Question

work done in bringing a mass from infinity to center of earth

Answer 1

Answer on Question #37760 - Physics - Other

Work done in bringing a mass from infinity to center of earth.

Solution.

The mass of object is $m$.

We use the model: Earth is homogeneous sphere. Earth's radius is $R_{e}$, density is $\rho$, mass is

$$M(R_{e}) = \frac{4\pi}{3} \rho R_{e}^{3}$$

We know that the gravitational potential at infinity is zero $\varphi(\infty) = 0$.

Potential inside the Earth is $\varphi(r) - \varphi(0) = \int_{0}^{r} \frac{GM(r')}{r'^2} dr' = \int_{0}^{r} \frac{G \frac{4\pi}{3} \rho r'^3}{r'^2} dr = G \frac{2\pi}{3} \rho r^2$, $r$ is

distance to Earth center, $G = 6.67 \cdot 10^{-11} \frac{m^3}{kg \cdot s^2}$, $M(r) = \frac{4\pi}{3} \rho r^3$ is mass of sphere, which has density $\rho$ and radius $r$.

If $r > R_{e}$ we choose $\varphi(r) = \frac{GM(R_{e})}{r}$. We need the continuity of potential at $r = R_{e}$. Whence,

$$\varphi(R_{e}) = \frac{GM(R_{e})}{R_{e}} = \frac{4\pi}{3} G \rho R_{e}^{2} = G \frac{2\pi}{3} \rho R_{e}^{2} + \varphi(0)$$ $$\varphi(0) = 2\pi G \rho R_{e}^{2}$$

The work is $A = \left| m(\varphi(0) - \varphi(\infty)) \right| = 2\pi G m \rho R_{e}^{2}$.

At surface of Earth the gravitational acceleration is $g = \frac{GM(R_{e})}{R_{e}^{2}} = \frac{4\pi}{3} G\rho R_{e}$.

Whence $A = \frac{3}{2} g m R_{e}$

Answer:

$$A = \frac{3}{2} g m R_{e}$$