# Integration and forces

1. May 13, 2016

### struggles

1. The problem statement, all variables and given/known data
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.

1. (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.]

2. (iii) Find an expression for the parcel’s position, x(t).

3. (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 .

4. (v) Show (using the approximations for R and g in (ii)) that it takes about an hour for the parcel to arrive in Australia.

2. Relevant equations

3. 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!

2. May 13, 2016

### SammyS

Staff Emeritus
Those limits of integration are incorrect.

The parcel goes from the surface of the earth, looks like x = R, the way you have thing set up.

The parcel starts at rest.

Then you want to find v at some arbitrary value of x. So the x integral goes from R to x. The v integral goes from 0 to v.

3. May 14, 2016

### struggles

Ok i still can't get this to work out:
so after integrating i get -gm/R(x2/2 -R2/2) = mv2/2
Rearranging i get v = √(gR - gx2/R) = dx/dt

Then rearranging to integrate again -
dx(gR - gx2/R)-1/2 = 1dt
and integrating
R/gx (gR - gx2/R)1/2

This isn't right but i've played around and cant get it to a reasonable answer

4. May 14, 2016

### SammyS

Staff Emeritus
Check that last integration.
The derivative of your result does not give the integrand.

You might try a trig substitution. Part (3) gives a clue that trig functions are involved.

5. May 14, 2016

### struggles

so eventually after making substitutions i get √r/√g ∫ du/√(1-u2) = √r/√g arcsin(x/r) (where u = x/r)
and rearranging get x = Rsin(t√g/√R). However the next part of the question ( and that at t=0 c should = r) implies that it should be cos. Any ideas or have i just made a slip with signs somewhere along the way (potentially the order in which i applied the limits in the first integration?) Thanks so much!

6. May 14, 2016

### SammyS

Staff Emeritus
What are the limits for that integration?

Also, I assume you have used r and R interchangeably.

7. May 14, 2016

### struggles

So the limits would be t is from 0 to t and x from R to x which gives √R2/g2(π/2 - arcsin(x/r)) = t.
Rearranging sin(π/2 - t√(g2/R2)) = cos(t√(g2/R2) = x! Thank you! think i've finally got there!!

8. May 14, 2016

### Ray Vickson

The differential equation for $x(t)$ is just that of simple harmonic motion, like that of a mass attached to a spring obeying Hook's law. Here, $x(t)$ is displacement from the earth's center, and can be < 0 as well as > 0.