# What will happen to the rotational speed of earth when all ice caps melt?

Since there is no external force applied to earth in its movement about its own axis, the product `I*omega =constant`  need to conserve (`omega` is the angular speed of earth around its axis).

Supposing all ice caps are polar, they do not contribute to the total momentum of inertia of earth. For a solid sphere of mass

`M_(earth) =5.97*10^24 kg` and radius `R_(earth)=6370 km =6370*10^3 m`

we have a moment of inertia

`I =(2/5)*M_(earth)*R_(earth)^2 = 9.69*10^37 kg*m^2`

Now suppose all ice caps melts and the water distributes uniformly on its surface. The total volume of ice caps on Earth is

`V_(ice) =30*10^6 km^3 =30*10^15 m^3 `

which gives a total mass of water of

`M_(water) = ro*V_(ice) =1000*30*10^15 =3*10^19 kg`

The moment of inertia of a hollow sphere (all water is distributed at the surface) of mass `M_(water)` and radius `R_(earth)` is

`I_(water) =(2/3)*M_(water)*R_(earth)^2 = 8.11*10^32 kg*m^2`

Therefore the total moment of inertia of the Earth with all ice caps melted will be

`I_("total") = I_(earth) +I_(water) =9.6900811*10^37 kg*m^2`

Therefore the relative change in the rotational speed of the earth after the all ice caps will melt is

`I_(earth)*omega_("initial") =I_("total")*omega_("final")`

`omega_("final")= I_(earth)/I_("total") *omega_("initial") =0.9999916*omega_("initial")`

and `(Delta(omega))/omega_("initial") =1-0.9999916 =8.4*10^-6 =0.000084 %"`

The Earth will slow down with 0.000084%.

If we take the initial length of one day

`T =24 hours =86400 seconds`

after all the ice caps melt one day will be longer with

`Delta(t) =0.726 seconds`