The mass of Earth is 5.972 \times 10^{24}\ \text{kg} and its orbital radius is an average of 1.496 \times 10^{11}\ \text{m}. Calculate its linear momentum.

Solution:

Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum \textbf{p} is defined to be

\textbf{p} = m \textbf{v}

where m is the mass of the system and \textbf{v} is its velocity.

For this problem, the mass has already been given to be m=5.972 \times 10^{24}\ \text{kg}. We now proceed to calculating the velocity. We can solve the velocity of the using the formula

\textbf{v} = \frac{d}{t}

where d is the total distance travelled, and t is the total time of travel. The total distance traveled is just the circumference of the orbit, and the total time of travel is 1 year for 1 full revolution.

\begin{align*}
\textbf{v} & = \frac{d}{t} \\
\textbf{v} & = \frac{2 \pi r}{t} \\
\textbf{v} & = \frac{2 \pi \left( 1.496\ \times 10^{11} \ \text{m} \right)}{365 \frac{1}{4} \ \text{days} \times \frac{24\ \text{hrs}}{\text{day}} \times \frac{3600\ \text{sec}}{1\ \text{hr}}} \\
\textbf{v} & = 29 785. 6783\ \text{m}/\text{s}
\end{align*}

Therefore, its linear momentum is

\begin{align*}
\textbf{p} & = m \textbf{v} \\
\textbf{p} & = \left( 5.972 \times 10^{24}\ \text{kg} \right) \left( 29 785. 6783\ \text{m}/\text{s} \right) \\
\textbf{p} & =  1.78 \times 10^{29}\ \text{kg} \cdot \text{m}/\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}