Free Fall Problem 200

Newtonian Mechanics - Easy Physics - Difficult Math

[Jamesbondhere]

(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.

[dhairyasarda]

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

[YannisAlepidis]

Congratulations for the publication of 200th problem. Best wishes for the publication of the 300th one.

[graemegeorge]

Greeting from Paris France. Thanks for your knowledge and enthusiasm in bringing to us. ❤

[ahmad-z-eina]

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

[elbekelbek]

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

[dyana]

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 😊

[STARWARS]

Sir please start an internship program for high school students so that we could learn more about physics.

[oldtvnut]

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.

[ulfhaller]

(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.

[KrishanKumar-jflu]

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

[mauriziobianconi]

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:
• arctan(sqrt(R2/R1-1))
Plugging in the values of R1 and R2, I get:
T=313426 s = 87h 3m

[allahcc]

t = sqrt(R2^3 / 2GM) * [arccos(sqrt(R1/R2)) + sqrt(R2*R1 - R1^2) / R2]; For the given values, t ~ 87 hours.

[_John_Sean_Walker]

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 =
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


Thus 20 hours.

[UPian_]

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

[sergejzukov]

Is there a mark scheme for this? I’d love to see the answer

[promitdutta]

Respected professor, how may I send my solution to this problem to you?

[Mayank]

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❤️🙏

[abhishekraj]

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

[complexgranola]

Hope you are doing well, professor. Greetings from CO!