Jtdylktuy

So my friend asked me this random maths question - heres the scenario - an alien fleet fires 6 billion bombs onto a perfectly sphircal flat planet that has a surface area 5 trillion. each bomb has a radius of 5. if all bombs fall the exact same distance apart what is the distance between each blast.

i worked it out by getting total are of explosion (6 X 78.5) = 471(X10^9)
5000 - 471 = 4529(X10^9)
4529(X10^9) / 6(X10^9) = 754.833M

now im 90% sure that im wrong because what i found is the area left not touched by explosion per bomb not the distance between said circles

Ah! Nothing suits a Black Friday better than contemplating destruction at the planetary level!

First, let's define coverage ratio $P$ as the fraction of the planetary surface area $A_s$ covered by the $N$ circular blasts of radius $r$, each having area $A_circ$:

$$bbox{A_circ = pi r^2} , quad bbox{P = frac{A_s}{N A_circ} = frac{A_s}{N pi r^2}}$$

In general, the best packing for spheres (or circular disks like here) is hexagonal close packing:

If you look at the blue illustration on the right, you'll see that an equilateral triangle with edge length $a = r + d + r$ covers one half of a blast disk. The area of an equilateral triangle with edge length $a$ is
$$bbox{A_triangle = frac{sqrt{3}}{4} a^2 = frac{sqrt{3}}{4}left(2 r + dright)^2}$$
This triangle represents the hexagonal close packing pattern of the planetary bombardment, if their coverage ratios $P$ are the same:
$$bbox{frac{A_s}{N A_circ} = P = frac{A_triangle}{frac{1}{2} A_circ}}$$
i.e.
$$bbox{frac{A_s}{N pi r^2} = frac{ frac{sqrt{3}}{4}left(2 r + dright)^2 }{frac{1}{2} pi r^2 }}$$
If we solve this for $d$, we get
$$bbox{d = sqrt{frac{2 A_s}{sqrt{3} N}} - 2 r approx 1.07457 sqrt{frac{A_s}{N}} - 2 r}$$

In OP's particular case, $A_s = 5,000,000,000,000$ (assuming short trillion), $r = 5$, $N = 6,000,000,000$ (assuming short billion). Then,
$$bbox[#ffffef, 1em]{d approx 21}$$

In the illustration, the green middle figure is intended as a reminder that if complete coverage is required, $d$ must be negative. Because each angle in an equilateral triangle is $60text{°}$, for full coverage we need
$$bbox{2 r + d = 2 r cos(30text{°})}$$
i.e.
$$bbox{d = r left ( sqrt{3} - 2 right ) approx -0.267949 r}$$
This means that to get a full coverage, i.e. every point on the surface within at least one blast radius $r$, you need $N$ bombs:
$$bbox{N = frac{2}{3sqrt{3}}frac{A_s}{r^2} approx 0.3849 frac{A_s}{r^2}}$$
In OP's case, that is about 76,980,000,000 bombs, or under 77 billion bombs; less than 13 times the suggested number of bombs, to utterly cover the planetary surface with glorious, wondrous hellfire!

Apologies, but I must now cut my contemplations short: The nice men have come to take me back to my padded room.