The correct answer is: A. 8.04 km/sec at a height of 285 km.
A satellite in a circular orbit around the Earth is moving at a speed that is just fast enough to keep it from falling back to Earth. The speed of a satellite in a circular orbit is given by the formula:
$$v = \sqrt{\frac{GM}{r}}$$
where $M$ is the mass of the Earth, $r$ is the radius of the orbit, and $G$ is the gravitational constant.
The radius of the Earth is 6371 km, and the radius of a circular orbit at a height of 285 km is 6656 km. Substituting these values into the formula, we get:
$$v = \sqrt{\frac{(6.67408 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2})(5.9722 \times 10^{24} \text{ kg})}{6656 \text{ km}}} = 8.04 \text{ km/s}$$
Therefore, the correct answer is A. 8.04 km/sec at a height of 285 km.
Option B is incorrect because the speed of a satellite in a circular orbit at a height of 37,400 km is 11.11 km/sec.
Option C is incorrect because the speed of a satellite that escapes the pull of the Earth is 11.26 km/sec.