Question

Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. mastering

Answer 1

The angular momentum of the Earth around the Sun is given by:

`L=m \omega r^2`
where
m is the Earth's mass

`\omega` is the Earth's angular velocity
r is the average distance of the Earth from the Sun

The Earth takes 365 days to make a complete revolution around the Sun, which corresponds to

`t=365 d \cdot 24 \cdot 60 \cdot 60 =3.15 \cdot 10^7 s`
A complete revolution corresponds to
`2 \pi rad`, therefore the Earth's angular velocity is

`\omega = (2 \pi rad)/(3.15 \cdot 10^7 s)=1.99 \cdot 10^(-7) rad/s`

The average distance of Earth from the Sun is 149.6 million km:

`r=149.6 Mkm = 149.6 \cdot 10^9 m`

And the Earth's mass is
`m=5.97 \cdot 10^(24) kg`, therefore its angular momentum is

`L=m \omega r^2 =(5.97 \cdot 10^(24) kg)(1.99 \cdot 10^(-7) rad/s)(149.6 \cdot 10^9 m)^2=`

`=2.66 \cdot 10^(40) kg m^2/s`