Question

Suppose that a hole were drilled to the center of the Earth and an evacuated pipe were inserted in the hole. Assume the radius of the Earth is R and the density is uniform with the total mass M. An object of mass m is dropped down the pipe?

Suppose that a hole were drilled to the center of the Earth and an evacuated pipe were inserted in the hole. Assume the radius of the Earth is R and the density is uniform with the total mass M. An object of mass m is dropped down the pipe. Find the velocity of the object when it reaches the center, and the time it takes for the object to reach the center.

Answer 1

• Time: `qquad qquad approx 21 " mins"`

• Velocity: `qquad approx 7.9 " kps"`

Explanation:

From Gauss' gravity law, gravitational flux is:

`Phi_g = oint_(S) mathbf {g} cdot d \ mathbf {A} =-4\pi GM_s`

Using a spherical Gaussian surface, a concentric sphere of radius `r` within the earth, and for uniform earth density `rho`, this becomes:

`g(r) \ mathbf (hat r) cdot 4 pi r^2 \ mathbf (hat r) = - 4\pi G(rho 4/3 pi r^3)`

`rho = M/(4/3 pi R^3)`

`implies g(r) = - ( G \ M)/( R^3) r`

It follows that:

`ddot r + ( G \ M)/( R^3) r = 0`

Solving as:

`r = alpha cos omega t + beta sin omega t`, where `omega^2 = ( G \ M)/( R^3)`

IV's are:

• `r_o = R`

• `dotr_o = 0`

`implies r = R cos omega t`

On arrival at the centre at time `tau`:

• `r = 0 implies omega tau = pi/2 implies tau = pi/2 sqrt( R^3 /(GM)) approx 21 " mins"`

And velocity:

`dot r = - omega R sin omega t`

`dot r(tau) = - sqrt( ( G \ M)/( R^3)) \ R \ sin (pi/2) = - sqrt( ( G \ M)/( R)) approx - 7.9 " kps"`