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 ]

Step-by-step solution

The gravitational acceleration at any distance r from the center of Earth is a g = GM E r 2 . On differentiating, we get da g = - 2 GM E r 3 dr , where da g is infinitesimal change in a due to differential change dr in r . ⇒ da g = 2 6.67 × 10 - 11 6 × 10 24 6 .67 3 × 10 6 3 × 1.50 = 1.8 × 10 - 6   m   s - 2 ∴  N = 1 . 8