Question

An astronaut whose height is $1.50cm$ floats "feet down" in an orbiting space shuttle at a distance $r=\sqrt[3]{6.67}\times 10^{6}m$ away from the centre of Earth. The magnitude of difference between the gravitational acceleration at her feet and at her head is found to be $N\times 10^{-6}m{s}^{-2}$. What is the value of $N$ ? $[M_E=6\times 10^{24}kg$ and $G=6.67\times 10^{-11}N{m}^{2}k{g}^{-2}]$

Answer 1

Answer: 180

The gravitational acceleration at any distance $r$ from the center of earth is
$a_g=\frac{G{M}_E}{r^2}$.
On differentiating, we get
$d{a}_g=-2\frac{G{M}_E}{r^3}dr$,
where $d{a}_g$ is infinitesimal change in a due to differential change $dr$ in $r$.
$\Rightarrow \left|d{a}_g\right|=2\frac{\left(6.67\times 10^{-11}\right)\left(6\times 10^{24}\right)}{{\left(\sqrt[3]{6.67}\times 10^6\right)}^3}\times 1.50$
$=180\times 10^{-6}m{s}^{-2}$
$\therefore N=180$