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3 years ago

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Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun

. Assume that the earth is a uniform sphere and that its path around the sun is circular. For comparison, the total energy used in the United States in one year is about 1.1 x 1020 J.

1 answer:

Dominik [7]3 years ago

4 0

Answer:

a. $K_{Axis}=2.574x10^{29}J$

b. $K_{Orbit}=2.6577x10^{33}J$

Explanation:

$K_{Axis}=\frac{1}{2}I*w^2$

$I_{Sphere}=\frac{2}{5}*m*r^2$

$w=\frac{2\pi }{T}$ , $T=24hrs*\frac{3600s}{1hr} =86400s$

radius earth = 6371 km

mass earth = 5,972*10^24 kg

a.

$K_{Axis}=\frac{1}{2}*\frac{2}{5}*m*r^2*(\frac{2\pi}{T})^2$

$K_{axis}=\frac{4\pi^2}{5}*5.98x10^{24}kg*(6.38x10^6m)^2*(\frac{1}{86400s})^2$

$K_{Axis}=2.574x10^{29}J$

b.

$T=1year*\frac{365day}{1year}*\frac{24hr}{1day}*\frac{3600s}{1hr}=31536000s$

$K_{Orbit}=\frac{1}{2}*I*w$

$I=m*r^2$

$K_{Orbit}=\frac{1}{2}*m*r^2*(\frac{2\pi}{T})^2$

$K_{Orbit}=\frac{4\pi^2}{5}*5.98x10^{24}*6.38x10^6m*(\frac{1}{31.536x10^6s})^2$

$K_{Orbit}=2.6577x10^{33}J$

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