Gravitational Acceleration and the Speed of Light

 

Gravitational Acceleration and the Speed of Light 

Pavle I. Premović

Laboratory for Geochemistry, Cosmochemistry&Astrochemistry, 

University of Niš, pavleipremovic@yahoo.com, Niš, Serbia 

Newton’s gravitational law states that the gravitational force between two astronomical objects is proportional to their masses and inversely proportional to the square of the distance between their centers. We assume that these two objects are spherical and the mass of one object is M and the mass of the other object is m and that M >> m.[1].

The gravitational force is given by

F = GMm/R² 

where G (= 6.67 × 10^-11 N m² kg^-2)[2] is the gravitational constant and R is the distance between their centers.

The gravitational force is attractive and causes the object of the mass m (hereinafter the m object) to deflect from a straight-line path to a circular path (orbit) around the object with the mass M (hereinafter the M object).  We know in this orbit the m object is exposed to the gravitational centripetal force: F_p = GMm/R² and the centrifugal force: F_f = mυ²/R. These two forces are balanced: F_p = F_f or GMm/R² = mυ²/R. For the m object to move one-dimensionally along the straight line towards the M object F_p > F_f or, after some algebra,

GM/R > υ2.

The mass of the M object can be presented by the following equation 

M = (4/3)πR_M³ρ    … (1)

where R_M and ρ are its radius and density. Introducing this term into the above inequality, we obtain, after some algebra,

υ² < 4GR_M³ρ/R.

When the m object reaches the object M then R = R_M. Plugging into this inequality R_M instead R we get, after some algebra,

 υ < 2R_M√(Gρ).

The maximum gravitational force F_max between these two objects would be near the surface of the M object. It can be mathematically expressed as

Fmax = GMm/RM2.

Newton’s second law of motion states that the acceleration of an object equals the net force acting on it divided by its mass. In this case, the maximum acceleration of the m object can be represented by the equation

a_max = (F_max/m) = GM/R_M²    … (3).

Combining this equation and eqn. (1) and a bit of algebra we get

 a_max ≈ 4GR_Mρ    … (4). 

Suppose now that the density of the M object is 1000 kg m^-3. (For comparison, the average density of the Earth is about 5500 kg m^-3). Plugging this value in eqn. (4) and the above values of G and ρ we obtain

amax ≈ 2.7 × 10-7RM.

Using this equation, we find that the m object would reach the gravitational acceleration of c × 1 sec^-1 ≈ 3 × 10⁸ m sec^-2 if the radius of the M object is about 10¹² km.[3,4] The Earth’s mass is about 6 × 10²⁴ kg and its radius is about 6371 km.[5] The Sun is the star at the center of the Solar System it has a mass of about 2 × 10³⁰ kg and a radius is about 7 × 10⁵ km. On average, the Sun’s density is about 1400 kg m^-3.
 
The Milky Way's mass is about 3 × 10⁴² kg although the mass of a galaxy is difficult to estimate with any accuracy. It has a density less than a billionth of a billionth of the density of air on Earth then according to eqn. (4): a_max is almost zero.
 
UY Scuti, the star near the center of the Milky Way, is the largest known star in the Universe its radius is about 1.2 × 10⁹ km and its density is probably about the density of the Sun. It appears that the gravitational acceleration, at least, in the Milky Way is much less than c × 1 sec^-1. 

As we pointed out above, gravitational acceleration is described as the object receiving an acceleration due to the gravitational force F acting on it or in equation F = ma. From the eqn. (1), we find that the gravitational acceleration of the m object is equal

a = GM/R².

As we have shown above, this acceleration is limited by the speed of light c and it must be a < c × sec^-1. It can be mathematically expressed as

GM/R² < c × sec^-1

or

M/R² < c × sec^-1/G.

Substituting the above values of c × sec^-1 (≈ 3 × 10⁸ m sec^-2) and G (= 6.67 × 10^-11 N m² kg^-2) into this inequality, we obtain

 M/R² < 4.5 × 10¹⁸ (expressed in kg m^-2).

After a bit of algebra, we have the following expression

R < 4.7 × 10^-10√M    … (5).

Venus has a nearly circular orbit with an eccentricity of 0.007. Its mass is 4.87 × 10²⁴ kg and it is much smaller than mass of the Sun equal to 1.989 × 10³⁰ kg. Using eqn. (5) we find that R < 670 km which is much smaller than the radius of the Sun is about 7 × 10⁵ km. 

Currently, it appears the Segue 2 in the Aries constellation is the one closest to the Earth orbiting the Milky Way. It is a dwarf galaxy that has an approximately round shape with a half-light radius of about 1.1 × 10¹¹ km. Its mass is about 2.2 × 10³⁶ kg. Using eqn. (5) we find that R < 8 × 10⁸ km which is much smaller than the radius of the Milky Way about 5 × 10¹⁷ km.

[1] In astronomy, we usually deal with a radius of a massive object much greater than the radius of a much smaller object (see below).

[2] To avoid confusion in further text, the SI units are given in italics.

[3] Of course, c is the speed of light and is equal to 2.99792 × 10⁸ m sec^-1 or approximately 3 × 10⁸ m sec^-1.   

[4] A simple calculation shows the m object would reach the acceleration of about 3 × 10⁹ m sec^-2 if R_M of the M object is about 10¹³ km.

[5] Just to remind you, near Earth's surface, the gravity acceleration is approximately 9.81 m sec^-2.

 

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