#### Answer

(a) At a height of $1.38\times 10^7~m$ above the earth, the free-fall acceleration is 10% of its value at the surface.
(b) The speed of a satellite orbiting at that height is 4.44 km/s

#### Work Step by Step

(a) We can write an expression for the free-fall acceleration $g$ at the earth's surface. Let $M_e$ be the mass of the earth. Let $R_e$ be the radius of the earth:
$g = \frac{G~M_e}{R_e^2}$
Let's suppose the free-fall acceleration at a height $h$ above the earth is $0.1~g$;
$0.1~g = \frac{G~M_e}{(R_e+h)^2}$
$(0.1)~(\frac{G~M_e}{R_e^2}) = \frac{G~M_e}{(R_e+h)^2}$
$(R_e+h)^2 = 10~R_e^2$
$R_e+h = \sqrt{10}~R_e$
$h = (\sqrt{10}-1)~R_e$
$h = (\sqrt{10}-1)(6.38\times 10^6~m)$
$h = 1.38\times 10^7~m$
At a height of $1.38\times 10^7~m$ above the earth, the free-fall acceleration is 10% of its value at the surface.
(b) We find the orbital radius $R$ as:
$R = R_e+h$
$R = (6.38\times 10^6~m)+(1.38\times 10^7~m)$
$R = 2.02\times 10^7~m$
We can use the orbital radius to find the orbital speed.
$v = \sqrt{\frac{G~M_e}{R}}$
$v = \sqrt{\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)}{2.02\times 10^{7}~m}}$
$v = 4.44\times 10^3~m/s$
$v = 4.44~km/s$
The speed of a satellite orbiting at that height is 4.44 km/s.