Question

An Earth-like planet has a mass of 5.00×1024kg and a radius of 7.00×106m. The planet’s angular velocity due to its rotation about its own axis is 8.00×10?5rad/s. (a) How long is a day on this planet? (b) An asteroid with a mass of one-fifth the mass of the planet is traveling radially toward the planet’s equator. Upon impact, it buries itself just below the planet’s surface. What is the planet’s angular velocity after this happens? (c) By what percentageis the length of a day increased after the asteroid’s arrival?

Answer 1

Length of day= number of rotation/ angular speed = 2*3.14 / 8*10^-5 = 78500s = 78500/3600= 21.8 hrs

Initial moment of inertia of planet with mass 5*10^24kg is = 2mr^2 / 5 = 9.8*10^37 kgm^2

Final moment of inertia with initial mass+ 1/5th of initial mass is = 2m1r^2/ 5 = 1.176*10^38 kgm^2

So planet's angular velocity after this happens is 8*10^-5 *9.8*10^37 / 1.176*10^38 = 6.67*10^-5 rad/s

Final length of a day = 2*3.14 / 6.67*10^-5 = 94248s

So, percentage increase in the length of a day after the asteroids arrival is = 94248- 78500 / 78500 = 20%