At what height above the earths surface is the acceleration due to gravity is 1% less than its value at the surface?

 India's Super Teachers for all govt. exams Under One Roof Enroll For Free Now CONCEPT: Gravitational acceleration: It is the acceleration gained by an object due to gravitational force.  ​Its SI unit is m/s2. It has both magnitude and direction, hence, it’s a vector quantity. Acceleration due to gravity is represented by g. The standard value of g on the surface of the earth at sea level is 9.8 m/s2. Acceleration due to gravity on the surface is given: g1 = GM/R2 Acceleration due to gravity above the surface is given: g2 = GM/(R + H)2 Where, R = Radius of the earth (6400 km), and H = height in meters CALCULATION: From the above discussion it's clear that $$\frac{{{g_2}}}{{{g_1}}} = \frac{{GM/{{(R + H)}^2}}}{{GM/{R^2}}}$$ $$\frac{{{g_2}}}{{{g_1}}} = \frac{{{R^2}}}{{{{(R + H)}^2}}}$$ $$\frac{{{g_2}}}{{9.8}} = \frac{{{{(6.4 \times {{10}^6})}^2}}}{{{{(6.4 \times {{10}^6} + 0.01 \times {{10}^6})}^2}}}$$ g2 = 9.77 m/s2 The correct option is 9.77 m/s2 Ace your Gravitation preparations for The gravitational constant with us and master Physics for your exams. Learn today! India’s #1 Learning Platform Start Complete Exam Preparation Daily Live MasterClasses Practice Question Bank Mock Tests & Quizzes Get Started for Free Download App Trusted by 3.4 Crore+ Students