Question

(a)

Calculate the rotational kinetic energy (in J) of Saturn on its axis. (Assume its mass to be 5.69 ✕ 10²⁶ kg, the period about its axis to be 10.2 h, and its diameter to be 1.21 ✕ 10⁵ km.)

J

(b)

What is the rotational kinetic energy (in J) of Saturn in its orbit around the Sun? (Assume its distance from the sun to be 1.43 ✕ 10⁹ km and its period about the sun to be 10800 days.)

J

Answer 1

Answer 2

SOLJTION :

a.

I = 2/5 MR^2  for solid sphere about its axis through c.g. 

Where, M = mass of sphere and R = radius of sphere

So,

I(saturn) 

= 2/5 * 5.69 * 10^(26) * (1.21*10^(5)*10^(3)/2)^2   kg.m^2 

= 8.330729*10^41 kg.m^2

Angular speed of Saturn, w 

= 2 pi/(10.2*3600) rad/s

= 1.7111 *10^(-4) rad/s

Rotaional K. E. Of Saturn

=1/2  I w ^2

=  1/2 * 8.330729*10^41 * (1.7111 *10^(-4) )^2  kg.m^2/s^2 or (N.m) or J

=  1.21956  * 10^(34)J (ANSWER).

b.

I (about sun)

= 2/5 MR^2 + M*Rs^2 

= 8.330729*10^41 + 5.69*10^(26) *  (1.21*10^(5)*10^(3)/2)^2 

= 2.91575515 * 10^(42) kg.m^2

w(about sun) 

= 2 pi/(10800*24*3600)

= 6.7335233 * 10(-9) rad/s

Rotational K.E. around sun

=  1/2 * 2.91575515 * 10^(42) * (6.7335233 * 10(-9) )^2

= 5.35415 * 10^(47) J ( ANSWER).