The mass of the Earth is 5.98 x 1024 kg.The radius of the Earth is 6.37 x 106 m.(a) What is Earth's angular velocity in its daily rotation? rad/sec(b) What is Earth's moment of inertia? kg.m2(c) What is Earth's rotational kinetic energy? JoulesFor the next 2 parts, assume you can approximate yourself as a point particle.(d) If your mass is 40 kg and you are standing on the equator, what is your personal rotational energy about the earth's north-south axis? Joules(e) If you were instead to stand on the North Pole, what would your personal rotational energy about the earth's north-south axis then be? (roughly speaking) Joules

(a) The Earth completes one rotation every 24 hours. To convert this to seconds, we multiply by 60 seconds/minute, 60 minutes/hour, and 24 hours/day:

24 hours/day * 60 minutes/hour * 60 seconds/minute = 86400 seconds/day

The angular velocity (ω) is the rate of change of the angle with respect to time. Since the Earth completes one full rotation (2π radians) in one day, we can calculate the angular velocity as:

ω = 2π radians / 86400 seconds = 7.27 x 10^-5 rad/sec

(b) The moment of inertia (I) for a solid sphere is given by the formula I = 2/5 * m * r^2, where m is the mass and r is the radius. Substituting the given values:

I = 2/5 * (5.98 x 10^24 kg) * (6.37 x 10^6 m)^2 = 9.70 x 10^37 kg.m^2

(c) The rotational kinetic energy (K) is given by the formula K = 1/2 * I * ω^2. Substituting the values we've calculated:

K = 1/2 * (9.70 x 10^37 kg.m^2) * (7.27 x 10^-5 rad/sec)^2 = 2.56 x 10^29 Joules

(d) Your personal rotational energy (K) at the equator can be calculated using the formula K = 1/2 * m * r^2 * ω^2, where m is your mass, r is the radius of the Earth, and ω is the angular velocity. Substituting the given values:

K = 1/2 * 40 kg * (6.37 x 10^6 m)^2 * (7.27 x 10^-5 rad/sec)^2 = 6.69 x 10^9 Joules

(e) If you were standing on the North Pole, your distance from the axis of rotation would be essentially zero, so your rotational energy would also be zero.