hannah has information about an object in circular orbit around earth. (m_{earth}=6.0\times10^{24}\text{…

Explanation:

Step1: Recall centripetal - force and gravitational - force formula

The gravitational force between the Earth and the orbiting object $F_g=\frac{Gm_{Earth}m}{r^2}$, and the centripetal force $F_c = \frac{mv^2}{r}$, where $v$ is the tangential speed. At circular - orbit equilibrium, $F_g = F_c$. So, $\frac{Gm_{Earth}m}{r^2}=\frac{mv^2}{r}$, and we can cancel out the mass $m$ of the orbiting object on both sides, getting $v=\sqrt{\frac{Gm_{Earth}}{r}}$. Here, $r$ is the distance from the center of the Earth to the orbiting object.

Step2: Analyze the required variable

From the formula $v=\sqrt{\frac{Gm_{Earth}}{r}}$, we know that to calculate the tangential speed $v$ of the orbiting object, we need the mass of the Earth ($m_{Earth}$), the gravitational constant ($G$), and the distance of the orbiting object to the Earth ($r$). We already know $m_{Earth}$ and $G$, so we need the distance of the orbiting object to the Earth.

Answer:

the distance of the orbiting object to Earth