Find the rotational kinetic energy of the earth about the sun due to its orbit about the sun. The mass of the earth M is 6 x 1024 kg, the orbital radius r is 1.5 x 1011 m and the rotational period T is 1 year. Treat the earth as a point mass in this problem.
change 1 year to seconds
KE= 1/2 mr^2*(2PI/period)^2
To find the rotational kinetic energy of the Earth about the Sun due to its orbit, we can use the formula:
Rotational Kinetic Energy (KE) = 1/2 * I * ω²
Where:
I is the moment of inertia,
ω is the angular velocity.
Since we are treating the Earth as a point mass, we can simplify the moment of inertia to be:
I = M * r²
Where:
M is the mass of the Earth,
r is the orbital radius.
First, let's calculate the moment of inertia using the given values:
M (mass of the Earth) = 6 x 10^24 kg
r (orbital radius) = 1.5 x 10^11 m
I = M * r²
= (6 x 10^24 kg) * (1.5 x 10^11 m)²
= 6 x 10^24 kg * 2.25 x 10^22 m²
= 13.5 x 10^46 kg * m² (moment of inertia)
Next, we need to calculate the angular velocity (ω). The angular velocity can be calculated using the formula:
ω = 2π / T
Where:
T is the rotational period of the Earth.
T (rotational period) = 1 year = 365 days
Note: We need to convert the period to seconds for consistency.
1 year = 365 days = 365 * 24 * 60 * 60 seconds
= 3.15576 x 10^7 seconds
Using the given period, we can calculate the angular velocity:
ω = 2π / T
= 2π / (3.15576 x 10^7 seconds)
= 1.99 x 10^-7 radians/second (angular velocity)
Now, we can substitute the values into the rotational kinetic energy formula:
Rotational Kinetic Energy (KE) = 1/2 * I * ω²
= 1/2 * (13.5 x 10^46 kg * m²) * (1.99 x 10^-7 radians/second)²
By simplifying the expression, we can find the value of the rotational kinetic energy of the Earth about the Sun due to its orbit.