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European Journal of Applied Sciences – Vol. 12, No. 2

Publication Date: April 25, 2024

DOI:10.14738/aivp.122.16694

Chanana, R. K. (2024). Universal Mass-Energy Equivalence Relation is Expressed by Mechanical Energy of a Falling Body on Earth

at Constant Acceleration Due to Gravity. European Journal of Applied Sciences, Vol - 12(2). 45-46.

Services for Science and Education – United Kingdom

Universal Mass-Energy Equivalence Relation is Expressed by

Mechanical Energy of a Falling Body on Earth at Constant

Acceleration Due to Gravity

Ravi Kumar Chanana

Self-Employed Independent Researcher, Gr. Noida-201310, India

ABSTRACT

In this research article, it is shown that a falling body on earth at constant

acceleration due to gravity will also exhibit the universal mass-energy equivalence

relation given as dE/E = dm/m, when the energy considered is mechanical energy.

Total mechanical energy is the sum of the potential and kinetic energies.

Keywords: Acceleration due to gravity, mass-energy equivalence, mechanical energy,

falling body.

SHORT COMMUNICATION

It has been shown that the big and small bodies travelling at constant velocity exhibit

universal mass-energy equivalence relation given as dE/E = dm/m. Here dE is the differential

energy, E is the total energy, dm is the differential mass and m is the total mass [1-2]. When

the energy is only the potential energy of the falling object, then the change in potential

energy due to the change in height is tolerated for only 1 to 2% change in height [3]. In this

article it is shown that a falling mass under constant acceleration due to gravity also exhibit

the universal mass-energy equivalence relation, when the energy is the total mechanical

energy. Consider a ball falling from a height h above the ground under constant acceleration

due to gravity g of 9.8 metres/sec2. The potential energy of the mass at a height h is given by

the equation E = mgh, where m is the mass and h are the height. The potential energy will be

in Joules, if the mass is in Kg, acceleration due to gravity g, is in meters/sec2 and height h is in

meters. Since the velocity of the ball is zero, therefore the total mechanical energy is also mgh.

Now, if the ball is dropped, it falls to an intermediate height h’ lower than h and reaches a

velocity v’ under constant acceleration g. The total mechanical energy will be 0.5mv’2 + mgh’ =

mgh of the original height h. That is, the total mechanical energy as the sum of the potential

and kinetic energy remains as mgh, as if the height h is invariable. Thus, differentiating this

equation once gives

dE = gh (dm) + mg (dh) (1).

Here, dm is a differential mass that can be achieved by say, taking a larger mass, dE is a

differential potential energy and dh is the differential height. Dividing this equation by the

original equation of E =mgh gives:

dE / E = dm/m + dh/h (2).

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Services for Science and Education – United Kingdom 46

European Journal of Applied Sciences (EJAS) Vol. 12, Issue 2, April-2024

Now, since the height for the mechanical energy is invariable, therefore dh is zero or dh/h is

zero. The mechanical energy of the falling ball under constant gravity g expresses the

universal mass-energy equivalence relation:

dE / E = dm/m (3).

The friction due to air on the falling body is neglected.

References

[1]. R.K. Channa, Linear model for the variation of semiconductor bandgap with high temperature for high

temperature electronics, IOSR-J. Electrical and Electronics Engg., 2021, 16(6), p. 5-8.

[2]. R.K. Chanana, Universal mass-energy equivalence for relativistic masses, International J. of Engg. and Science

Invention, 2023, 12(3), p. 35-36.

[3]. R.K. Chanana, Falling bodies on earth at constant acceleration due to gravity exhibit universal mass-energy

equivalence relation, European J. Appl. Sciences, 2023, 11(6), p. 270-271.