Is the Time to Reach Earth's Center Correct?

  • Thread starter Thread starter tanaygupta2000
  • Start date Start date
  • Tags Tags
    Gravitation Time
AI Thread Summary
The discussion focuses on the calculation of the time to reach the Earth's center, using the formula for gravitational acceleration at a depth 'd'. The correct formula for the time period of a complete journey is T = 2π sqrt(R/g), but for a half journey to the center, it simplifies to T = π sqrt(R/g). A sign error in the differential equation was noted, indicating that the solutions do not include sine and cosine functions. The journey from the surface to the center is identified as one-fourth of a complete cycle. Overall, the approach is mostly correct, with adjustments needed for the sign error.
tanaygupta2000
Messages
208
Reaction score
14
Homework Statement
A tunnel is dug from the surface of the earth through the center and opens at the other end. A ball is dropped from one end of the tunnel. The acceleration due to gravity on the earth’s surface is g and the radius of the earth is R. Assuming that the earth has a constant density, what is the time taken by the ball to reach the center of the earth?

Options:
(a) π√(R/g)
(b) 2π√(R/g)
(c) (π/2)√(R/g)
(d) (π/4)√(R/g)
Relevant Equations
Acceleration due to gravity inside the earth, g' = g(1 - d/R)
IMG_20200727_173222.jpg


The value of acceleration due to gravity at a depth 'd' inside the Earth is given by-
g' = g(1 - d/R)
which can also be written as
g' = g(x/R) from the diagram

so that x'' = (w2)x
where w2 = g/R is the angular frequency

Hence the time period T is given by
T = 2π sqrt(R/g)

but the question is asking only for the half journey
so the answer should be
T = π sqrt(R/g)
Is this correct?
 
Physics news on Phys.org
tanaygupta2000 said:
so that x'' = (w2)x
Watch the signs.
tanaygupta2000 said:
but the question is asking only for the half journey
What would one complete cycle involve?
 
IMG_20200727_205927.jpg
 
tanaygupta2000 said:
Is this approach correct ?
Looks good.
 
As @haruspex pointed out, you have a sign error. The differential equation ##\ddot x = \omega^2 x## does not have ##\sin \omega t## and ##\cos \omega t## as solutions, as you can check.

However, the rest of your work looks good. (Regarding your first approach in post #1, traveling from the surface to the center of the Earth is 1/4 of a cycle.)
 
  • Like
  • Informative
Likes tanaygupta2000 and PeroK
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top