Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. Assume that the earth is a uniform sphere and that its path around the sun is circular. For comparison, the total energy used in the United States in one year is about 1.1 x 1020 J.
To calculate the kinetic energy of the Earth due to its rotation about its own axis, we need to know the Earth's moment of inertia and its rotational speed.
Step 1: Determine the Earth's moment of inertia (I).
The moment of inertia of a uniform sphere rotating about its own axis is given by the formula:
I = (2/5) * M * R^2
Where:
- M is the mass of the Earth
- R is the radius of the Earth
The mass of the Earth is approximately 5.972 × 10^24 kg, and the radius of the Earth is approximately 6.371 × 10^6 m. Plugging these values into the formula:
I = (2/5) * (5.972 × 10^24 kg) * (6.371 × 10^6 m)^2
Step 2: Determine the Earth's rotational speed (ω).
The rotational speed of the Earth is the number of complete rotations it makes in a given time. It is commonly expressed as an angular velocity, measured in radians per second. The Earth takes approximately 24 hours to complete one rotation, so its rotational speed can be calculated by converting this time to seconds:
T = 24 hours = 24 * 60 * 60 seconds
The angular velocity (ω) is then given by:
ω = 2π / T
Plugging in the value of T:
ω = 2π / (24 * 60 * 60 seconds)
Now we have the moment of inertia (I) and the rotational speed (ω).
Step 3: Calculate the kinetic energy (KE) due to rotation.
The kinetic energy of an object rotating about its axis is given by the formula:
KE = (1/2) * I * ω^2
Plugging in the values of I and ω:
KE = (1/2) * (moment of inertia) * (angular velocity)^2
Now, to calculate the kinetic energy of the Earth due to its motion around the Sun:
Step 4: Determine the Earth's orbital radius (r) and orbital speed (v).
Assuming a circular path, the Earth's orbital radius is the average distance between the Earth and the Sun, which is approximately 149.6 × 10^6 km or 1.496 × 10^11 m.
To calculate the Earth's orbital speed, we'll use Kepler's third law of planetary motion, which states that the square of the period (T) of a planet's orbit is proportional to the cube of its average distance from the Sun (r):
T^2 = (4π^2 / GM) * r^3
Where:
- T is the orbital period (1 year)
- G is the gravitational constant (approximately 6.674 × 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the Sun (approximately 1.989 × 10^30 kg)
Simplifying the equation:
T^2 = (4π^2 / GM) * r^3
1^2 = (4π^2 / GM) * r^3
Solving for r:
r^3 = (GM / 4π^2)
r = (GM / 4π^2)^(1/3)
Plugging in the values of G and M, we can calculate r.
Step 5: Calculate the kinetic energy (KE) due to orbital motion.
The kinetic energy of an object in circular motion is given by the formula:
KE = (1/2) * m * v^2
Where:
- m is the mass of the Earth
- v is the orbital speed
The mass of the Earth is the same as before (5.972 × 10^24 kg), and the orbital speed can be calculated using the formula:
v = (2πr) / T
Plugging in the values of r and T, we can calculate v.
Now you have the kinetic energy due to rotation (KE_rotation) and the kinetic energy due to orbital motion (KE_orbital). You can add them together to find the total kinetic energy of the Earth. If the value exceeds the total energy used in the United States in one year (1.1 × 10^20 J), then the Earth's kinetic energy is greater than the energy used in the United States in one year.