Question

a neutron star with a moment of inertia of 18 x 10^37 kg m^2 is spinning with a period of 0.06 s. what is it's rotational kinetic energy (leave as x 10^37)?\

Answer 1

The neutron star's rotational kinetic energy is approximately 9.882 x 10³⁷ joules, calculated using its moment of inertia and period of rotation.

Step-by-step explanation:

To calculate the rotational kinetic energy of a neutron star, we can use the formula K = 0.5 × I × ω², where K is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity. The moment of inertia is given as 18 × 10³⁷ kg·m². To find ω, we use the relation ω = 2π/T, where T is the period of rotation. Replacing the values, we obtain:

ω = 2π/0.06 s = 104.7197551 rad/s

Now, substituting the values for I and ω, the rotational kinetic energy K is calculated as:

K = 0.5 × 18 × 10³⁷ kg·m² × (104.7197551 rad/s)²

K = 0.5 × 18 × 10³⁷ × 10965.88291934

K = 9.882294428 × 10³⁷ J

Therefore, the neutron star's rotational kinetic energy is approximately 9.882 × 10³⁷ J.