What is the rotational kinetic energy of the earth about the sun? assume the earth is a uniform sphere, mass of the earth is 5.97 x 10^24 kg, mass of the sun is 2.99x10^30 kg, radius of the earth is 6.37x10^3 km, radius of the sun is 6.96 x 10^5 km distance from the center of the sun to the center of the earth is 1.5x10^8 km

2.66x10^33 J

Explanation:

In order to do this, we first need to know the expression for kinetic energy:

E = 1/2 I*w²  (1)

I is moment of innertia

w is angular speed.

Moment of Innertia can be calculated using the following expression:

I = 2/5 M*R²  (2)

M is mass of earth, R is radius of earth

Replacing the data in expression (2) we have:

I = 2/5 * 5.97x10^24 * (6.37x10^6)²

I = 9.69x10^37 kg m²

Next, we need to calculate the angular speed of Earth over it's axis, this is easy, because we know the Earth rotates over it's own axis once a day, 24 hours (86400 s), and assuming Earth is a perfect sphere, we can calculate the speed:

w = 2π / 86400 = 7.27x10^-5 rad/s

next thing we need to do is calculate the rotational kinetic energy of earth on it's axis, using equation (1) so:

E = 1/2 * 9.69x10^37 * (7.27x10^-5)²

E = 2.56x10^29 J

Now that we have this value, we can finally calculate the rotational kinetic energy of earth about the sun. For that, we need to calculate again the angular speed of earth about the sun. The Earth rotates around the sun once a year, or 365 days, which is 3.1536x10^7 s, so the angular speed would be:

w = 2π/3.1536x10^7 = 1.99x10^-7 rad/s

finally the energy is the combination of the sun and earth so:

K = 1/2 (Ie + Me*Rorb²)wo²

I is innertia for earth

Me mass of earth

Rorb RAdius of orbit around the sun

wo is angular speed around the sun

Replacing the data we finally have:

K = 1/2 [9.69x10^37 + 5.97x10^24 * (1.5x10^11)²]*(1.99x10^-7)²

K = 2.66x10^33 J

the answer is to check all of them! they all apply well at least i think so!

50 inches should be the resultant vector