Question

Find the percentage decrease in the weight of a body when taken to a height of 16 km above the surface of earth. (R = 6400 km )

Answer 1

The value of acceleration due to gravity at a height $\sf{h}$ from the surface of the earth of radius $\sf{R}$ is,

$\longrightarrow\sf{g'=g\left(1+\dfrac{h}{R}\right)^{-2}\quad\quad\dots(1)}$

But if $\sf{h\ \textless\textless\ R,\quad\!\dfrac{h}{R}\ \textless\textless\ 1.}$ Hence we know that,

$\longrightarrow\sf{(1+x)^{-2}=1-2x,\quad\! x\ \textless\textless\ 1}$

Taking $\sf{x=\dfrac{h}{R},}$

$\longrightarrow\sf{\left(1+\dfrac{h}{R}\right)^{-2}=1-\dfrac{2h}{R}}$

Thus (1) becomes,

$\longrightarrow\sf{g'=g\left(1-\dfrac{2h}{R}\right)\quad\quad\dots(2)}$

Then percentage decrease in the value of acceleration due to gravity is,

$\longrightarrow\sf{\delta g=\dfrac{g'-g}{g}\times100}$

$\longrightarrow\sf{\delta g=\dfrac{g\left(1-\dfrac{2h}{R}\right)-g}{g}\times100}$

$\longrightarrow\sf{\delta g=\dfrac{g\left[\left(1-\dfrac{2h}{R}\right)-1\right]}{g}\times100}$

$\longrightarrow\sf{\delta g=\left[1-\dfrac{2h}{R}-1\right]100}$

$\longrightarrow\sf{\delta g=-\dfrac{200h}{R}\quad\quad\dots(3)}$

We know that, weight of a body,

$\longrightarrow\sf{W=mg}$

Hence percentage change in weight will be,

$\longrightarrow\sf{\delta W=\delta m+\delta g}$

Since mass of body remains constant, $\sf{\delta m=0.}$ Then,

$\longrightarrow\sf{\delta W=\delta g}$

From (3),

$\longrightarrow\sf{\delta W=-\dfrac{200h}{R}}$

According to the question,

• $\sf{h=16\ km}$

• $\sf{R=6400\ km}$

Hence,

$\longrightarrow\sf{\delta W=-\dfrac{200\times16}{6400}}$

$\longrightarrow\underline{\underline{\sf{\delta W=-0.5\,\%}}}$

Hence the percentage decrease is $\bf{0.5\,\%.}$ Negative sign indicates the decrease in the value.

Answer 2

• A body is taken 32 km above the surface of the earth
• Radius of earth = 6400 km
• Percentage decrease in weight of the body = ?

$\begin{gathered}\begin{gathered}\sf :\implies g' = g\bigg\lgroup 1 - \dfrac{h}{R}\bigg\rgroup\;\;-eq(1)\\\end{gathered}\end{gathered}$

• So then the % decrease in the value of acceleration due to gravity is gonna be,

$\begin{gathered}\begin{gathered}\sf :\implies \delta g = \dfrac{g - g'}{g} \times 100\\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\sf :\implies \delta g = \dfrac{g - g\bigg\lgroup 1 - \dfrac{h}{R}\bigg\rgroup}{g} \times 100\\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\sf :\implies \delta g = \dfrac{g\bigg\{ 1 - \bigg\lgroup 1 - \dfrac{h}{R}\bigg\rgroup\bigg\} }{g} \times 100\\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\sf :\implies \delta g = \bigg\{ 1 - 1 + \dfrac{h}{R} \bigg\} 100\\\end{gathered}\end{gathered}$

$\sf :\implies \delta g = \dfrac{100h}{R} \;\; -eq(2)$

• Finally, the % change in weight will be given by,

$\begin{gathered}\begin{gathered}\sf :\implies \delta W = \delta m + \delta g\\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\sf :\implies \delta W = \dfrac{100h}{R}\\\end{gathered}\end{gathered}$

$\displaystyle \underline{\bigstar\:\textsf{According to the Question :}}$

$\sf\dashrightarrow \delta W = \dfrac{100\times 32}{6400}$

$\dashrightarrow\underline{\boxed{\purple{\mathfrak {\delta w = 0.5 \%}}}}$