Fundamental Journal of Modern Physics

Vol. 8, Issue 2, 2015, Pages 141-145

Published online at http://www.frdint.com/

:esphras and Keywords hydrogen atom, Newton’s universal law of gravity.

Received July 3, 2015

© 2015 Fundamental Research and Development International

GRAVITATIONAL FORCE IN HYDROGEN ATOM

ANDIKA ARISETYAWAN

Universitas Pendidikan Indonesia

Jl DR Setyabudhi No. 229 Bandung

Indonesia

e-mail: andikaarisetyawan@upi.edu

Abstract

This is technical paper to prove mathematically that Newton’s Universal

Law of Gravity can be reshaped from elliptical orbit geometry in [1].

Thus, we can derive new formula for calculating gravitational force in a

case of Hydrogen atom. It enables us to quantize gravity for every electron

orbit.

1. Introduction

The problem of gravity in atomic scale is still mysterious in science. By all of

interactions in nature, only gravity could not be fully understood by scientists. This

paper is trying to solve how gravity works in quantum scale, especially in a case of

hydrogen atom.

2. Revisited Newton’s Universal Law of Gravity

Let us start from standard Newton’s Universal Law of gravity as follow:

.2

r

MmGF = (1)

Our current interpretation for standard Newton formula is that of if there are two

massive body Mm =1 and mm =2 separated with the distance r, then based on

ANDIKA ARISETYAWAN

142

equation (1), The bigger the masses of the objects, the bigger gravitational force

between them. This is not wrong interpretation, but actually, this is not fundamental

level to explain why gravity seems superior at macro and it seems inferior at micro

scale. The more we dig equation (1), the more new insight we get in this paper

3. Reshaping Newton’s Universal Law of Gravity from Elliptical Orbit

From [1], we have velocity for elliptical orbit as follows:

.2

2

20v

GMr

GMv

+

= (2)

By squaring both sides, we have

.2

2

20

2

vGMr

GMv

+

= (3)

Dividing it with r, we get

.2

2

20

2

+

=

vGMrr

GM

r

v (4)

Multiplying it with m, we get centripetal force as follows:

.2

2

20

+

=

vGMrr

GMmFs (5)

Now, suppose we will move an object from the earth surface to infinity distance. We

have from [1] that total energy of a two-body gravitational system is related by

,05.0 2=−

r

GMmmv (6)

.5.0 2

r

GMmmv = (7)

Then, we get

GRAVITATIONAL FORCE IN HYDROGEN ATOM

143

.22

r

GMv = (8)

Since it is on earth surface, we can write equation (8) into

.2

20v

GMrE = (9)

In which Er is radius of the earth from the center, 0v is escape or initial velocity

from the earth surface.

In general situation, I mean not only on earth surface, we can rewrite (9) as

follows

.2

20

0v

GMr = (10)

Substituting (10) into (5) we get

( )

.2

0rrr

GMmFs +

= (11)

If ,0rr = then

( )

.2

20

gs Fr

GMm

rrr

GMmF ==

+= (12)

In which centripetal force is equal to Newton’s universal law of gravity. Since

,0rr = then we can substitute (10) into (12) as follows

,4

2

40

2

20

GM

mv

v

GM

GMmFg =

= (13)

.2

.2

1202

0 r

EK

GM

vmvFg =

= (14)

In which, the force of gravity is proportional to kinetic energy and it is inversely

proportional to distance. It means, the force of gravity can be viewed as kinetic

energy that is needed to move an object from the surface of a gravitating body to the

certain point or orbit per unit distance.

ANDIKA ARISETYAWAN

144

4. Gravitational Formula for Hydrogen Atom

Let’s start from the problem of Hydrogen atom, it is very simple atom which

only consist of one proton and one electron. We knew that the force that keeping the

electron to stay in its orbit is Coulomb’s force or electro static force. We will use

Bohr Model as follows:

Figure 1. Hydrogen atom.

Based on [2], we have formula for the possible radius of r that can be allowed in

Hydrogen Orbit as follows:

.4

2

220

me

nrn

ℏπε= (15)

From the figure 1, equating (15) with radius of 0rr = in equation (14), we get:

.4

222

0

220

n

GMmev

ℏπε= 16)

We can express (16) into kinetic energy formula as follows:

.4 22

0

22

n

eGMmEK

ℏπε= (17)

Substituting (17) into (14), we get final formula for the magnitude of gravitational

force in Hydrogen orbit as follows:

22

2

220

22 2

4 n

GMmK

rn

eGMm

r

EKFg

ℏℏ

=πε

== (18)

with

GRAVITATIONAL FORCE IN HYDROGEN ATOM

145

.,3,2,1,8 0

2

…=πε

= nr

eK

In which equation (18) is a quantized formula for calculating gravitational force for

Hydrogen atom. K is kinetic energy for electron (see [2]), G is universal constant of

gravity, M is the mass of proton, m is the mass of electron, ℏ is the reduced planck

constant, and n is quantum number.

5. Result and Discussion

Based on (18), for every kinetic energy K of the electron orbit, there is

gravitational force which work to attract electron to stay in its orbit besides

Coulomb’s Force. But, we must remember that kinetic energy EK in equation (17)

has different meaning with kinetic energy K in equation (18). EK in equation (17) has

the same meaning with equation (14).

References

[1] A. Arisetyawan, The imaginary and real velocity of an orbiting body based on different

types of conics section, arXiv:1312.0967 2013.

[2] Krane and S. Kenneth, Fisika Modern (Modern Physics); Translated by Hans J.

Wospakrik, UI- Press, Sofia Niksolihin, Jakarta, 1992.