The rotational kinetic energy of the earth is 2.587 × 10²⁹ J
The rotational kinetic energy of the earth is given by K = 1/2Iω² where I = rotational inertia of earth = 2MR²/5 where M = mass of earth = 5.972 1× 0²⁴ kg and R = radius of earth = 6.4 × 10⁶ m. ω = angular speed of earth = 2π/T where T = period of earth = 24 hours = 24 × 60 × 60 = 86400 s = 8.64 × 10⁴ s
So, K = 1/2Iω²
K = 1/2 × 2MR²/5 × (2π/T)²
K = 4π²MR²/5T²
Substituting the values of the variables into the equation, we have
K = 4π²MR²/5T²
K = 4π² × 5.972 × 10²⁴ kg × (6.4 × 10⁶ m)²/[5(8.64 × 10⁴ s)²]
K = 4π² × 5.972 × 10²⁴ kg × 40.96 × 10¹² m²/[5(74.6496 × 10⁸ s)²]
K = 4π² × 244.61312 × 10³⁶ kgm²/373.248 × 10⁸ s²
K = 9656.93890286 × 10³⁶ kgm²/373.248 × 10⁸ s²
K = 25.87 × 10²⁸ kgm²/s²
K = 2.587 × 10²⁹ J
So, the rotational kinetic energy of the earth is 2.587 × 10²⁹ J
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