Question

At what height $(h<<R)$ above the surface of Earth the value of $g$ decreases by $2\%?$ (radius of the Earth is $6400\text{km}$)

• A. $32\text{km}$
• B. $64\text{km}$
• C. $128\text{km}$
• D. $1600\text{km}$

Answer 1

The correct option is B $64\text{km}$

Acceleration due to gravity above the surface of earth at a height h is given $$g^{\prime }=g(1-\frac{2h}{R_e})(h<<<R)...\left(1\right)$$Given,

$g^{\prime }-g=\frac{2hg}{R_e}$

$\frac{g^{\prime }-g}{g}=\frac{2h}{R_e}=2\%$

$\frac{2h}{R_e}=0.02$

$\Rightarrow h=0.01R_e$

Substituting the value of $R_e$,

$\Rightarrow h=0.01\times 6400$

$\therefore h=64\text{km}$

Hence, option (b) is the correct answer.

$\text{Key Concept:}$ Acceleration due to gravity at a height $h$ above the surface of the Earth is given by $$g^{\prime }=g(1-\frac{2h}{R_e})$$.This formula is only valid when $h<<R.$ $\text{Why this question:}$ Caution - This result can be used only when $h<<R$.