Question

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?

Answer 1

Part a)

`v_(cm) = 2.98 * 10^4 m/s`

Part b)

`K_(trans) = 2.68 * 10^(33) J`

Part c)

`\omega = 7.27 * 10^(-5) rad/s`

Part d)

`KE_(rot) = 2.6 * 10^(29) J`

Part e)

`KE_(tot) = 2.68 * 10^(33) J`

Step-by-step explanation:

Time period of Earth about Sun is 1 Year

so it is

`T = 1 year = 3.15 * 10^7 s`

now we know that angular speed of the Earth about Sun is given as

`\omega = (2\pi)/(T)`

`\omega = (2\pi)/(3.15 * 10^7)`

now speed of center of Earth is given as

`v_(cm) = r\omega`

`r = 1.5 * 10^(11) m`

`v_(cm) = (1.5 * 10^(11))((2\pi)/(3.15 * 10^7))`

`v_(cm) = 2.98 * 10^4 m/s`

Part b)

now transnational kinetic energy of center of Earth is given as

`K_(trans) = (1)/(2)mv^2`

`K_(trans) = (1)/(2)(6 * 10^(24))(2.98 * 10^4)^2`

`K_(trans) = 2.68 * 10^(33) J`

Part c)

Angular speed of Earth about its own axis is given as

`\omega = (2\pi)/(T)`

`\omega = (2\pi)/(24 * 3600)`

`\omega = 7.27 * 10^(-5) rad/s`

Part d)

Now moment of inertia of Earth about its own axis

`I = (2)/(5)mR^2`

`I = (2)/(5)(6 * 10^(24))(6.4 * 10^6)^2`

`I = 9.83 * 10^(37) kg m^2`

now rotational energy is given as

`KE_(rot) = (1)/(2)I\omega^2`

`KE_(rot) = (1)/(2)(9.83 * 10^(37))(7.27 * 10^(-5))^2`

`KE_(rot) = 2.6 * 10^(29) J`

Part e)

Now total kinetic energy is given as

`KE_(tot) = KE_(trans) + KE_(rot)`

`KE_(tot) = 2.68 * 10^(33) + 2.6 * 10^(29)`

`KE_(tot) = 2.68 * 10^(33) J`