Problem 6-12: The approximate total distance traveled by planet Earth since its birth


Taking the age of Earth to be about 4×109 years and assuming its orbital radius of 1.5 ×1011 m has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).


Solution:

First, we need to compute for the linear velocity of the Earth using the formula below knowing that the Earth has 1 full revolution in 1 year

v=rωv=r\omega

where r=1.5×1011 mr=1.5\times 10^{11} \ \text{m} and ω=2π rad/year\omega = 2\pi \ \text{rad/year} . Substituting these values, we have

v=rωv=(1.5×1011 m)(2π rad/year)v=9.4248×1011 m/year\begin{align*} v & = r \omega \\ \\ v & = \left( 1.5\times 10^{11} \ \text{m} \right)\left( 2 \pi \ \text{rad/year} \right) \\ \\ v & = 9.4248\times 10^{11} \ \text{m/year} \end{align*}

Knowing the linear velocity, we can compute for the total distance using the formula

Δx=vΔt\Delta x = v \Delta t

We can now substitute the given values: v=9.4248×1011 m/yearv = 9.4248\times 10^{11} \ \text{m/year} and Δt=4×109 years\Delta t = 4\times 10^{9} \ \text{years} .

Δx=vΔtΔx=(9.4248×1011 m/year)(4×109 years)Δx=3.7699×1021 mΔx=4×1021 m  (Answer)\begin{align*} \Delta x & = v \Delta t \\ \\ \Delta x & = \left( 9.4248\times 10^{11} \ \text{m/year} \right) \left( 4\times 10^{9} \ \text{years} \right) \\ \\ \Delta x & = 3.7699 \times 10^{21} \ \text{m} \\ \\ \Delta x & = 4 \times 10^{21} \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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