Question

At what altitude above the Earth's surface is the acceleration due to gravity equal to g/3?

Answer 1

This question involves the concepts of altitude and acceleration due to gravity.

The altitude above Earth's Surface where the acceleration due to gravity equals g/3 is "2.13 x 10⁶ m".

VARIATION IN "g" WITH ALTITUDE:

Variation in the value of the acceleration due to gravity with the change in altitude is given by the following formula:

$g'=g(1-\frac{2h}{R_E})$

where,

• g' = acceleration due to gravity at given altitude = g/3
• g = acceleration due to gravity on the surface of the Earth = 9.81 m/s²
• h = altitude = ?
• $R_E$ = Radius of Earth = 6.4 x 10⁶ m

Therefore,

$\frac{g}{3}=g(1-\frac{2h}{R_E})\\\\ \frac{2h}{R_E}=1-\frac{1}{3}\\\\ \frac{h}{R_E}=\frac{2}{3(2)}\\\\ h=\frac{R_E}{3}=\frac{6.4\ x\ 10^6\ m}{3}$

h = 2.13 x 10⁶ m

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