The gravitational acceleration of an object is inversely proportional to the square of its distance from the center of the Earth. During an airplane flight, a pilot increases their distance from the center of the Earth by $0.5\%$ (compared to being on the ground). What was the percent change in the gravitational acceleration on the pilot?
+14903 What was the percent change in the gravitational acceleration on the pilot?
Hello Guest!
G (basic value), W (percent value), p (percentage)
\(W=\frac{G\cdot p}{100}\\ p=\frac{100W}{G}\\ W=\frac{Gm_1m_2}{1.005^2r^2}\ ,\ G=\frac{Gm_1m_2}{r^2}\)
\(p=100\cdot \frac{Gm_1m_2}{1.005^2r^2}\cdot\frac{r^2}{Gm_1m_2}=\frac{100}{1.005^2}\\ \color{blue}p=99.00745\%\)
Gravity on the pilot is reduced by 100% - 99.007% = 0.993%.
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