Assume that the Earth is a solid sphere of constant density, with mass 5.98 × 1024 kg and radius 6370 km.
What is the moment of inertia of the Earth with respect to rotation about its axis, and what is the kinetic energy of this rotation?
The moment of inertia of a solid sphere with uniform density is given by:
I = (2/5) * M * R^2
Where M is the mass of the sphere and R is its radius. Plugging in the given values:
I = (2/5) * (5.98 × 10^24 kg) * (6370 km)^2
I = 9.98 × 10^37 kg * m^2
The kinetic energy of rotation can be calculated using the formula:
KE = (1/2) * I * ω^2
Where ω is the angular velocity of rotation. We don't know the exact value of ω for the Earth, but we can estimate it as 2π/24 hr, since the Earth rotates once every 24 hours. Converting this to radians per second:
ω = 2π / (24 * 60 * 60 s) = 7.27 × 10^-5 rad/s
Plugging in the values:
KE = (1/2) * (9.98 × 10^37 kg * m^2) * (7.27 × 10^-5 rad/s)^2
KE = 2.14 × 10^29 J
Therefore, the moment of inertia of the Earth with respect to rotation about its axis is 9.98 × 10^37 kg * m^2, and the kinetic energy of this rotation is 2.14 × 10^29 J.
To find the moment of inertia of the Earth with respect to rotation about its axis, we can use the formula:
I = (2/5) * M * R^2
where:
I is the moment of inertia,
M is the mass of the Earth, and
R is the radius of the Earth.
Given:
Mass of the Earth, M = 5.98 × 10^24 kg
Radius of the Earth, R = 6370 km = 6.37 × 10^6 m
Let's calculate the moment of inertia:
I = (2/5) * M * R^2
= (2/5) * (5.98 × 10^24 kg) * (6.37 × 10^6 m)^2
Calculating this expression gives us:
I ≈ 9.99 × 10^37 kg m^2
Now, let's calculate the kinetic energy of this rotation. The kinetic energy of a rotating object is given by:
KE = (1/2) * I * ω^2
where:
KE is the kinetic energy,
I is the moment of inertia, and
ω (omega) is the angular velocity.
Since we are given that the Earth is rotating about its axis, we can assume a reasonable value for the angular velocity, which is approximately 7.292 × 10^-5 rad/s.
Using this value, we can calculate the kinetic energy:
KE = (1/2) * I * ω^2
= (1/2) * (9.99 × 10^37 kg m^2) * (7.292 × 10^-5 rad/s)^2
Evaluating the expression gives us:
KE ≈ 2.31 × 10^29 J
Therefore, the moment of inertia of the Earth with respect to rotation about its axis is approximately 9.99 × 10^37 kg m^2, and the kinetic energy of this rotation is approximately 2.31 × 10^29 J.