Question

Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. Assume that the earth is a uniform sphere and that its path around the sun is circular. For comparison, the total energy used in the United States in one year is about 1.1 x 1020 J.

Answer 1

a.
`K_(Axis)=2.574x10^(29)J`

b.
`K_(Orbit)=2.6577x10^(33)J`

Step-by-step explanation:

`K_(Axis)=(1)/(2)I*w^2`

`I_(Sphere)=(2)/(5)*m*r^2`

`w=(2\pi )/(T)` ,
`T=24hrs*(3600s)/(1hr) =86400s`

radius earth = 6371 km

mass earth = 5,972*10^24 kg

a.

`K_(Axis)=(1)/(2)*(2)/(5)*m*r^2*((2\pi)/(T))^2`

`K_(axis)=(4\pi^2)/(5)*5.98x10^(24)kg*(6.38x10^6m)^2*((1)/(86400s))^2`

`K_(Axis)=2.574x10^(29)J`

b.

`T=1year*(365day)/(1year)*(24hr)/(1day)*(3600s)/(1hr)=31536000s`

`K_(Orbit)=(1)/(2)*I*w`

`I=m*r^2`

`K_(Orbit)=(1)/(2)*m*r^2*((2\pi)/(T))^2`

`K_(Orbit)=(4\pi^2)/(5)*5.98x10^(24)*6.38x10^6m*((1)/(31.536x10^6s))^2`

`K_(Orbit)=2.6577x10^(33)J`