91Ó°ÊÓ

Height to be calculated for free-fall acceleration to be $10\%$of its value at the surface.

Acceleration due to gravity at height h above the surface is given by : $g_h=\frac{g{R}^2}{(R+h{)}^2}$

Where, g = value at surface.

$g_h$= value at height h.

R = radius of earth.

h = height above the surface of earth.

Here,

$$\begin{gathered} g_h=10\%\text{of}g,\text{so}{g}_h=0.1g \\ R=6.37×{10}^6\mathrm{m} \end{gathered}$$

At a height of $1.38×{10}^{7}m$above the earth the free-fall acceleration is$10\%$of its value at the surface.

The height of the satellite is$1.38×{10}^{7}m.$

For a satellite to revolve around the planet the necessary centripetal force is provided by the gravitational force of attraction between the planet and the satellite.

Now, $F_c=\text{centripetal force}=\frac{m{v}^2}{r}$

Where orbital radius of satellite : r = R + h

And $F_g=\frac{GMm}{r^2}$

Here,

$$\begin{gathered} h=1.38×{10}^7\mathrm{m} \\ M=\text{mass of earth}=5.97×{10}^{24}\mathrm{kg} \\ G=6.67×{10}^{-11}\frac{{\mathrm{Nm}}^2}{{\mathrm{kg}}^2} \end{gathered}$$

Radius of earth$R=6.371×{10}^6\mathrm{m}$

Now comparing we get,

$$\begin{gathered} F_c=F_g \\ \frac{m{v}^2}{r}=\frac{GMm}{r^2} \\ v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{R+h}} \end{gathered}$$

Substituting values and solving we get,$v=4.44×{10}^3\mathrm{m}/\mathrm{s}.$

The speed of the satellite is$4.44×{10}^3\mathrm{m}/\mathrm{s}.$