There is a fancy website dedicated to working out the speed of light from the Quran.

Their base their calculations on a verse saying that affairs ascend to God in one day equating 1000 years of what we count, interpreting the latter as angels covering in one day the same distance the Moon would cover in 1000 lunar years,

and attempt to show that, in an Earth-Moon system isolated from the Sun's gravity, the distance covered by the Moon in 12,000 lunar revolutions would be equal to the one covered by light in 1 day.

Let's grant those tendencious interpretations for the sake of the argument.

One important claim for their calculation is that, in such an isolated Earth-Moon system, the Moon's new orbital radius R' would be lower:

allegedly, R' = R*cosø R being the present orbital radius and ø the angle covered by the Earth around the Sun in 1 lunar revolution, namely 26.92°.

Here's how they justify this formula:

In a local frame non-rotating with respect to stars the velocity of the moon is not a constant. NASA measured the instantaneous velocity of the moon at various points throughout its orbit. These measurements show that the velocity of the moon varies considerably (from 3470 km/hr up to 3873 km/hr); which means that the moon accelerates and decelerates continuously. The average lunar velocity is Vavg = 3682.8 km/hr (1.023 km/sec).

The lunar orbit relative to Earth is a low eccentricity ellipse, however we cannot use the equation for the perimeter of an ellipse. Why? Because Earth lies on the major axis of this ellipse; but since the direction of the axes change with respect to stars then when the moon returns to the same position with respect to stars this does not mean that it made an exact ellipse (a local frame non-rotating with respect to the ellipse is actually rotating with respect to stars). So most astronomers calculate the length of the lunar orbit in a local frame non-rotating with respect to stars by the following equivalent circle method:

L = V T = 2 π R

⇒ V = 2 π R / T

However this velocity is under the influence of the gravitational pull of the sun. We can vectorially calculate the velocity of the moon relative to Earth without the gravitational assistance of the sun and hence the isolated length of the lunar orbit: Displacement is a vector (has magnitude and direction) and from this displacement vector we get the velocity vector (magnitude and direction); and from this velocity vector we get the kinetic energy. If external work is done we end up with a resultant displacement vector, resultant velocity vector and resultant kinetic energy. In our case we have to backtrack.

The work done by the gravitational field of the sun by creating a net rotational force on the lunar orbit (twist) is:

In a local frame non-rotating with respect to sun the moon speeds up when it heads towards the sun and then slows down by the same amount when it heads away from the sun. However in a local frame non-rotating with respect to stars the sectors where the moon speeds up and slows down are actually moving forward inside the orbit by the same angle Earth orbits the sun. In this frame there is a rotational force around Earth. This means that the moon's orbital energy is the sum of the energy acquired from this twist plus the intrinsic energy of Earth-moon system (kinetic energy of Earth's spin transferred to the lunar orbit by ocean friction).

Today's lunar orbit is a very low eccentricity ellipse (very close to a perfect circle) but when the moon first formed it was a very high eccentricity ellipse. The eccentric ellipse back then had a point very close to Earth and another point very far out. When the moon was nearest to Earth inside the ellipse the gravitational forces were stronger and hence more energy was transferred to the moon when it was closer to Earth than when the moon was farther out inside the ellipse. This made the moon recede more when it was closer to Earth than when it was farther out inside the ellipse. This difference in recession rates smoothed out the differences between the closest and the farthest points in the orbit (this is why today's lunar orbit is very close to a perfect circle). So for each direction the recession has different magnitude. Since this recession has magnitude and direction then it is a displacement vector (R'). However in every direction this displacement vector is normal to the rotational force around Earth (always at right angles with the rotational force around Earth, 90°). Since the resultant R is the sum of two normal displacement vectors then those displacement vectors form a right triangle. We can use trigonometry to solve those displacement vectors:

By definition in a right triangle cosø = side adjacent / hypotenuse.

⇒ side adjacent = hypotenuse cosø

⇒ R' = R cosø

We can verify this triangle by the Pythagorean Theorem:

(R sinø)2 + (R cosø)2 = R2sin2ø + R2cos2ø = R2 (sin2ø + cos2ø) = R2 (1)

Hence cosø is the only solution to this restricted three-body problem.

So do you think their justification for R'=R*cosø holds water?