KE=1/2 I w^2 where w is 2PI/day(change to seconds) and I=2mr^2/5
so change a day to seconds, and calculate.
KE=.5*.4*5.98e24*(6.37e6)^2 * (2*PI/8.64e4)^2
Obtain the rotational kinetic energy of the earth due its daily rotation on its axis. Assume the Earth to be a uniform sphere; take m = 5:98 x 10^(24) kg; r = 6:37 x10^6 m
1 point
9.8 x 10^(29)
2.56 x 10^(29)
8.85 x 10^(25)
1.00 x 10^(29)
3 answers
2.56*10^29J
Correct! The rotational kinetic energy of the Earth due to its daily rotation on its axis is approximately 2.56 x 10^29 J.