The question is pretty long and wordy so apologies in advance!
Inside the Earth the gravitational field falls off linearly as one approaches the centre. An accurate description of motion in a very deep hole would therefore be to use Newton’s law, f = ma, but with the force being
f = −gmx/R
where m is the object’s mass, x is the distance of the object from the centre of the Earth, R is the radius of the Earth, and g is the usual acceleration due to gravity measured at the surface.
- (i) Suppose that a parcel destined for Australia is dropped from rest at ground level into a hole that goes through the centre of the Earth. Derive an equation for the speed of the subsequent motion v(x) (where v = x ̇). [Hint: Use x ̈ = vdv/dx.]
- (iii) Find an expression for the parcel’s position, x(t).
- (iv) Using the θ ≪ 1 approximation, cosθ ≈ 1 − θ2 + ..., show that when the displacement is small, (R − x) ≪ R, your expression for x(t) gives the usual result, x ≈ R − gt2/2 .
- (v) Show (using the approximations for R and g in (ii)) that it takes about an hour for the parcel to arrive in Australia.
The Attempt at a Solution
1) I tried -gmx/R = mx.. = mvdv/dx
I then separated the variables and integrated
-gm/R∫xdx = mvdv with the limits of x from 0-->2R and the limits for v 0-->v
this gave me -4gmR = mv2
Rearranging for v (I'm not sure about the sign) v = 2√gr
2) Assuming the above equation is correct (which I'm fairly cercain it isn't!) I then simply integrated with respect to time to get x = 2t√gr. This is obviously wrong as it doesn't work in the next part parts.
I'd really appreciate any help, thanks!