Universal Law of Gravitation: Significance, Value, & Celestial Forces

14 Dec 2023

Newton’s Universal Law and Its Cosmic Significance

The Universal Law of Gravitation, formulated by Sir Isaac Newton, is pivotal in understanding the fundamental forces governing the cosmos. This law not only elucidates planetary orbits but also enables advancements in astronomy, space exploration, and a profound comprehension of the universe’s intricate mechanics.

Gravity’s Influence and the Cosmic Dance and Earthly Mysteries

  • Help in Understanding Universe: The Universal Law of Gravitation is foundational in understanding the large-scale structure and behaviour of the universe, from why apples fall from trees to the intricate dances of galaxies.
  • Crucial Law: The Universal Law of Gravitation is crucial because it unifies the understanding of several phenomena. 
  • Few of the more common ones are listed as under:
    • Earthly Attraction: It explains why things fall towards the Earth.
    • Moon’s Orbit: The motion of the moon around the Earth is due to the gravitational attraction between them.
    • Planetary Motion: Planets orbit the Sun due to gravitational forces.
    • Tidal Phenomena: Tides in the oceans are significantly influenced by the gravitational forces exerted by the Moon and, to a lesser extent, the Sun.

Free Fall and Gravity: Forces That Shape our Descent

  • Objects are said to be in free fall when they are only influenced by the gravitational force of the Earth, with no other forces (like air resistance) acting on them.
  • Acceleration due to Gravity (g): Objects in free fall experience an acceleration towards the Earth due to its gravitational pull. This acceleration is denoted as g (unit: m s-2) and is termed as the “acceleration due to gravity”. 
  • Gravitational Force on a Falling Object: 

F=mg 

Where: 

m = mass of the object; 

g = acceleration due to gravity

  • Relation with Universal Gravitational Constant: 

mg = G (M × m / d2 ) or,

   g = G (M / d2 ), 

where: 

G = Universal gravitational constant, 

M = Mass of the Earth, 

R = Radius of the Earth

  • Variation with Latitude: Earth is not a perfect sphere; it’s slightly flattened at the poles and bulges at the equator. Thus, g is slightly greater at the poles than at the equator.

Calculating the Value of g:

  • Using Known Values: 

g = G (M / R2

Using values: 

g = 6.7×10−11 Nm2 / kg-2 × 6×1024kg / (6.4 × 106 m)2

After calculation, we get 

g =9.8 ms-2

Hence, the standard value for the acceleration due to gravity on the surface of Earth is 

g = 9.8 ms-2

Free Fall and Acceleration: Uncovering the  Universal Law Gravitational of Gravitational Motion

  • Galileo’s Experiment:
    • Galileo, in an experiment from the Leaning Tower of Pisa, demonstrated that all objects, irrespective of their mass, fall at the same rate in the absence of air resistance.
  • Equations of Motion in Free Fall:
    • With g as constant near Earth’s surface, equations of uniformly accelerated motion apply. Here, acceleration a is replaced by g:
    • v= u+gt
    • s = ut + ½  ​gt2
    • v2 = u2 + 2gs 
  • Here: u = Initial velocity, v = Final velocity, s = Distance covered in time t
  • The direction of g is taken positive when it’s in the direction of motion (downwards) and negative when opposing it (like in case of an object thrown upwards).

Example: A car falls off a ledge and drops to the ground in 0.5 s. Let g = 10 m s–2 (for simplifying the calculations). 

(i)   What is its speed on striking the ground?

(ii)  What is its average speed during the 0.5 s?

(iii) How high is the ledge from the ground?

Solution: 

Time, t = 1⁄2 second
Initial velocity, u = 0  m s–1
Acceleration due to gravity, g = 10 m s–2 

Acceleration of the car, a = + 10 m s–2  (downward) 

(i) speed v = a t
v = 10ms–2  × 0.5s 

= 5ms–1 

(ii) average speed = u + v / 2 

= (0ms–1 + 5ms–1) / 2 

= 2.5 m s–1
(iii) distance travelled, s = 1⁄2 a t2 

= 1⁄2 × 10ms–2 × (0.5s)2 = 1⁄2 × 10ms–2 × 0.25 s2 = 1.25 m 

Thus, 

(i) its speed on striking the ground = 5 m s–1 

(ii) its average speed during the 0.5 s = 2.5 m s–1 

(iii) height of the ledge from the ground = 1.25 m. 

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