an artificial satellite orbits earth at a speed of 7800 m/s and a height of 200 km above earth’s surface…

Explanation:

Step1: Recall gravity - acceleration formula

The acceleration due to gravity $g$ at a height $h$ above the Earth's surface is given by $g=\frac{GM}{(R + h)^2}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, $R$ is the radius of the Earth ($R\approx6371\ km = 6371000\ m$), and $h$ is the height of the satellite above the Earth's surface. At the Earth's surface ($h = 0$), $g_0=\frac{GM}{R^2}\approx9.8\ m/s^2$. Since $h>0$ (the satellite is at a height $h = 200\ km=200000\ m$ above the Earth's surface), the denominator $(R + h)^2>R^2$. So, $g=\frac{GM}{(R + h)^2}<\frac{GM}{R^2}=g_0$. Also, the gravitational force still acts on the satellite, so $g>0$.

Answer:

D. less than 9.8 m/s2 but greater than zero