To determine the best representation of the Earth's orbital speed around the Sun based on the given information that Earth travels about 2.6 million kilometers in one day, we can convert kilometers into meters and then calculate the speed in meters per day.
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Convert 2.6 million kilometers to meters: \[ 2.6 \text{ million kilometers} = 2.6 \times 10^6 \text{ kilometers} = 2.6 \times 10^6 \times 10^3 \text{ meters} = 2.6 \times 10^9 \text{ meters} \]
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Since the Earth travels approximately \(2.6 \times 10^9\) meters in one day, the speed in meters per day is: \[ \text{Speed} = 2.6 \times 10^9 \text{ meters per day} \]
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To convert \(\text{meters per day}\) into \(\text{meters per second}\), we use the fact that there are \(86400\) seconds in one day: \[ \text{Speed} = \frac{2.6 \times 10^9 \text{ meters per day}}{86400 \text{ seconds per day}} \approx 30138.9 \text{ meters per second} \] Rounding to two significant figures, that is approximately \(3.0 \times 10^4\) meters per second.
Given the options:
A. \(3 \times 10^{-4}\) meters per day
B. \(3 \times 10^{4}\) meters per second
C. \(3 \times 10^{-4}\) meters per second
D. \(3 \times 10^{4}\) meters per day
The best representation of Earth's orbital speed around the Sun is B. \(3 \times 10^4\) meters per second.