Question

At what height in km over the earth's pole the free fall acceleration decreases by one percent? (Assume the radius of the earth to be $6400km$)

• A. $32$
• B. $64$
• C. $80$
• D. $1.253$

Answer 1

The correct option is A $32$

Acceleration on the earth's surface is $g=\frac{GM}{R^2}$ and the acceleration at height h will be $g_h=\frac{GM}{(R+h{)}^2}$

So, $\frac{g_h}{g}=\frac{R^2}{(R+h{)}^2}$

According to question, $\frac{g_h}{g}=\frac{99}{100}$

so, $\frac{99}{100}=\frac{R^2}{(R+h{)}^2}$

or $\frac{R}{(R+h)}=0.995$ or $R=0.995R+0.995h$

or $h=\left(\frac{0.005}{0.995}\right)R=\left(\frac{0.005}{0.995}\right)\left(6400\right)=32.16\simeq 32$ km