Travelling Through the Earth

https://isaacphysics.org/questions/travelling_through_the_earth?board=60da2182-1587-4b35-8150-82bf3e30c03a&stage=a_level

For part c of this question, I managed to get the right answer, but I don't really understand why. The video hint talks about conserving energy but honestly that made me more confused so I just stuck with forces.

I worked out the component of the force that made the mass "fall" and I basically guessed w = sqrt(G*rho*4pi/3) but didnt really know how this worked out - why didnt R^3/r^3 come into play?

Help greatly appreciated.

(edited 10 months ago)

Reply 1

mosaurlodon

Working:

im pretty sure the pythagoras at the beginning has some usefulness but apparently not in my working out (prob because my working out is wrong)

math.jpg170.2KB

Reply 2

mqb2766

Original post by mosaurlodon

https://isaacphysics.org/questions/travelling_through_the_earth?board=60da2182-1587-4b35-8150-82bf3e30c03a&stage=a_level
For part b of this question, I managed to get the right answer, but I don't really understand why. The video hint talks about conserving energy but honestly that made me more confused so I just stuck with forces.
I worked out the component of the force that made the mass "fall" and I basically guessed w = sqrt(G*rho*4pi/3) but didnt really know how this worked out - why didnt R^3/r^3 come into play?
Help greatly appreciated.

Dont know where you get R^3/r^3 from and this would imply when r=0 then the gravitational acceleration was infinite etc? By symmetry, the acceleration must be zero at the center as there is equal mass attracting in all directions.

Part a) gives the justification for g=-4pi/3 G*rho*r, so
d^2 r/dt^2 = -4pi/3 G*rho*r
so shm where w^2 = ...

(edited 10 months ago)

Reply 3

mosaurlodon

Really sorry I meant part c

Reply 4

mqb2766

Original post by mosaurlodon

Really sorry I meant part c

Not worked it through, but a bit of simple trig should give the acceleration along the chord?

Reply 5

mosaurlodon

Ohhhh wait I think I see it
So g(r) = -G*rho*4pi/3*r where r is the distance from the centre
and we want g(r) * sin(theta)
sin(theta) = z/r
so g(z) =-G*rho*4pi/3*z
g(z) = -w^2*z

I was overcomplicating it massively, really you dont need pythagoras from the video at all, nor h, but they give it to you anyways? I just assumed my answered required the use of h.

Thank you

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