Question

Find the percentage change in the weight of a person when he moves 16 km above the earth's surface ?​

Answer 1

Answer:

Here d= 32 km R= 6400 km

weight of body at depth d is mg`

=mg(1-d/R)%

decrease in weight =( mg- mg`)/mg × 100= 32 / 6400×100 = 0.5%.

Explanation:

Answer 2

$\huge\sf\underline\red{Solution}$

$\sf{Weight\:of\:body = mg} \\ \\ \sf{Radius = 6400 \:km = 64 \times 10^5 \:m} \\ \\ \sf{Mass\:of\:Earth (m) = 5.98 \times 10^24 \:kg} \\ \\ \sf{ G = 6.673 \times 10^-11 \: Nm^2/kg^2} \\ \\ \implies\sf\pink{g(Near\: earth\: Surface) = GM/(R + h)^2} \\ \\ \sf{g = (5.98 \times 10^24 \times 6.673 \times 20^-11)/(64 \times 10^5)^2} \\ \\ \implies\bold\purple{9.8 m/s^2}$

Hence,

• Since h = 16 km and R = 6400 km

Then,

$\longrightarrow\sf{R+h = 6416 km = 6416 \times 10^3m} \\ \\ \longrightarrow\sf{g = (5.98 \times 10^13 \times 6.67)/(6416 \times 10^3)^2} \\ \\ \longrightarrow\sf{9.7 m/s^2}$

Hence % of decrease in weight :-

$\longrightarrow\sf{100-981.92.97} \\ \\ \longrightarrow\sf{- 1.02%}$

$\sf\underline\pink{Thanks}$