Question 4.6: The Centripetal Acceleration of the Earth (A) What is the ce...
The Centripetal Acceleration of the Earth
(A) What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun?
(B) What is the angular speed of the Earth in its orbit around the Sun?
(A) Conceptualize We will model the Earth as a particle and approximate the Earth’s orbit as circular (it’s actually slightly elliptical, as we discuss in Chapter 13).
Categorize The Conceptualize step allows us to categorize this problem as one of a particle in uniform circular motion.
Analyze We do not know the orbital speed of the Earth to substitute into Equation 4.21. With the help of Equation 4.22, however, we can recast Equation 4.21 in terms of the period of the Earth’s orbit, which we know is one year, and the radius of the Earth’s orbit around the Sun, which is 1.496 × 10^{11} m.
a_c=\frac{v^2}{r} (4.21)
T=\frac{2 \pi r}{v} (4.22)
Combine Equations 4.21 and 4.22:
a_c=\frac{v^2}{r}=\frac{\left(\frac{2 \pi r}{T}\right)^2}{r}=\frac{4 \pi^2 r}{T^2}Substitute numerical values:
a_c=\frac{4 \pi^2\left(1.496 \times 10^{11} m\right)}{(1 yr)^2}\left(\frac{1 yr}{3.156 \times 10^7 s}\right)^2=5.93 \times 10^{-3} m/s^2(B) Analyze
Substitute numerical values into Equation 4.23:
\omega=\frac{2 \pi}{T} (4.23)
\omega=\frac{2 \pi}{1 yr}\left(\frac{1 yr}{3.156 \times 10^7 s}\right)=1.99 \times 10^{-7} s^{-1}Finalize The acceleration in part (A) is much smaller than the free-fall acceleration on the surface of the Earth. An important technique we learned here is replacing the speed v in Equation 4.21 in terms of the period T of the motion. In many problems, it is more likely that T is known rather than v. In part (B), we see that the angular speed of the Earth is very small, which is to be expected because the Earth takes an entire year to go around the circular path once.