An astronaut whose height is 1.50 m floats "feet down" in an orbiting space shuttle at a distance r = 6 . 67 3 × 10 6 m away from the centre of Earth. The magnitude of difference between the gravitati

Question

An astronaut whose height is 1.50 m floats "feet down" in an orbiting space shuttle at a distance r = 6 . 67 3 × 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 × 10 - 6 m s - 2 . What is the value of N ? [ M E = 6 × 10 24 kg and G = 6 . 67 × 10 - 11 N m 2 kg - 2 ]

TestHub · Accepted Answer

The gravitational acceleration at any distance $r$ from the center of Earth is $a_{g} = \frac{{GM}_{E}}{r^{2}}$ .
On differentiating, we get ${da}_{g} = - 2 \frac{{GM}_{E}}{r^{3}} dr$ , where ${da}_{g}$ is infinitesimal change in a due to differential change $dr$ in $r$ .

$\Rightarrow |{da}_{g}| = 2 \frac{(6.67 \times 10^{- 11}) (6 \times 10^{24})}{{(\sqrt[3]{6 .67} \times 10^{6})}^{3}} \times (1.50)$

$= 1.8 \times 10^{- 6} m s^{- 2}$

$∴ N = 1.8$

Physics - Gravitation Question with Solution | TestHub

Doubts & Discussion