Question

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

Answer 1

Eu achei:
velocidade`=3xx10^4m/s`
aceleração centripeta`=6xx10^-3m/s^2`

Explanation:

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=1.5xx10^11m`.
So:
in `1` year, `363` days, each day `24` hours, each hour `60` minutes and each minute `60` seconds, or:
`365xx24xx60xx60=3.15xx10^7m~~3xx10^7` seconds.

Describing a distance of `2pir=2*3.14*1.5xx10^11=9.42xx10^11m~~9xx10^11m`

Velocity will be: `v="distance"/"time"=(9xx10^11)/(3xx10^7)=3xx10^4m/s`

Centripetal acceleration will be: `a_c=v^2/r=(3xx10^4)^2/(1.5xx10^11)=6xx10^-3m/s^2`