Question
The universal gravitational constant is approximately$$G=6.67 \times 10^{-11} \mathrm{~m}^{3} / \mathrm{kg} \cdot \mathrm{s}^{2}$$and the semimajor axis of the Earth's orbit is approximately$$a=149.6 \times 10^{6} \mathrm{~km}$$Estimate the mass of the Sun in $\mathrm{kg}$.
Step 1
Given: $$ a = 149.6 \times 10^{6} \mathrm{~km} $$ To convert kilometers to meters, we multiply by \( 10^3 \): $$ a = 149.6 \times 10^{6} \times 10^{3} \mathrm{~m} = 149.6 \times 10^{9} \mathrm{~m} $$ Show more…
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The astronomical unit (AU, equal to the mean radius of the Earth's orbit) is $1.4960 \cdot 10^{11} \mathrm{~m},$ and a year is $3.1557 \cdot 10^{7}$ s. Newton's gravitational constant is $G=$ $6.6743 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} .$ Calculate the mass of the Sun in kilograms. (Recalling or looking up the mass of the Sun does not constitute a solution to this problem.)
The astronomical unit (AU, equal to the mean radius of the Earth's orbit) is $1.4960 \cdot 10^{11} \mathrm{~m}$, and a year is $3.1557 \cdot 10^{7} \mathrm{~s}$. Newton's gravitational constant is $G=6.6738 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} .$ Calculate the mass of the Sun in kilograms. (Recalling or looking up the mass of the Sun does not constitute a solution to this problem.)
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