The scientists on the far-away planet of Zakron 5 have discovered that their planet’s orbit will decay, causing it to fall into and be incinerated by its sun. The data shows that the inhabitants only have approximately 10 million years left. The alarmed Zakronians look for a new planet, but there are none available that are big enough to accommodate their vast population. In desperation, they turn to you, a young Earth scientist. You tell them about a theoretical planet design you have read about. It would involve building concentric circles of rock around a star to form a CD shaped planet, perfect for habitation.

 

 

 

 

 

 

The Zakronians are overjoyed by your suggestion and plan to begin construction right away. They decide to use rock from a nearby asteroid belt with an average density of 5,000 kg/m³. However, they want to know how thick to build the disc so that the acceleration due to gravity for the majority of the planet (anywhere not too close to the inner or outer edge) will be 7 m/s². What is your answer?

 

 

According to Newton’s law of universal gravitation, the force of gravity one mass exerts on another mass is:

 

F= (Gm₁m₂/D²)

 

Where G is the universal gravitational constant and D is the distance between the centers of mass.

 

In this case:

 

Force on person (or Zakronian) = Gm_personm_rock/D²

 

Newton’s second law of motion, åF = ma, reduces the above expression by canceling m_person when converting to acceleration. Therefore:

 

Acceleration due to gravity = Gm_rock/D²

 

 

SIDE VIEW

 

 

 

 

 

 

In order to integrate the planet, it is easiest to think of it as a series of concentric rings about the subject. When we integrate each ring, the vectors pulling horizontally will cancel each other out, leaving our vertical gravity vector. Since the planet is huge, these rings go on effectively into infinity, but as the rings get farther and farther away, their potency decreases very rapidly (for reasons to be seen later). Each of these rings is made up of wedges that can be integrated easily using polar coordinates.

 

TOP VIEW

 

 

 

 

 

 

 

The area from a function to the origin in polar coordinates:

 

 

 

 

 

 

 

 

 

 

 

 

The area of each wedge, dA, is equal to the linear measure of the arc, Rdq, times dR. Therefore:

 

dA =   R dq dr

A =   R dr dq

A =   ½R² dq

 

This is the fundamental integral for polar coordinates. (Incidentally, by plugging r = R, 0 = a, 2p = b, this integral proves that the area of a circle with radius r does indeed have an area of pr²)

 

All we need, however, is the volume of each ring, dV. That is:  

 

2pTRdR

 

The total acceleration vector for each ring is, as given above, Gm/D². Since the mass of each ring is just dVr (where r is the density, 5,000 kg/m³):

 

dP = 2pGTrRdR/D²

 

R

However, the actual pull of gravity, dZ, is only the y-component of dP. Fortunately, a relationship between dP, dZ, D, and T can be found with a little geometry:

 

.5T

Z

 

D

  

 

From similar triangles, it can be said that:

 

(dP/dZ) = (2D/T)

 

dP = 2dZD/T

 

When we plug this back into our equation, it yields:

 

2dZD/T = 2pTqGRdR/D²

 

Or:

 

dZ = pqT²GRdR/D³

 

Since D, as observable from the side view, is just (R² / .25T²)^½:

 

 dZ = pqT²GRdR(R² + .25T²)^(-3/2)

 

We can now (finally!) do the integral. As noted above, the planet is so huge compared to the subject standing on it that the upper limit is infinity.

 

Z =  pqGT²R(R² + .25T²)^(-3/2) dR

 

If u is defined as (R² + .25T²), we can substitute:

 

Z = pqGT²  u^(-3/2) ½ du

 

(This is because du = 2RdR and therefore ½du = RdR)

 

Z = ½pqGT² [-2u^-½]

 

When we substitute back:

 

Z = ½pqGT² [-2(R² + .25T²)^-½]

 

Plugging infinity in for R (fortunately for us) returns 0. Therefore:

 

Z = ½pqGT² (-(-2(.25T²)^-½))

 

Z = pqGT² ((.25T²)^-½)

 

Z = pqGT² ((.25T²)^-½)

 

Z = 2pqGT

 

ASTOUDING! A messy, complicated integral reduces to a simple, linear relationship between the thickness of the planet and it gravitation! Once the calculus is over with, answering the Zakronians proves easier than first expected:

 

7 = 2p(5,000)(6.67259 e^-11)T

 

T = 7/(2.096e^-6)

 

T = 3,339,287 km

 

The fact that we were integrated out to infinity did not matter because the gravitation pull of each ring got successive smaller and smaller. The reason for this is twofold. First, the rings were getting farther and farther away from the subject and gravity varies inversely as the square of the distance. Secondly, since the rings were moving farther out horizontally, the vertical component of each successive ring got small very quickly.

 

This problem has parallels to the "real" world. An electron near a positively charged plate effectively "thinks" that the plate goes on forever.