Question

question 50
(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 × 10^6 s. assume that the orbit of the
moon is circular.
what is the magnitude of the centripetal force acting on the moon in its ancient orbit? the mass of the moon is 7.4 × 10^22 kg.
○ 3.6 × 10^24 n
○ 1.3 × 10^23 n
○ 5.6 × 10^26 n
○ 2.4 × 10^22 n
○ 4.4 × 10^20 n

Explanation:

Step1: Find the angular velocity formula

The angular velocity \(\omega=\frac{2\pi}{T}\), where \(T\) is the orbital period.

Step2: Find the centripetal force formula

The centripetal force \(F = m\omega^{2}r\), substituting \(\omega=\frac{2\pi}{T}\) into it, we get \(F = m(\frac{2\pi}{T})^{2}r\).

Step3: Convert units

The radius \(r = 24400\space km=24400\times10^{3}\space m\), the mass \(m = 7.4\times 10^{22}\space kg\), the period \(T = 23400\space s\).

Step4: Calculate the centripetal force

$$
LATEXBLOCK0
$$

First calculate \((\frac{2\pi}{23400})^{2}\approx(\frac{6.28}{23400})^{2}\approx(2.68\times 10^{- 4})^{2}=7.18\times 10^{-8}\)

Then \(F = 7.4\times 10^{22}\times7.18\times 10^{-8}\times24400\times 10^{3}\)

\(7.4\times7.18\approx53.13\), \(53.13\times24400\approx1.3\times10^{6}\)

\(F\approx1.3\times 10^{23}\space N\)

Answer:

\(1.3\times 10^{23}\space N\)