Question

On a planet similar to Earth a vertical hole is drilled from the planet’s surface to its centre.
An object is dropped from the planet’s surface and falls toward the planetary centre.

Assume that the free-fall acceleration due to gravity varies linearly from g=8.57m/s^2 at the surface to zero at the centre (as it would do if the planet were of uniform density). The planet's radius is 6490km, neglect air resistance and effects due to planets rotation.

a) What is the objects velocity when it reaches the planetary centre?

b) How long does the object take to reach the planetary centre from instant it is dropped?

Answer 1

acceleration varies lineary with distance, and assuming distance to be zero at origin the aceleration( and so the force) is in the opposite direction with the increasing values of distance, so
F=-kx
so you're dealing with simple harmonic motion

Answer 2

Like this? \[a(x)=\frac{ -8.57x }{ 6490000 } +8.57\]? Acceleration as a function of distance? I'm still confused as to how I can obtain the velocity.

Answer 3

at surface area of the Earth acceleration is not zero. Your plot should have looked like this:
then
\[a(x)=-\frac{ g }{ R }x\]
a is \[\ddot x\]
so
\[\ddot x+\frac{ g }{ R }x=0\]It's the equation of Simple Harmonic Motion with period
\[T=2\pi \sqrt{\frac{ R }{ g }}\]
the time to reach the centre is T/4

Answer 4

I get part B (time taken to reach centre), but still stuck on A

Answer 5

solution due to that differential equation:
\[x(t)=x_{\max}\cos(\omega t )\]
\[v(t)=\dot x(t)=-\omega x_{\max}\sin(\omega t)\]
x max is simply R. We know that at the center of the Earth the velocity reaches its maximum so this amplitude of that sinusoidal will be the velocity at centre

\[v=\omega R= \sqrt{Rg}\]