Question

Calculate the kinetic energy that the earth has because of its rotation about its own axis...

Assume that the earth is a uniform sphere and that its path around the sun is circular.
(a) Calculate the kinetic energy that the earth has because of its rotation about its own axis. For comparison, the total energy used in the United States in one yearis about 9.33 109 J.


J

(b) Calculate the kinetic energy that the earth has because of its motion around the sun.


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Answer #2

Part A)

The KE of the rotation of the earth about its own axis is KE = .5Iω2

The moment of inertia of a sphere is 2/5MR2

The mass of the earth is 5.98 X 1024 kg

The radius of the earth is 6.38 X 106 m

Angular velocity is 2π radians every 24 hours, which is 2π/86400 sec = 7.27 X 10-5 rad/s

KE = (.5)(2/5)(5.98 X 1024)(6.38 X 106)2(7.27 X 10-5)2

KE = 2.57 X 1029 J

Part B)

The earth is 1.496 X 1011 m from the sun

It makes one trip around in 1 year

The distance of travel is (2π)(1.496 X 1011 m) = 9.40 X 1011 m

The time required is one year, which is 31557600 seconds

v = d/t

v = (9.40 X 1011)/(31557600)

v = 2.98 X 104 m/s

KE = .5mv2

KE = (.5)(5.98 X 1024)(2.98 X 104)2

KE = 2.66 X 1033 J

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Answer #3
a) KE(angular) = 1/2 IW^2 = 1/2 2/5 Mr^2 W^2; where I = 2/5 Mr^2 for a solid sphere is the inertia of Earth with radius r, mass M, and spinning at W = 2 pi radians/24 hr.

b) KE(linear) = 1/2 Mv^2 = 1/2 M (wR)^2; where v = wR is the tangential velocity of the Earth as it revolves around the Sun at a distance R ~ 93 million miles and an angular velocity w = 2 pi radian/year.

You need to look up M, Earth's mass and radius r, and convert both w and W into rad/sec. Then you can do the math.

The physics is this...there are two basic types of kinetic energy: rotational from spinning around the mass' own axis and linear from moving linearly with a linear velocity.

NOTE: We can also do Earth's motion around the Sun as angular if we treat the Earth as a point mass M In which case KE(angular) = 1/2 Iw^2 = 1/2 MR^2 w^2 = 1/2 Mv^2 since v = wR. This results because moment of inertia I = MR^2 for a point mass rotating around an external axis.
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