Question

# For the purposes of this problem, treat the Earth as a solid, uniform sphere with mass...

For the purposes of this problem, treat the Earth as a solid, uniform sphere with mass 5.97×1024 kg and radius 6.37×106 m, and assume that the Earth's orbit around the sun is circular with a radius of 1.5×1011 m.

(A) What is the angular kinetic energy of the Earth due to its orbit around the sun?

(B) What is the magnitude of the Earth's angular momentum due to its orbit around the sun?

(C) What is Earth's angular kinetic energy due to its rotation around its axis?

(D) What is the magnitude of the Earth's angular momentum due to its rotation around its axis?

(E) Which of the following best explains where the Earth's angular kinetic energy and momentum came from?

-The solar system formed from a massive cloud of gas and dust, which was slowly rotating. As the cloud collapsed under its own gravitational pull, the cloud started to spin faster, just as an ice skater pulling his arms in will spin faster. Because all of the material that accreted to form the planet was rotating, the planet was rotating as well.

-As the Earth formed, it experienced a series of collisions with asteroids and comets. These asteroids and comets hit the ball of rock that was forming into the planet off-center. Over time, the off-center collisions gradually caused the planet to rotate faster.

-As the Moon orbits around the Earth, it creates tides on the Earth. Over time the tides have caused the Earth to rotate faster and faster.

-Sheer force of will.

(a) rotational kinetic energy is given as

K.E = 1/2 * I * w2

where I = m * r2 =  5.97e24 * 1.5e112 = 1.343e47 Kg.m2

w = 2 * pi / T

where T is time taken to complete one orbit = 365 days = 3.153e7 seconds

so,

K.E = 2.665e33 J

--------------------------------------------------

(b) angular momentum

L = Iw

L = 1.343e47 * 1.992e-7

L = 2.675e40 Kg.m2 / s

----------------------------------------------

(c) K.E ( about rotation axis)

rotational kinetic energy is given as

K.E = 1/2 * I * w2

where I = 2/5 * m * r2 = 2/5 * 5.97e24 * 6.37e62 = 9.689e37 Kg.m2

w = 2 * pi / T

where T is time taken to complete one rotation = 24 hours = 86400 seconds

so,

K.E = 2.56e29 J

----------------------------------------------

(d) angular momentum about rotation axis

L = 9.689e37 * 7.27e-5

L = 7e33 Kg.m2 /s

----------------------------------------------