Question

(II) Estimate the kinetic energy of the Earth with respect to the Sun as the sum of two terms, (a) that due to its daily rotation about its axis, and (b) that due to its yearly revolution about the Sun. (Assume the Earth is a uniform sphere with\({\bf{mass}} = {\bf{6}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{24}}}}{\bf{ kg}}\),\({\bf{radius}} = {\bf{6}}{\bf{.4 \times 1}}{{\bf{0}}^{\bf{6}}}{\bf{ m}}\)and is\({\bf{1}}{\bf{.5 \times 1}}{{\bf{0}}^{\bf{8}}}{\bf{ km}}\)from the Sun.)

Short answer

1. The kinetic energy of the Earth due to its daily rotation about its axis is \(2.6 \times {10^{29}}{\bf{ }}{\rm{J}}\).
2. The kinetic energy of the Earth due to its yearly rotation about the Sun is \(2.67 \times {10^{33}}{\rm{ J}}\). The total kinetic energy of the Earth with respect to the Sun is \(2.67 \times {10^{33}}{\rm{ J}}\).

Step-by-step solution

Identification of given data 

The given data can be listed below as:

• The mass of the Earth is\({m_E} = 6.0 \times {10^{24}}{\bf{ }}{\rm{kg}}\).
• The distance between the Sun and the Earth is\(R = 1.5 \times {10^8}{\rm{ km}}\left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right) = 1.5 \times {10^{11}}{\rm{ m}}\).
• The radius of the Earth is \({r_E} = 6.4 \times {10^6}{\rm{ m}}\).
• The angular velocity of the Earth due to its daily rotation about its axis is\({\omega _d} = 1{\rm{ rev/day}}\).
• The angular velocity of the Earth due to its yearly revolution about the Sun is \({\omega _y} = 1{\rm{ rev/year}}\).

Understanding the motion of the Earth with respect to Sun 

The Earth is revolving in an orbit around the Sun. The energy possessed by the Earth while revolving is rotational kinetic energy. The force which acts between the Sun and the planet Earth is gravitational force. Due to this force, elliptical motion of the Earth takes place around the Sun.

(a) Determination of the rotational kinetic energy of the Earth due to its daily rotation about its axis

The diagram can be represented as:

Here,\({r_S}\)is the radius of the Sun, S is the Sun, and E is the Earth.

The rotational kinetic energy of the Earth due to its daily rotation can be expressed as:

\({K_d} = \frac{1}{2}{I_s}\omega _d^2\) … (i)

Here, \({\omega _d}\) is the angular speed of the Earth due to its daily rotation and \({I_s}\) is the moment of inertia of the Earth (sphere).

The angular velocity of the Earth due to its daily rotation about its axis is:

\({\omega _d} = \left( {\frac{{1{\rm{ rev}}}}{{1{\rm{ day}}}}} \right)\left( {\frac{{2\pi {\rm{ rad}}}}{{1{\rm{ rev}}}}} \right)\left( {\frac{{1{\rm{ day}}}}{{24{\rm{ h}}}}} \right)\left( {\frac{{1{\rm{ }}{\rm{h}}}}{{3600{\rm{ s}}}}} \right) = 7.27 \times {10^{ - 5}}{\rm{ rad/s}}\)

Assume the Earth to be a uniform sphere. The moment of inertia of the Earth when considered as a sphere can be expressed as:

\({I_s} = \frac{2}{5}{m_E}r_E^2\)

Substitute the values in equation (i).

\(\begin{align}{K_d} &= \frac{1}{2} \times \frac{2}{5}{m_E}r_E^2 \times \omega _d^2\\{K_d} &= \frac{1}{5}{m_E}r_E^2 \times \omega _d^2\end{align}\)

Substitute the values in the above equation.

\(\begin{align}{K_d} &= \frac{1}{5} \times 6.0 \times {10^{24}}{\bf{ }}{\rm{kg}} \times {\left( {6.4 \times {{10}^6}{\rm{ m}}} \right)^2} \times {\left( {7.27 \times {{10}^{ - 5}}{\rm{ rad/s}}} \right)^2}\\ &= 2.6 \times {10^{29}}{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ J}}}}{{1{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}}}} \right)\\ &= 2.6 \times {10^{29}}{\bf{ }}{\rm{J}}\end{align}\)

Thus, the kinetic energy of the Earth due to its daily rotation about its axis is \(2.6 \times {10^{29}}{\bf{ }}{\rm{J}}\).

(b) Determination of the rotational kinetic energy of the Earth due to its yearly revolution about the Sun 

The rotational kinetic energy of the Earth due to its yearly rotation can be expressed as:

\({K_y} = \frac{1}{2}{I_p}\omega _y^2\) … (ii)

Here,\({\omega _y}\)is the angular speed of the Earth due to its yearly rotation and\({I_p}\)is the moment of inertia of the Earth (point mass).

The angular velocity of the Earth due to its yearly revolution about the Sun is:

\({\omega _y} = \left( {\frac{{1{\rm{ rev}}}}{{1{\rm{ year}}}}} \right)\left( {\frac{{2\pi {\rm{ rad}}}}{{1{\rm{ rev}}}}} \right)\left( {\frac{{1{\rm{ year}}}}{{365{\rm{ day}}}}} \right)\left( {\frac{{1{\rm{ day}}}}{{24{\rm{ h}}}}} \right)\left( {\frac{{1{\rm{ }}{\rm{h}}}}{{3600{\rm{ s}}}}} \right) = 1.99 \times {10^{ - 7}}{\rm{ rad/s}}\)

Assume the Earth to be a point mass. The moment of inertia of the Earth when treating it as a point mass can be expressed as:

\({I_p} = {m_E}{R^2}\)

Substitute the values in equation (ii).

\(\begin{align}{K_y} &= \frac{1}{2} \times {m_E}{R^2} \times \omega _y^2\\{K_y} &= \frac{1}{2} \times {m_E}{R^2} \times \omega _y^2\end{align}\)

Substitute the values in the above equation.

\(\begin{align}{K_y} &= \frac{1}{2} \times 6.0 \times {10^{24}}{\bf{ }}{\rm{kg}} \times {\left( {1.5 \times {{10}^{11}}{\rm{ m}}} \right)^2} \times {\left( {1.99 \times {{10}^{ - 7}}{\rm{ rad/s}}} \right)^2}\\ &= 2.67 \times {10^{33}}{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ J}}}}{{1{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}}}} \right)\\ &= 2.67 \times {10^{33}}{\rm{ J}}\end{align}\)

Thus, the kinetic energy of the Earth due to its yearly revolution about the Sun is \(2.67 \times {10^{33}}{\rm{ J}}\).

Determination of the total kinetic energy of the Earth with respect to the Sun 

The total kinetic energy of the Earth with respect to the Sun can be expressed as:

\(K = {K_d} + {K_y}\)

Substitute the values in the above equation.

\(\begin{align}K &= 2.6 \times {10^{29}}{\bf{ }}{\rm{J}} + 2.67 \times {10^{33}}{\rm{ J}}\\ &= 2.67 \times \times {10^{33}}{\rm{ J}}\end{align}\)

Thus, the total kinetic energy of the Earth with respect to the Sun is \(2.67 \times \times {10^{33}}{\rm{ J}}\).