astronomers believe the moon used to be much closer to the earth (mass = 5.98 x 10^24 kg). when it was at…

Explanation:

Step1: Identify the centripetal - force and gravitational - force formula

The centripetal force $F_c = \frac{mv^2}{r}$ and the gravitational force $F_g=\frac{GMm}{r^2}$, where $m$ is the mass of the moon, $v$ is the orbital speed, $r$ is the distance between the Earth and the moon, $M = 5.98\times 10^{24}\text{ kg}$ is the mass of the Earth, and $G=6.67\times 10^{- 11}\text{ N}\cdot\text{m}^2/\text{kg}^2$ is the gravitational constant. At equilibrium, $F_c = F_g$.

Step2: Equate the two forces and solve for $v$

$\frac{mv^2}{r}=\frac{GMm}{r^2}$. Canceling out the mass of the moon $m$ on both sides, we get $v=\sqrt{\frac{GM}{r}}$.

Step3: Substitute the given values

We know that $G = 6.67\times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$, $M = 5.98\times 10^{24}\text{ kg}$, and $r = 1.92\times 10^{8}\text{ m}$. $v=\sqrt{\frac{6.67\times 10^{-11}\times5.98\times 10^{24}}{1.92\times 10^{8}}}$. First, calculate the numerator: $6.67\times 10^{-11}\times5.98\times 10^{24}=6.67\times5.98\times10^{-11 + 24}=39.8866\times10^{13}\approx3.99\times 10^{14}$. Then, $v=\sqrt{\frac{3.99\times 10^{14}}{1.92\times 10^{8}}}=\sqrt{2.078125\times 10^{6}}$. $v\approx1441.57\approx1440\text{ m/s}$.

Answer:

$1440$