a ball is dropped in a tunnel made through the earth. time taken for it to reach the centre is
Answers
Answered by
1
A few people have already mentioned that under totally unrealistic, very Physics 101 conditions (no friction or air resistance, perfectly spherical Earth with uniform density throughout), what you throw in will exhibit simple harmonic motion. I think it's a cool derivation, so I'll present it here.
Let G,R,MG,R,M be, respectively, the gravitational constant, Earth's radius, and Earth's mass. Since gravity follows an inverse-square law (ok, not really, but hey, this is Physics 101), we can apply Gauss's Law to a concentric spherical Gaussian surface with radius r≤Rr≤R (for those familiar with Gauss's Law in classical E&M, I'm using mass, 4πG4πG, and gravitational acceleration analogously to charge, ϵ0ϵ0, and electric field). Thus, we find that the gravitational acceleration aa at distance rr from the center of the Earth is
G(rR)3M(1r2)G(rR)3M(1r2).
Simplifying and adding signs where appropriate, we have
a=−(GMR3)r=−ω2ra=−(GMR3)r=−ω2r. But this is simple harmonic motion, for which the period of oscillation is given as T=2πωT=2πω. Plugging in values gives us a period of 84.35 minutes, meaning the object will emerge at the other end of the hole in roughly 42 minutes (what a coincidence!).
hope its helps you
do mark brainliest
Let G,R,MG,R,M be, respectively, the gravitational constant, Earth's radius, and Earth's mass. Since gravity follows an inverse-square law (ok, not really, but hey, this is Physics 101), we can apply Gauss's Law to a concentric spherical Gaussian surface with radius r≤Rr≤R (for those familiar with Gauss's Law in classical E&M, I'm using mass, 4πG4πG, and gravitational acceleration analogously to charge, ϵ0ϵ0, and electric field). Thus, we find that the gravitational acceleration aa at distance rr from the center of the Earth is
G(rR)3M(1r2)G(rR)3M(1r2).
Simplifying and adding signs where appropriate, we have
a=−(GMR3)r=−ω2ra=−(GMR3)r=−ω2r. But this is simple harmonic motion, for which the period of oscillation is given as T=2πωT=2πω. Plugging in values gives us a period of 84.35 minutes, meaning the object will emerge at the other end of the hole in roughly 42 minutes (what a coincidence!).
hope its helps you
do mark brainliest
Similar questions