Question

the earths radius (to sea level) is 6,371 km and its mass is $5.972\times10^{24}kg$. mount everest rises 8,848 m above sea level. how much does a 60.3 kg woman weigh at sea level? how much does she weigh on the summit of mount everest? (use the general force of gravity equation each time.)

Explanation:

Step1: Calculate weight at sea level

The formula for gravitational force \(F = G\frac{Mm}{r^{2}}\), where \(G = 6.67\times10^{- 11}\space N\cdot m^{2}/kg^{2}\), \(M = 5.972\times10^{24}\space kg\), \(m = 60.3\space kg\), \(r = 6371\times10^{3}\space m\)

$$F_{1}=G\frac{Mm}{r^{2}}=6.67\times 10^{-11}\times\frac{5.972\times 10^{24}\times60.3}{(6371\times 10^{3})^{2}}$$

$$F_{1}=6.67\times 10^{-11}\times\frac{5.972\times 10^{24}\times60.3}{4.058\times10^{13}}$$

$$F_{1}=6.67\times 10^{-11}\times\frac{3.601\times10^{26}}{4.058\times10^{13}}$$

$$F_{1}=6.67\times 10^{-11}\times8.874\times10^{12}$$

$$F_{1}=592.9\space N$$

Step2: Calculate weight on Everest summit

The height of Everest \(h = 8848\space m\), so \(r_{2}=6371\times10^{3}+8848=6379848\space m\)

$$F_{2}=G\frac{Mm}{r_{2}^{2}}=6.67\times 10^{-11}\times\frac{5.972\times 10^{24}\times60.3}{(6379848)^{2}}$$

$$F_{2}=6.67\times 10^{-11}\times\frac{5.972\times 10^{24}\times60.3}{4.07\times10^{13}}$$

$$F_{2}=6.67\times 10^{-11}\times\frac{3.601\times10^{26}}{4.07\times10^{13}}$$

$$F_{2}=6.67\times 10^{-11}\times8.848\times10^{12}$$

$$F_{2}=590.2\space N$$

Answer:

At sea level, the woman weighs \(592.9\space N\). On the summit of Mount Everest, she weighs \(590.2\space N\)