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

The charge of an ion with 55 protons and 52 electrons is +3. the reason is because protons are positive, and there are three more than the amount of negative electrons.

Answer8.0m;