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20 days ago

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The Earth is 1.5 × 1011 m from the Sun and takes a year to make one complete orbit. It rotates on its own axis once per day. It

can be treated approximately as a uniform-density sphere of mass 6 × 1024 kg and radius 6.4 × 106 m (actually, its center has higher density than the rest of the planet, and the Earth bulges out a bit at the equator). Using this crude approximation, calculate the following: (a) What is vCM ? (b) What is Ktrans ? (c) What is ω, the angular speed of rotation around its own axis? (d) What is Krot? (e) What is Ktot?

1 answer:

Maru [3.3K]20 days ago

6 0

Answer:

Part a)

$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

$KE_{rot} = 2.6 \times 10^{29} J$

Part e)

$KE_{tot} = 2.68 \times 10^{33} J$

Explanation:

The Earth takes 1 year to orbit around the Sun

therefore

$T = 1 year = 3.15 \times 10^7 s$

Now we know that angular speed of the Earth around the Sun is defined as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{3.15 \times 10^7}$

The linear speed of the Earth’s center can be calculated as

$v_{cm} = r\omega$

$r = 1.5 \times 10^{11} m$

$v_{cm} = (1.5 \times 10^{11})(\frac{2\pi}{3.15 \times 10^7})$

$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

The translational kinetic energy of the Earth's center can be computed as

$K_{trans} = \frac{1}{2}mv^2$

$K_{trans} = \frac{1}{2}(6 \times 10^{24})(2.98 \times 10^4)^2$

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

Angular rotational speed of the Earth on its axis can be calculated as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{24 \times 3600}$

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

Now, the moment of inertia of the Earth about its axis

$I = \frac{2}{5}mR^2$

$I = \frac{2}{5}(6 \times 10^{24})(6.4 \times 10^6)^2$

$I = 9.83 \times 10^{37} kg m^2$

The rotational kinetic energy can then be determined as

$KE_{rot} = \frac{1}{2}I\omega^2$

$KE_{rot} = \frac{1}{2}(9.83 \times 10^{37})(7.27 \times 10^{-5})^2$

Part e)

$KE_{rot} = 2.6 \times 10^{29} J$ Finally, total kinetic energy can be calculated as

$KE_{tot} = KE_{trans} + KE_{rot}$

$KE_{tot} = 2.68 \times 10^{33} + 2.6 \times 10^{29}$

$KE_{tot} = 2.68 \times 10^{33} J$

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