An object of mass $50 \mathrm{~kg}$ falls from a height $H_A = R_E$ to a lower height $H_B = \frac{R

Question

An object of mass $50 \mathrm{~kg}$ falls from a height $H_A = R_E$ to a lower height $H_B = \frac{R_E}{2}$ above the earth's surface. The change in its weight corrected to the nearest integer is ($R_E$ is the radius of the earth, $g = 10 \mathrm{~m/s}^2$)

Professor PC Thomas Classes · Answer

(a) : Weight at a height $h$ is $W_h = mg \left(\frac{R_E}{R_E+h}ight)^2$. Given surface weight $mg = 50 	imes 10 = 500 \mathrm{~N}$. At height $H_A = R_E$, $W_A = 500 	imes \left(\frac{R_E}{R_E+R_E}ight)^2 = 500 	imes \left(\frac{1}{2}ight)^2 = 125 \mathrm{~N}$. At height $H_B = R_E/2$, $W_B = 500 	imes \left(\frac{R_E}{R_E + 0.5R_E}ight)^2 = 500 	imes \left(\frac{2}{3}ight)^2 = 500 	imes \frac{4}{9} = \frac{2000}{9} \approx 222.22 \mathrm{~N}$. The change in weight is $W_B - W_A = 222.22 - 125 = 97.22 \mathrm{~N}$. Rounded to the nearest integer, it is $97 \mathrm{~N}$.