Question

An astronomer of 65 kg of mass hikes from the beach to the observatory atop the mountain in Mauna Kea, Hawaii (altitude of 4205 m). By how much (in newtons) does her weight change when she goes from sea level to the observatory?

Answer 1

Answer:

$0.845\ \text{N}$

Explanation:

g = Acceleration due to gravity at sea level = $9.81\ \text{m/s}^2$

R = Radius of Earth = 6371000 m

h = Altitude of observatory = 4205 m

Change in acceleration due to gravity due to change in altitude is given by

$g_h=g(1+\dfrac{h}{R})^{-2}\\\Rightarrow g_h=9.81\times(1+\dfrac{4205}{6371000})^{-2}\\\Rightarrow g_h=9.797\ \text{m/s}^2$

Weight at sea level

$W=mg\\\Rightarrow W=65\times 9.81\\\Rightarrow W=637.65\ \text{N}$

Weight at the given height

$W_h=mg_h\\\Rightarrow W_h=65\times 9.797\\\Rightarrow W_h=636.805\ \text{N}$

Change in weight $W_h-W=636.805-637.65=-0.845\ \text{N}$

Her weight reduces by $0.845\ \text{N}$.