Question

Calculate the acceleration due to gravity inside Earth as a function of the radial distance r from the planet’s center. (Hint: Imagine that a mine shaft has been drilled from the surface to Earth’s center and an object of mass m has been dropped down the shaft to some radial position r 6 RE from the center, where RE is the radius of Earth. What is the vector sum of the forces exerted on the object by all of Earth’s mass that lies at any distance d 7 r from the center? d 6 r from the center?

Answer 1

Answer:

Explanation:

Volume of the earth is,

$V=\frac{4}{3}\pi R_E^3$

Volume of the sphere is,

$V=\frac{4}{3}\pi r^3$

The density of the earth is,

$\rho =\frac{M_E}{V}\\\\=\frac{M_E}{\frac{4}{3}\pi R_E^3}$

The mass of the sphere is,

$M=V\rho\\\\=\frac{4}{3}\pi r^3\frac{M_E}{\frac{4}{3}\pi R_E^3}\\\\=\frac{M_Er^3}{R_E^3}$

Consider a mass $m$ at r<$R_E$

Expression for the force is,

$F=\frac{GMm}{r^2}\\\\=\frac{G(\frac{M_Er^3}{R_E^3})m}{r^2}\\\\=\frac{GM_Erm}{R_E^3}$

From newtons second law,

$F=mg\\\\\frac{GM_Erm}{R_E^3}=mg\\\\g=\frac{GM_Er}{R_E^3}$