sin α = mvd sin 90° α = 30° => sin α = ½, sin 90° = 1 d = the distance from the center of Earth to apogee. d = 15R => ½ v₀R = vd => v₀/v = d/(½R) = 15R/(½R) = 2·15 Answer (b): v₀/v = 30 (c) The total energy of the spacecraft immediately after launch: Answer (c): Eᵢ = Kᵢ + Uᵢ = ½mv₀² - GMm/R (d) The total energy of the spacecraft when it is farthest away from the planet: E₀ = K₀ + U₀ = ½mv² - GMm/d Answer (d): E₀ = K₀ + U₀ = ½mv² - GMm/(15R)
Addition. Since both angular momentum and total energy are conserved we can set the equations in (c) and (d) equal and find out the actual velocities: v₀/v = 30 and ½mv² - GMm/d = ½mv² - GMm/(15R) gives v ≈ 360 m/s and v₀ ≈ 10810 m/s.
@@lecturesbywalterlewin.they9259 oh my god the great professor walter lewin replied to my comment.Sir i topped the class 12 examination CBSE in physics and chemistry which subject should i major in?
Hi sir I am from India I want to research in physics suggest me some topic I have not any one who help me in that work so sir please suggest me some topics in which I will research
Sir, I am 13 years old now, and I started watching your lectures from when I was 11. As an Indian ,I am very thankful to you and your lectures sir! Thank you!
I'm 15 sis
I hope you understand that your work as a teacher has truly inspired people into studying physics, and emphasize how wonderful it truly is. I studied your lectures in depth when I first started, and every single one still amaze me. Now I'm in highschool and studying quantum mechanics and nuclear physics, without you I would have reached this point in my knowledge. I truly thank you for your impact on my life, and other aspiring scientists.
I am delighted to hear this
Hii, sir I am Rishabh from India.
I am in class 11 but still I watch you because the you explain is outstanding. ❤❤ YOU WILL BE ONE OF THE GREATEST TEACHER OF ALL TIME. ❤
All the best
Good afternoon sir
This side parth from india
I am very lucky to be an era where such great teachers like you exist who make such tough concepts very easy
I start you all your 94 lectures
And now my conceptual knowledge and thinking ability towards tough concepts has increased thank you very much sir!
I hope i will meet you one day because now all i can say is thank you ❤
So nice of you
Your videos always put a smile on our face and the videos are full of knowledge !
And I really love your art quizes Sir!❤
Hi sir. I just wanted to tell you that your lectures have been extremely helpful to me. In our schools, teachers do not show interest in teaching physics, i mean by using real life examples. You really gave me the passion for physics and i thank you very much.
YOU ARE TRULY THE GREATEST
thank you kindly
An absolute genius ✨
Hi Professor, I was a bit confused by the concepts of Electric fields, dipoles, and forces. However, the 8.02x lectures have strengthened my core understanding of these concepts and induced my love for physics to a greater extent. Thank you, Professor, for helping millions of students by creating such wonderful lectures that one truly longs for!
Wishing you excellent health and happiness!
great!
Good evening sir 🙏🏻🙏🏻, your lectures is very helpful for me 👍🏻👍🏻💖💖💖, love from India 🇮🇳🇮🇳🇮🇳
If it wasnt for your lectures and bi-weekly/monthly physics problem i wouldnt have made it into the physics program at KSU.
Thank you for everything. Thank you for being honest.
you are welcome
(a) By total energy conservation,
K_i + U_i = K_f + U_f
The spacecraft could escape the planet's gravity when it is at infinite distance i.e., U_f = 0
K_i + U_i ≥ 0
1/2 *m(v_0)² - GMm/R ≥ 0
Therefore, (Vo)min = (2GM/R)^(1/2)
(b) torque t = dL/dt, since there is no external force t = 0 hence L = constant
L at R = m(Vo)R sin(30)
L at 15R = m(V)15R
We get, Vo/V = 30
(c) By conservation of total energy,
1/2 * m(Vo)² - GmM/R = 1/2 *m(V)² - GmM/15R
On solving,
We find that V = ((28/15)(GmM/899R))^(1/2)
Therefore
TE = (59/899)GmM/R or 0.065(GmM/R)
(d) Since TE is the same, TE at farthest distance is also (59/899)* GmM/R
You will always be one of the best physics teachers sir , i always love people who teach physics interactively like you
Lots of love from India ❤
so nice of you
Sir your lectures givee me vibes like Einstein sir is teaching me ❤️❤️❤️
1) Energy is conserved, therefore by conservation of energy
We get the escape velocity v = √(GM/R)
2) Taking center of planet as the axis, only force on the planet (gravitational force) passes through the axis always. So angular momentum is conserved.
mVoR sin 30° = mV× 15R
So Vo/V = 30
3) Total Energy = 1/2 m Vo^2 - GMm/R
4) Energy is conserved so, it is still the same as 3rd answer
Hi sir I am from Pakistan I wish you will be fine I'm watching all your lectures your method is very beautiful and perfect ❤❤❤
(a) sqrt(2GM/R) {By conservation of energy}
This velocity is the escape velocity for the planet and does not depend upon the angle of projection
(b) By conservation of angular momentum, mv0Rsin30 = mv(15R) {at max distance velocity is perp to radius vector}
v0/v = 30
(c) Initial energy of the spacecraft is 0.5mv0^2 - GMm/R = 450mv^2 - GMm/R {v0 = 30v}
Final energy of the spacecraft (At r=15R) is 0.5mv^2 - GMm/(15R)
Equating both by conservation of energy: 450mv^2 - GMm/R = 0.5mv^2 - GMm/(15R)
So 449.5 mv^2 = 14GMm/(15R)
So mv^2 = 14GM / (6742.5 R)
Now initial energy = 450mv^2 - GMm/R = 442.5/6742.5 (-GMm/R) = 0.06563 (-GMm/R)
(d) Total energy when spacecraft is at farthest point is the same, i.e., 0.06563 (-GMm/R)
PS: We can see that the orbit will be bounded since total energy is negative.
Solutions:
A: V0 = sqrt(2*G*M/R)
B: V/v0 = sqrt(188790)/13485, approx 3.222%
C: At launch, Etotal = -59/899*G*M*m/R, approx -0.0656*G*M*m/R
D: At apsis, Etotal = -59/899*G*M*m/R, approx -0.0656*G*M*m/R
Negative sign indicates its path is bounded, and will have an elliptical orbit if allowed to continue as a freely moving body.
Supporting reasoning:
In part A, the rocket drifts away indefinitely, which is the escape velocity. Energy changes are independent of launch angle, since gravity is conservative.
Initial GPE = -G*M*m/R
Final GPE = 0, for infinitely far away
KE needed to achieve final GPE:
1/2*m*V0^2 = G*M*m/R
Everybody brought rocket mass to the party:
1/2*V0^2 = G*M/R
Solve for V0:
V0 = sqrt(2*G*M/R)
In part B, r = 15*R at its apsis point. To find its speed at apsis, use conservation of angular momentum and conservation of energy to lock-down the two unknowns (initial velocity vi and apsis velocity V).
Initial & apsis angular momentum
Launch: L = vi*R*sin(30 deg)
Apsis: L = V*15*R
Equate and solve for va:
V = vi/30
Also apply conservation of energy, for an additional constraint.
1/2*m*vi^2 - G*M*m/R = 1/2*m*V^2 - G*M*m/(15*R)
Everybody brought rocket mass to the party:
1/2*vi^2 - G*M/R = 1/2*V^2 - G*M/(15*R)
Combine equations and solve for V
V = 2*sqrt(7/13485)*sqrt(G*M/R)
Compare V to V0:
V/V0 = sqrt(188790)/13485 = 3.222%
Parts C and D are equal, by conservation of energy. Calculate total energy at apsis as answer to both parts.
Etotal = 1/2*m*V^2 - G*M*m/(15*R)
Etotal = 14/13485*G*M*m/R - G*M*m/(15*R)
Etotal = -59/899*G*M*m/R
My initial thought to solve this problem, was to construct an ellipse with a focus at the origin, a circle centered at the origin for the planet, tangent and radial lines at launch, and the angle. Then, adjust the parameters of the ellipse, and find curvature at the apsis, such that the centripetal acceleration equals acceleration of gravity at the apsis. The results are consistent with the above result, but a lot more challenging to solve in closed form.
Context-1
a) It's basically the escape velocity required to get out of its gravitation which √2(GM/R). Any higher velocity than this would provide a non zero kinetic energy at infinity.
Context-2:
b) Since no other external force exists and force is along radius vector. Torque=dL/dt is conserved along the radius vector. Hence Vosin30R=Vsin15R. V is Vo/30.
c)T.E is conserved assuming no energy loss outside the system
KE+PE is conserved. Applying conservation of total energy and find V or Vo in standard terms. Total energy=-885/(15*899)GMm/r.
d) Same as above
Good evening sir ❤
Awesome explanation sir❤
Thanks for liking
Looks Like Sandeep Maheswari in Thumbnail
A) the second speed: (2GM/R)^(1/2)
B)imagine there is no speed when reaching the 15R; all the way, gravity equal to GMm/r^2, where r is the distance from the center; the work done equal to integral (GMm/r^2)dr from R to 15R which is GMm(1/R-1/15R); the initial speed is {2GM/(1/R-1/15R)}^0.5
C) D) the work done by the gravity above
Sir here is your proud student . Sir I'm selected for International Scientific Physics Olympiad from Team India which is gonna held in MIPT Russia. Thank you so much sir for making me capable for this huge glory 😢
All the best
Thank you so much sir , I wish if I could have a 1:1 Video Call session with you for guidance . That was also my dream since 8th Grade . If you could accept it it would be a pleasure and dream come true moment for me . If you could think of it then please 🙏 🙂
@@Cyber_ghost7 30% of my 1.7 million subscribers want to know my email address, my home phone and they want to make video calls with me. *All are off limit. Email + address top secret*
Professor please make a video on what diet you follow to live long and healthy like you.
eat yogurt every day but *never on Fridays*
Looking handsome, Prof!
at age 30 I was handsome - but that fades away with age
Hello sir im from india currently pursuing Bachelor's in CS i was a formal JEE aspirant at present preparing for Btech sems and other examinations i will be honoured to get a reply from you ❤😇
Hi sir hope this text finds you fine. Sir I'm from India and recently our NEET examination was conducted, despite I was able to solve enough paper to get selection I would probably miss that. I don't know how i filled my booklet number wrong and then i messed up with the question answer sequence. And now I'm not able to overcome from this trauma. Whenever feel alone or sit in corner, some fear overwhelmes me. Need some words from your side, perhaps they might help.
one of my close Indian friends also failed the NEET examination. Yet he ended up at some of Indians best colleges (but not IT). I gave a talk at his college in 2014. After his degree there he was accepted at a Univeristy in Sweden where he got a degree in Physics and he will soon get his Dr degree at a university in Switzerland.
@@lecturesbywalterlewin.they9259 will save this text lifetime. Thanku ❤
My frnd did the same mistake man filled wrong roll no.
a) V0=sqrt(2MG/R)
b) V0/V=30
c) 1/2 mV0^2 - mMG/R
d) 1/2 mV0^2/900 -mMG/(15R).
Keep going. Thank you.
a) G * m * M / R = 1/2 * m * v0² or v0 = √(2 * G * M / R)
b) G * m * M / (15*R) = 1/2 * m * v² or v =√(2/15 * G * M / R) and v0 / v = √15
c) and d) Energy conservation: Eafter launch = Efar = - G * m * M / (15 * R) + 1/2 * m * v² = 0
Physics is difficult but i tried to understand but i also a NEET aspirants then physics seems so difficult please professor give advice to understand how physics become easy.
Sir, I read your book 'For the love of physics' finished it yesterday and I just want to say thank you for writing that wonderful book:)
and I am so sorry for your loss from WW2....
PS: Can you give me some of the best physics book for Indian Institute of science I just got in 11th grade(India)
thank you
ask your teachers
@@lecturesbywalterlewin.they9259 Sir please help me you know they wont help I am following all your lectures just require some good books please
Are your lectures enough for jee ?
Hello sir ! I'm from India.... Sir I had a doubt in class 10 physics which I did asked to my teachers but wasn't satisfied with the answers...
" Sir does convex mirror form a real image ? If yes , then how ?"
My teachers said that yes , if the object is virtual , but how is that practically possible?
Sir pls reply as I have been confused over this question since a long period of time....🙂
Concave mirrors can produce both real and virtual images depending on the distance from the mirror to the object and the curvature of the mirror, while convex mirrors produce only virtual images.
@@lecturesbywalterlewin.they9259 sir so there is no such case of convex mirror producing virtual images?
Sir I'm reading in class 9th and I have a question that....when I was seeing a Hypergolic fuel Nitrogen tetroxide and I came to a equation of this showing ΔH^θ what is it actually?....Enthalpy?
use google
Standard enthalpy ig . Enthalpy taken when it's 1mole product formed ig
In standard conditions too
(a) The minimum speed for escape velocity
occurs when the spacecraft can reach infinity
with zero speed. Thus both kinetic and potentional
energy becomes zero.
=> ½mv₀² - GMm/R = 0
Answer (a): v₀ = √[2GM/R]
(b) Conservation of angular momentum:
mv₀R sin α = mvd sin 90°
α = 30° => sin α = ½, sin 90° = 1
d = the distance from the center of Earth to apogee.
d = 15R
=> ½ v₀R = vd => v₀/v = d/(½R) = 15R/(½R) = 2·15
Answer (b): v₀/v = 30
(c) The total energy of the spacecraft
immediately after launch:
Answer (c):
Eᵢ = Kᵢ + Uᵢ = ½mv₀² - GMm/R
(d) The total energy of the spacecraft
when it is farthest away from the planet:
E₀ = K₀ + U₀ = ½mv² - GMm/d
Answer (d):
E₀ = K₀ + U₀ = ½mv² - GMm/(15R)
Addition.
Since both angular momentum and total energy are conserved
we can set the equations in (c) and (d) equal and find out the
actual velocities:
v₀/v = 30 and
½mv² - GMm/d = ½mv² - GMm/(15R)
gives v ≈ 360 m/s and v₀ ≈ 10810 m/s.
looks handsome
I was handsome at age 30
The thumbnail is total clickbait. It's not a Malcolm Gladwell video.
this picture of me was taken in 1966 when I was 30 yr old
@@lecturesbywalterlewin.they9259 Oh sorry. The resemblance is striking!
@@lecturesbywalterlewin.they9259You were surely handsome back then 😆
Isnt the thumbnail guy sandeep maheshwari?
I was 30 yr old when this picture was taken of me during my first month at MIT
@@lecturesbywalterlewin.they9259 oh my god the great professor walter lewin replied to my comment.Sir i topped the class 12 examination CBSE in physics and chemistry which subject should i major in?
Hi sir I am from India I want to research in physics suggest me some topic I have not any one who help me in that work so sir please suggest me some topics in which I will research
black holes
Ok sir@@lecturesbywalterlewin.they9259
tell me the truth from which university you graduated
thanks
a. V_crit = sqrt(2GM/R)
b. V0/V = 30
c. E = - (59/899) * (GMm/R)
d. same
Sorry for your tragic loss.
a)Vo = sqrt(2MG/R(sin30°)^2)
b)Vo/V = sqrt(1/15R - (1/4)sin30°)/(sin30°)^3
c) Etot = (1/2)mVo^2(sin30°)^2-mMG/R
d)Etot = (1/2)mVo^2(sin30°)^2-mMG/15R
Sir what time is it in the US
I am from india now its 10:00 pm here
use google
@@lecturesbywalterlewin.they9259 😂
@@lecturesbywalterlewin.they9259 can he use bing instead?
a) v min = sqrt(2MG/R)
b) v₀/V = 30
c) = d) TE = -59/899 (mMG/R)