Question

Calculate the rotational kinetic energy (in j) of mercury on its axis. Assume its mass to be 3.59 ✕ 10²³ kg, the period about its axis to be 1420 h, and its diameter to be 4880 km.

Answer 1

The rotational kinetic energy of mercury on its axis is 1.88 x 10²⁸ J.

Step-by-step explanation:

The rotational kinetic energy of a rotating rigid body or system of particles is given by the equation K = 1/2Iw², where K is the rotational kinetic energy, I is the moment of inertia, and w is the angular velocity.

To calculate the rotational kinetic energy of mercury on its axis, we need to find the moment of inertia and the angular velocity. The moment of inertia, I, can be calculated using the formula for a solid sphere I = 2/5MR², where M is the mass and R is the radius. In this case, the radius is half of the diameter given, so R = (4880 km / 2) * 1000m/km = 2.44 x 10⁶ m.

To find the angular velocity, we first need to convert the period about the axis to seconds. The given period is 1420 hours, so we multiply by 60 minutes/hour and 60 seconds/minute to get 1420 x 60 x 60 = 5,112,000 seconds. Dividing 2π radians by the period gives us the angular velocity w = 2π/(5,112,000 s).

Now we can substitute the values into the equation for rotational kinetic energy:

K = 1/2(2/5)(3.59 x 10²³ kg)(2.44 x 10⁶ m)²(2π/(5,112,000 s))² = 1.88 x 10²⁸ J.