Question

The radius of the earth's very nearly circular orbit around the sun is $1.5\times {10}^{11}\text{m}$. Find the magnitude of the earth's (a) velocity, (b) angular velocity, and (c) centripetal acceleration as it travels around the sun. Assume a year of 365 days.

Short answer

Part (a) The orbital velocity of the earth around the sun is $3.0\times {10}^4\text{m}/\text{s}$.

Part (b) the angular velocity of the earth around the sun is $2.0\times {10}^{-7}\text{rad}/\text{s}$.

Part (c) The centripetal acceleration of the earth is$6.0\times {10}^{-3}\text{m}/{\text{s}}^2$.

Step-by-step solution

Part (a) Step 1. Given information

The radius of the earth's very nearly circular orbit around the sun is $1.5\times {10}^{11}\text{m}$, the number of days in a year is 365 days.

Part (a)  Step 2. Explanation 

The total distance travelled by the earth in 1 year is,

$\begin{array}{l}d=2\pi r\\ d=2\pi \left(1.5\times {10}^{11}\text{m}\right)\\ =9.424\times {10}^{11}\text{m}\end{array}$

The velocity of the earth is

$v=\frac{d}{t}$

$v=\frac{9.424\times {10}^{11}\text{m}}{\left(365\text{days}\times \frac{24\text{h}}{1\text{dyy}}\times \frac{60\text{min}}{1\text{h}}\times \frac{60\text{s}}{1\text{min}}\right)}$

$\begin{array}{l}=29885.77\text{m}/\text{s}\\ \approx 3.0\times {10}^4\text{m}/\text{s}\end{array}$

Therefore, the orbital velocity of the earth around the sun is $3.0\times {10}^4\text{m}/\text{s}$.

Part (b) Step 1. Given information

The radius of the earth's very nearly circular orbit around the sun is$1.5\times {10}^{11}\text{m}$, the number of days in a year is 365 days.

From the solution of the part (a), the orbital velocity of the earth around the sun is $29885.77\text{m}/\text{s}$.

Part (b)  Step 2. Explanation 

The angular velocity of the earth is,

$\omega =\frac{v}{r}$

$\omega =\frac{29885.77\text{m}/\text{s}}{1.5\times {10}^{11}\text{m}}$

$\begin{array}{l}=1.99\times {10}^{-7}\text{rad}/\text{s}\\ =2.0\times {10}^{-7}\text{rad}/\text{s}\end{array}$

Thus, the angular velocity of the earth around the sun is $2.0\times {10}^{-7}\text{rad}/\text{s}$.

Part  ( c) Step 1. Given information

The radius of the earth's very nearly circular orbit around the sun is $1.5\times {10}^{11}\text{m}$, the number of days in a year is 365 days.

Part  ( c)   Step 2. Explanation

The centripetal acceleration of the earth is,

$a_c=\frac{v^2}{r}$

$a_c=\frac{{(29885.77\text{m}/\text{s})}^2}{\left(1.5\times {10}^{11}\text{m}\right)}$

$\begin{array}{l}=5.95\times {10}^{-3}\text{m}/{\text{s}}^2\\ \approx 6.0\times {10}^{-3}\text{m}/{\text{s}}^2\end{array}$

Thus, the centripetal acceleration of the earth is $6.0\times {10}^{-3}\text{m}/{\text{s}}^2$.