Physics

Physics

Problem

A laboratory determines the local value of gg on Earth to be 9.80 m s−29.80\,\text{m s}^{-2}. Using Earth’s mass 5.97×1024 kg5.97\times10^{24}\,\text{kg} and radius 6.37×106 m6.37\times10^{6}\,\text{m}, compute the experimental value of the universal gravitational constant GG. a) 6.67×10−10 N m2kg−26.67\times10^{-10}\,\text{N m}^{2}\text{kg}^{-2} b) 6.67×10−11 N m2kg−26.67\times10^{-11}\,\text{N m}^{2}\text{kg}^{-2} c) 4.00×10−11 N m2kg−24.00\times10^{-11}\,\text{N m}^{2}\text{kg}^{-2} d) 6.67×10−12 N m2kg−26.67\times10^{-12}\,\text{N m}^{2}\text{kg}^{-2}
Intuitive explanation

The key is recognizing that surface gravity gg is just Newton's law of universal gravitation applied at Earth's surface—the gravitational force per unit mass. Rearrange g=GMR2g = \frac{GM}{R^2} to solve for GG directly using the measured gg, Earth's mass MM, and radius RR. This is a straightforward algebraic inversion that lets you extract a fundamental constant from everyday measurements.

Intuitive explanation

The key is recognizing that surface gravity gg is just Newton's law of universal gravitation applied at Earth's surface—the gravitational force per unit mass. Rearrange g=GMR2g = \frac{GM}{R^2} to solve for GG directly using the measured gg, Earth's mass MM, and radius RR. This is a straightforward algebraic inversion that lets you extract a fundamental constant from everyday measurements.

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