Question

It is sometimes said that if the entire population of China stood on chairs and jumped off simultaneously, it would alter the rotation of the Earth. Fortunately, physics gives us the tools to investigate such speculations. a) Calculate the moment of inertia of the Earth about its axis. For simplicity, treat the Earth as a uniform sphere of mass \(m_{\mathrm{E}}=5.977 \cdot 10^{24} \mathrm{~kg}\) and radius \(6371 \mathrm{~km}\). b) Calculate an upper limit for the contribution by the population of China to the Earth's moment of inertia, by assuming that the whole group is at the Equator. Take the population of China to be 1.30 billion people, of average mass \(70.0 \mathrm{~kg}\) c) Calculate the change in the contribution in part (b) associated with a \(1.00-\mathrm{m}\) simultaneous change in the radial position of the entire group. d) Determine the fractional change in the length of the day the change in part (c) would produce.

Short answer

In this exercise, we calculated the moment of inertia of the Earth about its axis to be approximately \(9.696 \cdot 10^{37}\ \mathrm{kg}\cdot \mathrm{m}^2\). We also determined the maximum contribution to the Earth's moment of inertia by the population of China at the equator to be \(3.669 \cdot 10^{34}\ \mathrm{kg}\cdot \mathrm{m}^2\). When considering a 1-meter simultaneous jump by China's population, the change in the moment of inertia was calculated to be approximately \(1.145 \cdot 10^{28}\ \mathrm{kg}\cdot \mathrm{m}^2\). This change resulted in a fractional change in the length of a day of approximately \(1.181 \cdot 10^{-10}\), which is very small and negligible.

Step-by-step solution

Moment of inertia of a uniform sphere

The formula for the moment of inertia of a sphere rotating about its axis is given by \(I_{sphere} = \frac{2}{5}MR^2\), where \(M\) is the mass of the sphere and \(R\) is its radius. For the Earth, we are given \(m_{\mathrm{E}}=5.977 \cdot 10^{24} \mathrm{~kg}\) and the radius \(R_{\mathrm{E}} = 6371 \mathrm{~km}\).

Calculate the moment of inertia of the Earth

Now, we will plug the values for the Earth's mass and radius into the formula for the moment of inertia of a sphere, obtaining: $$ I_{\mathrm{E}} =\frac{2}{5}m_{\mathrm{E}}R_{\mathrm{E}}^2 = \frac{2}{5}(5.977 \cdot 10^{24}\mathrm{kg})(6371 \cdot 10^3 \mathrm{m})^2 $$

Calculate the numerical value of the Earth's moment of inertia

Performing the calculations, we find: $$ I_{\mathrm{E}} \approx 9.696 \cdot 10^{37}\ \mathrm{kg}\cdot \mathrm{m}^2 $$ So, the moment of inertia of the Earth about its axis is approximately \(9.696 \cdot 10^{37}\ \mathrm{kg}\cdot \mathrm{m}^2\). #b) Calculate an upper limit for the contribution by the population of China to the Earth's moment of inertia#

Moment of inertia of a point mass

The formula for the moment of inertia of a point mass is given by \(I_{point} = mR^2\), where \(m\) is the mass of the point mass and \(R\) is the distance from the rotation axis. Since we are assuming that the whole population of China is at the Equator, we can use the Earth's radius as the distance for each person.

Calculate the total mass of the population

The population of China is given as 1.30 billion people, with an average mass of 70.0 kg. Multiplying the average mass by the total number of people to obtain the total mass: $$ m_{\mathrm{China}} = 1.30 \cdot 10^9 \mathrm{~people} \times 70.0\mathrm{~kg} = 9.10 \cdot 10^{10}\mathrm{~kg} $$

Calculate the maximum contribution to the Earth's moment of inertia

Now, we will use the formula for the moment of inertia of a point mass to find the maximum contribution from the population of China to the Earth's moment of inertia, considering them all at the Equator: $$ I_{\mathrm{China}} = m_{\mathrm{China}}R_{\mathrm{E}}^2 = (9.10 \cdot 10^{10}\mathrm{kg})(6371 \cdot 10^3 \mathrm{m})^2 $$

Calculate the numerical value of the maximum contribution of China's population to Earth's moment of inertia

Performing the calculations, we find: $$ I_{\mathrm{China}} \approx 3.669 \cdot 10^{34}\ \mathrm{kg}\cdot \mathrm{m}^2 $$ So, the maximum contribution to the Earth's moment of inertia by the population of China is \(3.669 \cdot 10^{34}\ \mathrm{kg}\cdot \mathrm{m}^2\). #c) Calculate the change in contribution in part (b) associated with a \(1.00-\mathrm{m}\) simultaneous change in the radial position of the entire group#

Calculate the moment of inertia for the changed radial position

In this step, we want to calculate the moment of inertia for the changed radial position by adding a simultaneous 1-meter vertical jump. This means that the new moment of inertia would be \(I_{\mathrm{China,new}} = m_{\mathrm{China}} (R_{\mathrm{E}} + 1\mathrm{m})^2\), where \(m_{\mathrm{China}}\) is the total mass of the population of China and \(R_{\mathrm{E}}\) is the radius of the Earth.

Calculate the change in the moment of inertia

Now, we will calculate the change in the moment of inertia due to the 1-meter simultaneous jump: $$ \Delta I = I_{\mathrm{China,new}} - I_{\mathrm{China,old}} = m_{\mathrm{China}} ((R_{\mathrm{E}} + 1\mathrm{m})^2 - R_{\mathrm{E}}^2) $$

Calculate the numerical value of the change in moment of inertia

Plugging the values for \(m_{\mathrm{China}}\), \(R_{\mathrm{E}}\), and \(1\mathrm{m}\) into the equation above, we find: $$ \Delta I \approx 1.145 \cdot 10^{28}\ \mathrm{kg}\cdot \mathrm{m}^2 $$ So, the change in the moment of inertia associated with a 1-meter simultaneous jump by the population of China is approximately \(1.145 \cdot 10^{28}\ \mathrm{kg}\cdot \mathrm{m}^2\). #d) Determine the fractional change in the length of the day the change in part (c) would produce#

Calculate the fractional change in moment of inertia

The fractional change in moment of inertia due to the jump can be calculated using the following relationship: $$ \frac{\Delta I}{I_{\mathrm{E}}} = \frac{1.145 \cdot 10^{28}\ \mathrm{kg}\cdot \mathrm{m}^2}{9.696 \cdot 10^{37}\ \mathrm{kg}\cdot \mathrm{m}^2} \approx 1.181 \cdot 10^{-10} $$ The fractional change in moment of inertia is approximately \(1.181 \cdot 10^{-10}\).

Calculate the fractional change in the length of the day

According to the conservation of angular momentum, the fractional change in the length of the day due to the change in the Earth's moment of inertia should be equal to the fractional change in moment of inertia. Therefore, the fractional change in the length of a day due to the \(1.00-\mathrm{m}\) simultaneous jump by the population of China would be approximately \(1.181 \cdot 10^{-10}\). This value is quite small and negligible, hence the jump would not significantly alter the Earth's rotation.

Key concepts

Angular Momentum Conservation

The principle of angular momentum conservation is central to understanding rotational dynamics and can be applied to various systems, from atoms to astronomical objects. It states that if no external torque acts on a system, the total angular momentum of the system remains constant over time.

In our scenario involving the Earth's rotation, if nothing else in the universe exerts a twisting force on the Earth, the angular momentum stays the same. Mathematically, the angular momentum, denoted as L, is the product of the moment of inertia (I), which depends on the mass distribution relative to the rotation axis, and the angular velocity (ω), which is the rate of rotation. The conservation law can be written as:
\[L = I \times \text{ω} = \text{constant}\]

When the population of China hypothetically jumps, they slightly affect the Earth's distribution of mass. However, due to angular momentum conservation, the Earth's rotation (angular velocity) would have to adjust to maintain a constant angular momentum, leading to an infinitesimally small alteration in the length of the day. This notion is crucial when studying rotational mechanics and can explain phenomena such as why spinning ice skaters can speed up their rotations simply by tucking in their arms.

Uniform Sphere Inertia

The concept of uniform sphere inertia is central to solving the given textbook problem. The moment of inertia is a measure of an object's resistance to changes in its rotation rate. It is affected by the object's mass and how that mass is distributed with respect to the rotation axis.

For a solid uniform sphere, like the simplified model of the Earth used in the problem, the moment of inertia is calculated using the formula:\[I_{\text{sphere}} = \frac{2}{5}MR^2\]

Where \(M\) is the mass of the sphere and \(R\) is the radius. The uniform mass distribution implies that this formula gives an accurate representation of how the mass contributes to the rotational resistance. In this case, the resistance to rotation would be consistent across all points of the sphere.

Understanding this concept allows us to not only calculate such moments of inertia but also to appreciate the factors that can affect an object's rotational behavior, including size, shape, and mass distribution.

Earth Rotation Alteration

Talking about altering the Earth's rotation might seem like the stuff of science fiction, but it's based on real physical principles. The effect of the entire population of China jumping off chairs and onto the ground might seem significant, but physics shows us that this would have an imperceptible impact on Earth's rotation.

This is because the Earth is incredibly massive, and while the mass and the distribution of the population are sizable, they pale in comparison to the Earth's mass. Plus, the average distance from the Earth's axis (the radius) ensures any changes in mass distribution would have a minuscule effect on the rotation.

To bring the concept home, let's imagine we could alter the Earth's rotation. Such a change would be tangible in the length of a day, defined as one complete rotation around its axis. But according to the calculations from the textbook problem, the fractional change in the length of the day caused by the population's jump is around \(1.181 \times 10^{-10}\), an exceedingly small number. This exercise helps students grasp the relative scale and mass distribution's role in rotational dynamics and the immense forces required to cause any significant alteration in celestial bodies like Earth.