Question

With what speed would mail pass through the center of Earth if falling in a tunnel through the center?

Short answer

Answer:

The speed of the mailpass through the center of Earth if falling in a tunnel through the centeris$7.9\times {10}^{6}m/s$

Step-by-step solution

Listing the given quantities

Radius of the Earth R

Understanding the concept of the gravitational acceleration

Here,we apply the work-energy theorem to the object in question. It starts from a point at the surface of the Earth with zero initial speed and arrives at the center of the Earth with final speed The corresponding increase in its kinetic energy, is equal to the work done on it by Earth's gravity

Formula:

$KE=\frac{1}{2}mv{f}^2$

Calculation of the gravitational acceleration of a particle at points from the center of the planet 

$\int Fdr =\int (-Kr)dr$Thus,

$$\begin{gathered} \frac{1}{2}mv{f}^2=\int Fdr \\ =\int (-Kr)dr \end{gathered}$$

where R is the radius of Earth. Solving for the final speed, we obtain$vf=R\sqrt{\frac{K}{m}}$.

$ag=g=9.8m/s^2$on the surfaface of the earth

$$\begin{gathered} ag=\frac{GM}{R^2} \\ =\frac{G\left(4\pi \frac{R^3}{3}\right)\rho }{R^2}, \end{gathered}$$

Where$\rho$ is Earth's average density.

$$\begin{gathered} \frac{K}{m}=\frac{4\pi G\rho }{3} \\ =\frac{g}{R} \end{gathered}$$

$$\begin{gathered} vf=R\frac{K}{m} \\ =R\frac{g}{R} \\ =\sqrt{gR} \\ =\sqrt{(9.8m/s^2)(6.37\times {10}^{6}m)} \end{gathered}$$

$7.9\times {10}^{6}m/s$

The speed of the mailpass through the center of Earth if falling in a tunnel through the center is$7.9\times {10}^{6}m/s$