Question

The height above the surface of earth at which the gravitational field intensity is reduced to 1% of its value on the surface of earth is:

• A. $100R_e$
• B. $10R_e$
• C. $99R_e$
• D. $9R_e$

Answer 1

The correct option is D $9R_e$

$E\propto \frac{1}{r^2}$
$\frac{E_2}{E_1}={\left(\frac{r_1}{r_2}\right)}^2$
or $\frac{1}{100}={\left(\frac{R_e}{R_e+h}\right)}^2$
or $h=9R_e$