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`