Question

At what height above the surface of earth acceleration due to gravity reduces by 1%?
- R7​

Answer 1

Answer:

Let the height be x .

Now gravitational acceleration is inversely proportional to square of distance from centre of earth .

g reduced 1% or remain 99% .

Let radius of earth be r.

$g \: \: \alpha \: \: \frac{1}{ {d}^{2} } \\ \\ \frac{g}{0.99g} = \frac{ \frac{1}{ {r}^{2} } }{ \frac{1}{(r + x) {}^{2} } } \\ \\99 (r + x) {}^{2} = 100 {r}^{2} \\ \\ \sqrt{99} (r + x) = 10r \\ \\ x = \frac{10 - 3 \sqrt{11} }{3 \sqrt{11} } r$

r = 6400 km

$x \approx32.4 \: kms$