Question

The radius of the earth's very nearly circular orbit around the sun is 1.5 x 10^11 m. find the magnitude of the earth's velocity as it travels around the sun. assume a year of 365 days. find the magnitude of the earth's angular velocity as it travels around the sun. assume a year of 365 days. find the magnitude of the earth's centripetal acceleration as it travels around the sun. assume a year of 365 days.

Answer 1

In 1 year, 365 days, each day 24 hours, each hour 60 minutes and each minute 60 seconds.

OR

$1 \ year = 365 \ days = 365 \times 24 \times 60 \times 60 = 3.15 \times 10^7 seconds$.

As the radius of the earth's very nearly circular orbit around the sun is 1.5 x 10^11 m.

Therefore, the magnitude of the earth's velocity

$v = \frac{2\pi r}{T} = \frac{2\times3.14\times 1.5 \times 10^{11} m}{3.15 \times 10 ^7 \ s} =2.99 \times 10^4 m/s \approx 3 \times 10^4 m/s$.

The magnitude of the earth's angular velocity is

$\frac{v}{r} = \frac{3 \times 10^4 m/s}{1.5 \times 10^{11} m} = 2 \times 10^{-7} rad/s$.

The magnitude of the earth's centripetal acceleration is

$a_{c} =\frac{v^2}{r} = \frac{(3 \times 10^4 m/s)^2}{1.5 \times 10^{11} m} = 6 \times 10^{-3} \ m/s^2$