Question

(problem reference m.7)
approximately one billion years ago, the moon orbited the earth much closer than it does today. the radius of the orbit was only 24 400 km. the orbital period was only 23 400 s. today, the average radius is 385 000 km; and the present period is 2.36 × 10⁶ s. assume that the orbit of the moon is circular.
what is the magnitude of the centripetal force acting on the moon in its current orbit? the mass of the moon is 7.4 × 10²² kg.
6.2 × 10²⁴ n
2.0 × 10²⁰ n
6.3 × 10²⁰ n
5.4 × 10²² n
9.0 × 10²⁰ n

Explanation:

Step1: Convert radius to meters

The radius \(r = 385000\space km=385000\times10^{3}\space m = 3.85\times 10^{8}\space m\)

Step2: Find the angular velocity formula

The period \(T = 2.36\times 10^{6}\space s\). The angular velocity \(\omega=\frac{2\pi}{T}\)

Step3: Write centripetal force formula

Centripetal force \(F = m\omega^{2}r\). Substitute \(\omega=\frac{2\pi}{T}\) into it, we get \(F=m\times(\frac{2\pi}{T})^{2}\times r\)

Step4: Substitute values

Given \(m = 7.4\times 10^{22}\space kg\), \(T = 2.36\times 10^{6}\space s\), \(r = 3.85\times 10^{8}\space m\)

$$
LATEXBLOCK0
$$

$$
LATEXBLOCK1
$$

Answer:

C. \(6.3\times 10^{20}\space N\)