Question

The radius of the earth very nearly circular orbit around the sun is $1.50\times {10}^{11}m$. What is the magnitude of the earth‘s velocity and centripetal acceleration as it travels around the sun?

Answer 1

Consider that in one year the Earth travels the entire circumference of the circle (ok, it should be an ellipse but let us assume it is a circle) of radius

$r=1.5\times {10}^{11}m$

So, In1 year, 363 days, each day 24 hours, each hour 60 minutes and each minute 60 seconds, or: $365\times 24\times 60\times =3.15\times {10}^7\approx 3\times {10}^{7}sec=T$

Describing a distance of

$D=2\pi r=2\times 3.14\times 1.5\times {10}^{11}m=9.42\times {10}^{11}\approx 9\times {10}^{11}m$

Velocity will be= $v=\frac{D}{T}=3\times {10}^4\frac{m}{s}$

Centripetal acceleration will be $=\frac{v^2}{r}=6\times {10}^{-3}\frac{m}{s^2}$