Answer
(a) $L = 2.67\times 10^{40}~kg~m^2/s$
(b) $L = 7.07\times 10^{33}~kg~m^2/s$
Work Step by Step
(a) We can find the speed of the earth.
$v =\frac{d}{t} = \frac{2\pi~r}{t}$
$v = \frac{(2\pi)(1.50\times 10^{11}~m)}{(365.3~days)(24~h)(3600~s)}$
$v = 2.986\times 10^{4}~m/s$
We can find the angular momentum of the earth around the sun.
$L = mvr$
$L = (5.97\times 10^{24}~kg)(2.986\times 10^4~m/s)(1.50\times 10^{11}~m)$
$L = 2.67\times 10^{40}~kg~m^2/s$
It is reasonable to model the earth as a particle because the earth's radius is very small compared to the distance from the sun.
(b) We can find the angular speed of the earth as it rotates on its axis.
$\omega = \frac{2\pi~rad}{(24~h)(3600~s)}$
$\omega = 7.27\times 10^{-5}~rad/s$
We can find the angular momentum of the earth as it rotates on its axis.
$L = I\omega$
$L = \frac{2}{5}mR^2\omega$
$L = \frac{2}{5}(5.97\times 10^{24}~kg)(6.38\times 10^6~m)^2(7.27\times 10^{-5}~rad/s)$
$L = 7.07\times 10^{33}~kg~m^2/s$
