# Why Gauss's law for gravity can be applied to this problem?

Consider a spherical and homogeneous celestial body with mass $$M$$ and radius $$R$$ crossed from pole to pole by a straight cylindrical tunnel of negligible diameter. At one of these poles a point object is released from rest at $$t = 0$$. Since $$G$$ is the universal gravitation constant, this object will reach the other pole after the time interval given by:

Answer: $$\pi \left(\frac{R^3}{GM}\right)^{\frac{1}{2}}$$

One of the most viable ways to solve this problem is using Gauss's law for gravity. But why? Could someone give me some explanation of why Gauss's law for gravity can be applied to this problem? Why is it often used instead of Newton's law?

• I don't know I never used this technique to solve this instead the solution is simple The particle will undergo SHM and the value for the time interval will be T/2. Where T is the time period which can be easily calculated. Sep 27 at 17:13

In suitable units the field equation that you have to solve is $$\begin{equation} \nabla \cdot \mathbf{G} = \rho \end{equation}$$ where $$\mathbf{G}$$ is the gravitational field and $$\rho$$ is the density of the earth. Now integrate this up to some radius $$R < R_\text{earth}$$, $$\begin{equation} \int_{S_2(R)} d^3x \; \nabla \cdot \mathbf{G} = \int_{S_2(R)} d^3x \; \rho \end{equation}$$ here is where we can make use of Gauss' law to write $$\begin{equation} \int d \mathbf{S} \cdot \mathbf{G} = \frac{4}{3} \pi R^3 \rho \end{equation}$$ The symmetry of the problem suggests that $$\mathbf{G}$$ must always point in a radial direction. This is important. For general problems, this method does not work and you have to go back to a Green's function approach. But here things are easier and we can solve the integral directly, $$\begin{equation} 4\pi R^2 |\mathbf{G}| = \frac{4}{3}\pi R^3 \rho. \end{equation}$$ The equation of motion of your falling body is now $$\begin{equation} m \ddot{R} + m \frac{R \rho}{3} = 0, \end{equation}$$ which we recognize as an harmonic oscillator with $$\omega^2 = \frac{ \rho}{3}$$. Upon replacing $$\rho = 3M/4\pi R^3$$ and switching to SI-units, you obtain from this the correct period time.