1. Problem statement: Find the orbital speed of Earth in a circular orbit around the Sun given the gravitational constant, the Sun's mass, and Earth's orbital radius. 2. Formula: The formula for circular orbital speed is $$v = \sqrt{\frac{GM}{r}}$$ 3. Important rules: We assume a circular orbit so the formula applies, all quantities must be in SI units to keep units consistent, and $r$ is the distance from the center of the Sun to the orbiting body. 4. Given values: Use $G = 6.674\times10^{-11}\ \text{m}^3\text{kg}^{-1}\text{s}^{-2}$ $M = 1.989\times10^{30}\ \text{kg}$ $r = 1.496\times10^{11}\ \text{m}$ 5. Substitute the numerical values into the formula: $$v = \sqrt{\frac{6.674\times10^{-11}\times1.989\times10^{30}}{1.496\times10^{11}}}$$ 6. Multiply the constants in the numerator to simplify the radicand: $$v = \sqrt{\frac{1.327124\times10^{20}}{1.496\times10^{11}}}$$ 7. Show cancellation of powers of ten when simplifying the fraction: $$v = \sqrt{\frac{1.327124\times\cancel{10^{20}}}{1.496\times\cancel{10^{11}}}}$$ 8. After cancelling powers of ten we get: $$v = \sqrt{\frac{1.327124}{1.496}\times10^{9}}$$ 9. Compute the division inside the radicand and simplify: $$v = \sqrt{8.876\times10^{8}}$$ 10. Take the square root to find the speed: $$v \approx 2.979\times10^{4}\ \text{m/s}$$ 11. Final answer: The Earth's orbital speed is approximately $2.979\times10^{4}\ \text{m/s}$.