Calculate the effective value of g, the acceleration of gravity, at (a) 6400 m, and (b) 6400 km, above the Earth’s surface.

The distance above the Earth’s surface is $r_1=6400\mathrm{m}$.

The distance above the Earth’s surface is $r_2=6400\mathrm{km}$.

In the relation of acceleration due to gravity, r is the distance from the centre of the planet, which will be equal to the sum of radius of the Earth and the distance above the Earth’s surface.

The relation of acceleration due to gravityis given by,

$\begin{array}{l}g=G\frac{m}{R^2}\\ g=G\frac{m}{{\left(r_E+r_1\right)}^2}\end{array}$

Here, G is the gravitational constant, m is the mass of the Earth, $r_E$is the radius of the Earth and R is the distance from the centre of the planet.

On plugging the values in the above relation.

$\begin{array}{l}g=\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\frac{\left(5.98\times {10}^{24}\mathrm{kg}\right)}{{\left(6.38\times {10}^6\mathrm{m}+6400\mathrm{m}\right)}^2}\\ g=9.75\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, $g=9.75\mathrm{m}/{\mathrm{s}}^2$is the required value of acceleration due to gravity.

The relation of acceleration due to gravityis given by,

$\begin{array}{l}g\text{'}=G\frac{m}{R^2}\\ g\text{'}=G\frac{m}{{\left(r_E+r_2\right)}^2}\end{array}$

On plugging the values in the above relation.

$\begin{array}{l}g\text{'}=\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\frac{\left(5.98\times {10}^{24}\mathrm{kg}\right)}{{\left(6.38\times {10}^6\mathrm{m}+6400\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\right)}^2}\\ g\text{'}=2.44\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, $g\text{'}=2.44\mathrm{m}/{\mathrm{s}}^2$is the required value of acceleration due to gravity.