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The acceleration due to gravity of Earth at a height above the surface of Earth is 1 mm/s2. The distance of this location from the centre of Earth is (Assume g = 10 m/s2, radius of earth = 6400 km)


3200 km


7650 km


8640 km


9600 km

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Correct Answer - Option 4 :

9600 km

The correct answer is option 4) i.e. 9600 km


  • Acceleration due to gravity on the surface of Earth of mass M and radius Re is denoted by g.
    • It has an approximated uniform value of 9.8 m/s2 on the surface of Earth.

The acceleration due to gravity at a depth 'd' below the surface of Earth is given by 

\(⇒ g' = g(1- \frac{d}{R_e})\)

  • The acceleration due to gravity at a height 'h' above the surface of Earth is given by 

\(⇒ g'' = g(1+ \frac{h}{R_e})^{-2}\)

\(⇒ g'' = g(1- \frac{2h}{R_e})\)   for h << Re


Given that:

g'' = 1 mm/s2 = 10-3 m/s2

\(⇒ g'' = g(1- \frac{2h}{R_e})\)

\(⇒ 10^{-3}= 10(1- \frac{2h}{6400})\)

⇒ h = 3200 km 

Therefore, distance from the centre of earth = R + h = 6400 + 3200 = 9600 km.

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