Answer
The meteor's minimum speed 40 km above the Earth's surface would be 11.1 km/s
Work Step by Step
We can assume that kinetic energy and gravitational potential energy when the meteor is far away are both equal to zero. We can use conservation of energy to find the minimum speed when the meteor is 40 km above the Earth's surface:
$KE+U_g = 0$
$\frac{1}{2}mv^2-\frac{G~M_E~m}{R} = 0$
$\frac{1}{2}mv^2 = \frac{G~M_E~m}{R}$
$\frac{1}{2}v^2 = \frac{G~M_E}{R}$
$v = \sqrt{\frac{2G~M_E}{R}}$
$v = \sqrt{\frac{(2)(6.67\times 10^{-11}~m^3/kg~s^2)~(5.97\times 10^{24}~kg)}{(6.38\times 10^6~m)+(4.0\times 10^4~m)}}$
$v = 11.1\times 10^3~m/s = 11.1~km/s$
The meteor's minimum speed 40 km above the Earth's surface would be 11.1 km/s.
