How much work is needed to move a body of mass m = 1 kg from…

How much work is needed to move a body of mass $m = 1\ \mathrm{kg}$ from the centre of the Earth to its surface? The Earth's mean radius is $R_Z = 6370\ \mathrm{km}$ and its mean density $\rho = 5500\ \mathrm{kg/m^3}$.

Step 1: Idea

Inside the Earth the pull is not $\gamma Mm/r^2$. A uniform spherical shell exerts no net force on a body inside it, so at distance $r$ from the centre only the matter within radius $r$ attracts the body. As the body rises, more and more mass lies below it, and the force grows. Because the force changes along the way, the work must be found by integration.

Step 2: Force at distance $r$

The mass inside radius $r$ is $M(r) = \tfrac43\pi r^3\rho$. It acts as if it were at the centre: $F(r) = \gamma\frac{M(r)\,m}{r^2} = \gamma\frac{m}{r^2}\cdot\tfrac43\pi r^3\rho$.

\[ F(r) = \tfrac43\pi\gamma\rho m\,r. \]

The force is proportional to $r$: zero at the centre, largest at the surface.

Step 3: Work

We must pull outward with a force equal to $F(r)$ (moving slowly, without gaining speed). The work is the integral of this force over the path: $A = \int_0^{R_Z}\tfrac43\pi\gamma\rho m\,r\,dr = \tfrac43\pi\gamma\rho m\cdot\frac{R_Z^2}{2}$.

\[ A = \tfrac23\pi\gamma\rho m R_Z^2. \]

\[ A = \tfrac23\pi \cdot 6.67 \cdot 10^{-11} \cdot 5500 \cdot 1 \cdot (6.37 \cdot 10^6)^2. \]

\[A \approx 3.12 \cdot 10^7\ \mathrm{J}\]

A neat way to see this: the force grows linearly from $0$ to $mg$, so the average force is $\tfrac12 mg$ and the work is $\tfrac12 mgR_Z$, the area of the triangle in the diagram.

Check

The surface value $g $ $= \tfrac43\pi\gamma\rho R_Z $ $= 9.79\ \mathrm{m/s^2}$, so $ A $ $= \tfrac12 mgR_Z $ $= 0.5 \cdot 9.79 \cdot 6.37 \cdot 10^6 $ $= 3.12 \cdot 10^7\ \mathrm{J}\ \checkmark $