If there is 4% increase in orbital by how much its time period will increase?

Kepler's 3rd law of planetary motion is exactly
what we need in order to answer this one:

      (orbital period)² / (orbital radius)³ =
       the same number for all bodies orbiting the sun

Let's call that number 'K' just for convenience.

So we know that  T₀² / R₀³  =  K

We're going to be looking for 'T', so let's rearrange the equation now.
Multiply each side by  R₀³ .
Now it says
                             T₀²  =  K R₀³

Now, take the square root of each side, and we have
 
                             T₀  =   √ (K R₀³)  .

Now the radius is increased to  (1.04 R₀).
We want to find the new T .  

                            T  =   √ K · (1.04 R₀)³

                                =    √ K · 1.124864 R₀³

Pull that decimal out of the radical, by taking its square root:

                               =  1.0606 √K · R₀³       

                           T  =   1.0606  T₀

The orbital time has increased by 6%  .


I suspect we probably could have said that since T  varies
as  R^1.5 power, we should look for  (1.04)^1.5 power.

             (1.04)^1.5  =  1.0606  <== bada-bing

Is there any opitions? if you have the periodic table, the noble gases are 8a family, the halogen are the 7a family, and the nonmetal are  carbon, nitrogen, phosphorus, oxygen, sulfur, selenium, fluorine, chlorine, bromine, iodine and astatine.