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

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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

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