This manuscript focuses on Newton’s universal gravitational constant (G).

Newton’s constant G is one of the universal constants with the greatest uncertainty we currently encounter, and finding it accurately remains an experimental challenge. In recent years, attempts to find G have been unsuccessful; all experiments and data differ from one another, but surely some of the experimental data or measurements of G are correct. The margin of error in experiments can be as low as 0.01 (1%) or even 0.02 (2%).

In my opinion, we can calculate the constant G with high accuracy or precision through theoretical reasoning. My task in this manuscript is to find G with high precision and apply it to several equations; for this, I will use a small and simple Newtonian equation. Determining the value of G theoretically is not complicated, since simplicity often leads to the correct answer.

And after this brief introduction, I will try to shed some light on the great unknown and mysterious value of Newton’s constant G.

Thanks to NASA scientists, we know the Earth’s mass with considerable accuracy. They calculated this using the Moon’s movement, specifically its orbital period around the Earth. This data, the Earth’s mass, is crucial for accurately determining the value of the gravitational constant G.

According to NASA, the Earth has a mass equal to M ⊕ = 5.9722 × 10 24   kg .

Using the Earth’s mass and two other pieces of information, we can find the value of G with great precision.

These pieces of information are:

The gravitational intensity at the Earth’s surface, measured in a laboratory:

The Earth’s mean radius, whose exact value is:

With these three highly precise data points, we can calculate G with maximum accuracy, and we will call this new value “G_c” (correct G). The Newtonian equation is exactly:

And this data is the most precise value for Newton’s constant that logic can offer.

It is a mathematical certainty.

The currently used value of G is G = 6.67430 × 10 − 11 N ⋅ m 2 / kg 2 .

What happens if we try to find the mass of the Earth using the current value of G? The following occurs:

It’s clear that this data is incorrect; therefore, the current value of G is obsolete and useless.

The relative margin of error E_r, between G and G_c, is equal to:

It is fortunate that Earth’s gravitational intensity can be accurately measured in the laboratory.

And once we have found the new and correct Newton’s universal gravitational constant, we will proceed to make several adjustments to some data considered correct [1].

Quantum Mechanics

According to the following equation, we can find a value for G.

The formula is

where ℓ p is the Planck length ℓ p = 1.616255 × 10 − 35   meters , c is the speed of light c = 299792458   m / s , and ℏ is the reduced Planck constant ℏ = 1.054571817 × 10 − 34   J ⋅ s .

The equation results in the currently accepted constant G, but we’ve already seen what happened with this value of G when calculating the Earth’s mass; G is incorrect. What we’ve done is adapt Planck’s constants to Newton’s constant, not the other way around. The incorrect value, by elimination, is the Planck length ℓ p . We’ll quantify it by including G_c, and we’ll call it ℓ p x .

The equation is as follows

Between ℓ p and ℓ p x , there is a relative difference E r = ( ℓ p − ℓ p x ) / ℓ p x .

And of course, the Planck time t p is also affected, since t p = ℓ p / c .

Therefore, the new value of t p is equal to:

The mass of the Sun.

Another example I want to give is the mass of the Sun using the correct equation, G_c.

The famous equation that solves it would be the following:

M ⊙ is the mass of the Sun, to be deduced.

M ⊕ is the mass of the Earth, 5.9722 × 10²⁴ kg.

π is the number pi.

a is the semi-major axis of the Earth-Sun ellipse, 149,597,870,700 meters.

G_c is Newton’s constant, correct, 6.665021314 × 10⁻¹¹ N·m²/kg².

T is the period of revolution of the Earth around the Sun, 31,558,146 seconds.

The mass of the Sun has a value of M ⊙ = 1.991178368 × 10 30   kg .

This data has a relative difference of 0.001043787 (1 thousandth).

Calculated with G gives a mass equal to M ⊙ = 1.988410202 × 10 30   kg , this data is short, in my opinion [1][3].

EscapeVelocity( v e )

The famous escape velocity is the speed an object must reach to escape the gravitational pull of a sphere forever.

The escape velocity of Earth, using the current gravitational constant G, is equal to:

Using the new, correct Newton’s constant G_c, the equation for Earth’s escape velocity is:

Between the two results, we find an absolute margin of error E a = v e − v ′ e E a = 7.77827   m / s .

And a relative margin of error E r = 0.000695831165 ( 6.9583 × 10 − 04 ) .

This demonstrates that G_c is more accurate [1].

I am referring to the Earth’s radius, R_T = 6,371,000 meters; although this value may appear to be rounded—either up or down—it is not; this figure represents the radius the Earth would have if it were a perfect sphere, and it is well measured and exact.

Out of curiosity, I’d like to know what the Earth’s radius would be if calculated using the current value of the gravitational constant G,

We found an absolute margin of error of E_a = 4433.14029 meters, that is, a significant discrepancy. This result suggests that the current value of G is incorrect. The three data points presented at the beginning of the theoretical discussion are exact, or at the very least highly precise. This demonstrates that Newton’s constant G_c is highly precise.

There is no more accurate method than the one presented here for calculating the value of G_c in a purely theoretical manner.

And now we must continue to wait for the experimental results, and everything seems to indicate that we will need a great deal of patience.

Sometimes, as physicists, we complicate our lives so much that we forget the fundamentals, the basics, simplicity, and coherence. In this short paper, I have attempted to shed light on the subject by identifying the “correct G constant”—G_c—and treating it as the most accurate possible value for Newton’s constant based on theory; I hope I have succeeded.

To this end, I have presented three clear examples—and no more, since the idea has already been made clear. I have provided the exact data, without rounding, and included the decimal places I deemed appropriate.

Personally, I consider G_c to be correct, exact, or nearly exact, and I believe I have demonstrated this; the equation to deduce it has always been right there, right under our noses—the three exact data points: the Earth’s mass, the Earth’s mean radius, and the exact gravitational intensity at the Earth’s surface as measured in the laboratory.