Question

an undiscovered planet, many lightyears from earth, has one moon in a periodic orbit. this moon takes 1.455 × 10⁶ seconds (about 17 days) on average to complete one nearly circular revolution around the unnamed planet. the distance from the center of the moon to the surface of the planet is 235.0 × 10⁶ m and the planet has a radius of 3.10 × 10⁶ m. calculate the moon’s centripetal acceleration ( a_c ). ( a_c = ) (\boxed{}) ( \text{m/s}^2 )

Explanation:

Step1: Find the orbital radius \( r \)

The orbital radius \( r \) is the sum of the distance from the moon to the planet's surface and the planet's radius.
Given: distance from moon to surface \( d = 235.0\times 10^{6}\, \text{m} \), planet radius \( R = 3.10\times 10^{6}\, \text{m} \)
So, \( r = d + R = (235.0\times 10^{6}+ 3.10\times 10^{6})\, \text{m}=238.1\times 10^{6}\, \text{m} \)

Step2: Find the angular velocity \( \omega \)

The period \( T = 1.455\times 10^{6}\, \text{s} \). The formula for angular velocity is \( \omega=\frac{2\pi}{T} \)
\( \omega=\frac{2\pi}{1.455\times 10^{6}\, \text{s}} \approx \frac{6.2832}{1.455\times 10^{6}}\, \text{rad/s}\approx 4.318\times 10^{-6}\, \text{rad/s} \)

Step3: Calculate centripetal acceleration \( a_c \)

The formula for centripetal acceleration is \( a_c = \omega^{2}r \)
Substitute \( \omega = 4.318\times 10^{-6}\, \text{rad/s} \) and \( r = 238.1\times 10^{6}\, \text{m} \)
\( a_c=(4.318\times 10^{-6})^{2}\times238.1\times 10^{6} \)
First, calculate \( (4.318\times 10^{-6})^{2}=18.645\times 10^{-12} \)
Then, \( 18.645\times 10^{-12}\times238.1\times 10^{6}=18.645\times238.1\times 10^{-6} \)
\( 18.645\times238.1\approx 4440.37 \)
So, \( a_c\approx 4440.37\times 10^{-6}\, \text{m/s}^2\approx 0.00444\, \text{m/s}^2 \) (or more accurately, let's do the calculation with more precision)

Alternative way using \( a_c=\frac{4\pi^{2}r}{T^{2}} \)
\( r = 238.1\times 10^{6}\, \text{m} \), \( T = 1.455\times 10^{6}\, \text{s} \)
\( a_c=\frac{4\pi^{2}\times238.1\times 10^{6}}{(1.455\times 10^{6})^{2}} \)
\( (1.455\times 10^{6})^{2}=2.117\times 10^{12} \)
\( 4\pi^{2}\times238.1\times 10^{6}\approx 4\times 9.8696\times238.1\times 10^{6}\approx 39.4784\times238.1\times 10^{6}\approx 9.400\times 10^{9} \)
Then \( a_c=\frac{9.400\times 10^{9}}{2.117\times 10^{12}}\approx 0.00444\, \text{m/s}^2 \) (or approximately \( 4.44\times 10^{-3}\, \text{m/s}^2 \))

Answer:

\( \approx 4.44\times 10^{-3}\, \text{(or } 0.00444\text{)} \)