Question

The period of oscillation of an object in a frictionless tunnel running through the center of the Moon is \(T=2 \pi / \omega_{0}=6485 \mathrm{~s},\) as shown in Example 14.2. What is the period of oscillation of an object in a similar tunnel through the Earth \(\left(R_{\mathrm{E}}=6.37 \cdot 10^{6} \mathrm{~m} ; R_{\mathrm{M}}=1.74 \cdot 10^{6} \mathrm{~m} ; M_{\mathrm{E}}=5.98 \cdot 10^{24} \mathrm{~kg}\right.\) \(\left.M_{\mathrm{M}}=7.35 \cdot 10^{22} \mathrm{~kg}\right) ?\)

Short answer

To summarize, the period of oscillation for an object going through a frictionless tunnel in Earth is approximately 5068 seconds, while the period of oscillation for an object going through a similar tunnel in Moon is 6485 seconds. The relationship between the period of oscillation and the radius and mass of a celestial body allowed us to find the period of oscillation for Earth using the known masses and radii of Earth and Moon.

Step-by-step solution

Given information

We have the period of oscillation for an object going through a tunnel at the center of the Moon:\(T_{M}=6485 \mathrm{~s}.\) We also have Earth's and Moon's mass and radius: \(R_{\mathrm{E}}=6.37 \cdot 10^{6}\mathrm{~m}\) , \(R_{\mathrm{M}}=1.74 \cdot 10^{6}\mathrm{~m}\) , \(M_{\mathrm{E}}=5.98 \cdot 10^{24}\mathrm{~kg}\) , \(M_{\mathrm{M}}=7.35 \cdot 10^{22}\mathrm{~kg}.\)

Find the relationship between period of oscillation and radius and mass

The period of oscillation formula in a frictionless tunnel is given by: \(T=2 \pi / \omega_{0}\) To find the relationship between \(T\), \(R\), and \(M\), we can use the equation for the gravitational field inside a sphere with uniform density: \(g(r)=-\cfrac{4}{3}\pi G \rho r\) where \(\rho\) is the uniform density of the sphere and \(G\) is the gravitational constant. We can replace \(\rho\) in terms of mass and radius of the celestial body: \(\rho = \cfrac{M}{\cfrac{4}{3}\pi R^{3}}\) Now, substituting the expression for \(\rho\) into the equation for \(g(r)\), we get: \(g(r)=-\cfrac{GM}{R^{3}}r\) Since the centripetal force equals the gravitational force, we have: \(\cfrac{mv^{2}}{r} = \cfrac{GMm}{R^{3}}r\) Simplifying and solving for \(v\), we find: \(v = \sqrt{\cfrac{GM}{R^2}}\) Now, we know that the angular frequency \(\omega_{0}\) is related to the velocity by: \(\omega_{0} = \cfrac{v}{R}\) Substituting the expression for \(v\) into the equation for \(\omega_{0}\), we get: \(\omega_{0} = \cfrac{\sqrt{\cfrac{GM}{R^2}}}{R}\) Finally, we can substitute this expression for \(\omega_{0}\) into the period of oscillation formula to find the relationship between \(T\), \(R\), and \(M\): \(T = \cfrac{2\pi}{\omega_{0}} = \cfrac{2\pi R}{\sqrt{\cfrac{GM}{R^2}}}\)

Calculate the period of oscillation for Earth

Now that we have derived the relationship between the period of oscillation, radius, and mass, we can determine the period of oscillation for a tunnel going through Earth. For Earth, we have the mass \(M_{\mathrm{E}}=5.98 \cdot 10^{24}\mathrm{~kg}\) and the radius \(R_{\mathrm{E}}=6.37 \cdot 10^{6}\mathrm{~m}\). Substituting these values into the formula, we get: \(T_{E} = \cfrac{2\pi R_{\mathrm{E}}}{\sqrt{\cfrac{GM_{\mathrm{E}}}{R^2_{\mathrm{E}}}}}\) Plugging in the values, and using the gravitational constant \(G = 6.67430 \times 10^{-11} \mathrm{m^3 kg^{-1} s^{-2}}\): \(T_{E} = \cfrac{2\pi \cdot 6.37 \cdot 10^{6} \mathrm{m}}{\sqrt{\cfrac{6.67430 \times 10^{-11} \mathrm{m^3 kg^{-1} s^{-2}} \cdot 5.98 \cdot 10^{24} \mathrm{kg}}{(6.37 \cdot 10^{6} \mathrm{m})^2}}}\) Once you've calculated the result, you find: \(T_{E} \approx 5068 \mathrm{~s}\) So, the period of oscillation for an object going through a tunnel in Earth is approximately 5068 seconds.

Key concepts

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. SHM can be observed in various physical systems, such as the oscillation of a mass on a spring or the motion of a pendulum for small angles.

In the context of an object oscillating through a frictionless tunnel through a celestial body like the Moon or Earth, the motion is simple harmonic because the gravitational force provides the restoring force necessary for SHM. The object's motion through the tunnel exhibits characteristics of SHM, with the gravitational pull being maximal at the ends of the tunnel and zero at the center, similar to the forces experienced by a pendulum at the extremities of its swing and at the bottom of its arc.

Gravitational Field Inside a Sphere

The gravitational field inside a spherical object with uniform density, such as a planet, can be determined by using the principles of classical physics. Inside the sphere, the gravitational field strength varies linearly with distance from the center, meaning the closer to the center, the weaker the gravitational force becomes. This relationship is described by the equation:

\(g(r) = -\frac{4}{3}\pi G \rho r\)

where \(g(r)\) is the gravitational field at a distance \(r\) from the center of the sphere, \(G\) is the gravitational constant, \(\rho\) is the density of the sphere, and \(r\) is the radial distance. For the object in the tunnel, this linear variation in the gravitational field results in a force that acts like a restoring force in SHM, pulling the object back towards the center each time it tries to move away, thus contributing to the oscillatory motion.

Relationship Between Period, Radius, and Mass

To comprehend the oscillatory dynamics of an object passing through a celestial body like the Moon or Earth, we can analyze the relationship between the period of oscillation (\(T\)), the body's radius (\(R\)), and its mass (\(M\)). The period of oscillation is the time it takes for one complete cycle of motion.

The derived formula from classical mechanics for the period of oscillation inside a tunnel through a spherical object is given by:

\(T = \frac{2\pi R}{\sqrt{\frac{GM}{R^2}}}\)

This formula demonstrates that the period of oscillation is directly proportional to the radius and inversely proportional to the square root of the mass of the celestial body. This relationship is critical for understanding how changes in the dimensions or mass of Earth or the Moon would affect the period of an object's oscillation through them. For instance, a larger radius or a smaller mass would result in a longer period, and vice versa.