Question

Uranus has a mass of 8.68 1025 kg and a radius of 2.56 107 m. Assume it is a uniform solid sphere. The distance of Uranus from the Sun is 2.87 1012 m. (Assume Uranus completes a single rotation in 17.3 hours and orbits the Sun once every 3.08 104 Earth days.)
(a) What is the rotational kinetic energy of Uranus on its axis?

_____________J

(b) What is the rotational kinetic energy of Uranus in its orbit around the Sun?

_____________J

Answer 1

Answer

Given,

Mass of the Uranus, M = 8.68 x 10²⁵ Kg

Radius of Uranus, R = 2.56 x 10⁷ m

Distance of Uranus, D = 2.87 x 10¹² days

a) Rotational Kinetic energy of the Uranus

moment of inertia of the Uranus

  $I = \dfrac{2}{5}MR^2$

  $I = \dfrac{2}{5}\times 8.68\times 10^{25}\times (2.56\times 10^7)^2$

         I = 22.75 x 10³⁹ kg.m²

  Angular speed

     $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{17.3\times 3600}$\

    $\omega = 1 \times 10^{-4}$

 Rotational Kinetic energy

     $KE = \dfrac{1}{2}I\omega^2$

     $KE = \dfrac{1}{2}\times 22.75\times 10^{39}\times (10^{-4})^2$

      $KE = 11.38\times 10^{31}\ J$

b) Rotational Kinetic energy of Uranus in its orbit around sun

moment of inertia of the Uranus

  $I = \dfrac{2}{5}MR^2+ Ma^2$

  $I = 22.75\times 10^{39}+ 8.68\times 10^{25}\times (2.87\times 10^{12})^2$

      I = 7.15 x 10⁵⁰ kg.m²

Angular speed

     $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{3.08\times 10^4\times 3600\times 24}$\

    $\omega =2.36\times 10^{-9}$

 Rotational Kinetic energy

     $KE = \dfrac{1}{2}I\omega^2$

     $KE = \dfrac{1}{2}\times 7.15\times 10^{50}\times (2.36\times 10^{-9})^2$

      $KE = 1.99\times 10^{33}\ J$