Chapter 13: Q88P (page 384)

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

Short Answer

Expert verified

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:

$K E = \frac{1}{2} m v f^{2}$

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

$\int F d r = \int (- K r) d r$ Thus,

$\frac{1}{2} m v f^{2} = \int F d r = \int (- K r) d r$

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

$a g = g = 9.8 m / s^{2}$ on the surfaface of the earth

$a g = \frac{G M}{R^{2}} = \frac{G (4 π \frac{R^{3}}{3}) ρ}{R^{2}},$

Where $ρ$ is Earth's average density.

$\frac{K}{m} = \frac{4 π G ρ}{3} = \frac{g}{R}$

$v f = R \frac{K}{m} = R \frac{g}{R} = \sqrt{g R} = \sqrt{(9.8 m / s^{2}) (6.37 \times 10^{6} m)}$

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

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

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

Short Answer

Step by step solution

Listing the given quantities

Understanding the concept of the gravitational acceleration

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

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