Tag Archives: earth mass orbital radius and linear momentum

College Physics by Openstax Chapter 8 Problem 6

The mass of Earth is 5.972×1024 kg5.972 \times 10^{24}\ \text{kg} and its orbital radius is an average of 1.496×1011 m1.496 \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 p\textbf{p} is defined to be

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

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

For this problem, the mass has already been given to be m=5.972×1024 kgm=5.972 \times 10^{24}\ \text{kg}. We now proceed to calculating the velocity. We can solve the velocity of the using the formula

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

where dd is the total distance travelled, and tt 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.

v=dtv=2πrtv=2π(1.496 ×1011 m)36514 days×24 hrsday×3600 sec1 hrv=29785.6783 m/s\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

p=mvp=(5.972×1024 kg)(29785.6783 m/s)p=1.78×1029 kg⋅m/s  (Answer)\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*}