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
To find the rotational kinetic energy of the Earth about the Sun due to its orbit, we can use the formula:
Rotational kinetic energy = 0.5 * I * ω²
where I is the moment of inertia and ω is the angular velocity.
The moment of inertia for a point mass rotating about an axis at a distance r is given by:
I = M * r²
where M is the mass of the object and r is the distance from the axis.
In this case, the mass of the Earth, M, is given as 6 x 10^24 kg and the orbital radius, r, is given as 1.5 x 10^11 m.
Calculating the moment of inertia:
I = (6 x 10^24 kg) * (1.5 x 10^11 m)²
I = 1.35 x 10^47 kg*m²
Next, we need to calculate the angular velocity, ω.
The angular velocity, ω, is related to the rotational period, T, by the formula:
ω = 2π / T
where π is the mathematical constant approximately equal to 3.14159.
In this case, the rotational period, T, is given as 1 year. To convert it to seconds:
1 year = 365.25 days (considering the leap year) = 365.25 * 24 hours * 60 minutes * 60 seconds
T = 3.15576 x 10^7 seconds
Calculating the angular velocity:
ω = 2π / T
ω = 2π / (3.15576 x 10^7 seconds)
Now, we can calculate the rotational kinetic energy:
Rotational kinetic energy = 0.5 * I * ω²
Rotational kinetic energy = 0.5 * (1.35 x 10^47 kg*m²) * (2π / (3.15576 x 10^7 seconds))²
Calculating the result:
Rotational kinetic energy ≈ 1.54 x 10^29 Joules
Therefore, the rotational kinetic energy of the Earth about the Sun due to its orbit is approximately 1.54 x 10^29 Joules.
To find the rotational kinetic energy of the earth about the sun, we need to calculate the moment of inertia and angular velocity.
The moment of inertia (I) of a point mass rotating about an axis is given by the formula:
I = m * r^2
where m is the mass of the point mass and r is the distance from the axis of rotation.
In this case, we are treating the earth as a point mass rotating about the sun. So the moment of inertia will be:
I = M * r^2
where M is the mass of the earth and r is the orbital radius.
Next, we need to calculate the angular velocity (ω), which is the rate at which the earth rotates about the sun. The angular velocity can be found using the formula:
ω = 2π / T
where T is the rotational period of the earth, which is 1 year in this case.
Now we have both the moment of inertia and angular velocity, we can calculate the rotational kinetic energy (KE) using the formula:
KE = (1/2) * I * ω^2
Substituting the values we have:
KE = (1/2) * M * r^2 * (2π / T)^2
Now let's plug in the given values:
M = 6 x 10^24 kg
r = 1.5 x 10^11 m
T = 1 year
Calculating the rotational kinetic energy using the formula:
KE = (1/2) * (6 x 10^24 kg) * (1.5 x 10^11 m)^2 * (2π / 1 year)^2
KE ≈ 2.65 x 10^29 Joules
Therefore, the rotational kinetic energy of the earth about the sun due to its orbit is approximately 2.65 x 10^29 Joules.
(1/2) I w^2
I = m r^2
= 6*10^24 * 1.5*10^11 * 1.5*10^11
w = 2 pi/T
= 2 pi/(1 yr * 365 d/yr *24hr/d*3600s/hr)