Acceleration Due to Gravity | Formula, Derivation, Important Questions & Answers, JEE, NEET & CBSE Class 11 Notes

Introduction

The force of attraction exerted by the Earth on a body is called gravitational pull or gravity. When a force acts on a body, it produces acceleration. Therefore, a body under the effect of gravitational pull must accelerate. The acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity (g).


Formula and Derivation of Acceleration Due to Gravity(g)

Consider a body of mass m lying on the surface of the Earth. The gravitational force on the body is given by:

F = (G × m × M) / r2

Where,

  • M = mass of the Earth
  • R = radius of the Earth
  • G = universal gravitational constant

Let’s consider an object of mass m, on which the acceleration due to gravity g is acting, now suppose F is the force acting on it and is given by:

F = mg

where F is the force acting on the object, g is the acceleration due to gravity and m is the mass of the object.

Using Newton’s second law, force can also be expressed as:

F = mg

Equating both equations:

mg = (G × m × M)/R2

Solving for g:

g = GM/R2

Hence, the formula of acceleration due to gravity is evaluated as shown above.

Note: It depends on the mass and radius of the earth.

This helps us understand the following:

  • All bodies experience the same acceleration due to gravity, irrespective of their mass.
  • Its value on earth depends upon the mass of the earth and not the mass of the object.

Acceleration Due to Gravity – Formula, Unit and Values

Acceleration Due to Gravity (g)
Symbolg
Dimensional FormulaM0L1T-2
SI Unitms-2
Formulag = GM/R2
Values of g in SI9.806 ms-2
Values of g in CGS980 cm s-2

Key Points About Acceleration Due to Gravity

  • It depends on the mass, radius, and density of the planet.
  • It is independent of the mass, shape, and density of the object.
  • The greater the value of M/R2, the greater the value of g.
  • It is a vector quantity directed towards the center of the planet.
  • Dimensional formula: [g]=[LT−2]
  • Standard value on Earth: g=9.8m/s2 at mean sea level.

Factors Affecting Acceleration Due to Gravity

  1. Shape of the Earth: Due to its oblate shape, g is greater at the poles and less at the equator.
  2. Height Above Earth’s Surface: As altitude increases, g decreases.
  3. Depth Below Earth’s Surface: As depth increases, g decreases.
  4. Axial Rotation of the Earth: Due to centrifugal force, g is slightly less at the equator.

Multiple Choice Questions (MCQs)

Q1. What is the SI unit of acceleration due to gravity?

A) m/s
B) m/s²
C) cm/s
D) N/kg

Answer: B) m/s²
Explanation: The SI unit of acceleration due to gravity is meters per second squared (m/s²).

Q2. What is the approximate value of acceleration due to gravity on Earth?

A) 8.9 m/s²
B) 10.2 m/s²
C) 9.8 m/s²
D) 6.7 m/s²

Answer: C) 9.8 m/s²
Explanation: The standard value of g on Earth’s surface is taken as 9.8 m/s².

Q3. Where is the acceleration due to gravity the highest on Earth?

A) Equator
B) Poles
C) Tropics
D) None of the above

Answer: B) Poles
Explanation: Due to the oblate shape of the Earth, g is higher at the poles than at the equator.


Frequently Asked Questions (FAQs)

Q1. Why does the acceleration due to gravity vary with height and depth?

As we move away from the Earth’s surface (either above or below), the effect of gravity weakens due to the inverse square relation with distance.

Q2. Does acceleration due to gravity depend on the mass of the falling object?

No, acceleration due to gravity is independent of the mass of the object. All objects fall at the same rate in the absence of air resistance.

Q3. How does the rotation of the Earth affect the acceleration due to gravity?

The Earth’s rotation creates a centrifugal force that reduces the effective value of g at the equator.


Solved Example

Question: Calculate the gravitational field (gravitational acceleration) of an object having mass 5 times that of the earth. (g of earth = 9.8 m/s2, radius of earth = 6.3 × 10 m, G = 6.67 × 10-11 m = 6 × 1024 kg)

Solution:

Gravitational acceleration is given by, 

g = Gm/R2 

Then,

g = (6.67 × 10-11 ) × 5(6 × 1024 )/(6.3 × 106)

g = 49 m/s2


Test Your Knowledge

Quiz – Acceleration due to gravity

Acceleration Due to Gravity Quiz

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Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.