Question

An astronaut whose height is $1.50 \,m$ floats "feet down" in an orbiting space shuttle at $a_g$ 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 $N \times 10^{-5} \, ms ^{-2}$. What is the value of $N$ ?
[mass of earth $=6 \times 10^{24} \,kg$ and $\left.G=6.67 \times 10^{-11} \,N m^2 \,k g^{-2}\right]$

Answer 1

The gravitational acceleration at any distance $r$ from the centre of the earth is $a_g=\frac{G M_E}{r^2}$
On differentiating, we get where $d a_g$ is an infinitesimal change due to differential change $d a_g$ in
$\Rightarrow\left|d a_g\right|=2 \frac{\left(6.67 \times 10^{-11}\right)\left(6 \times 10^{24}\right)}{\left(\sqrt[3]{6.67} \times 10^6\right)^3} \times(1.50)=18 \times 10^{-5}$