Calculate the value of $'g'$ on the surface of the earth. The mass of the earth is $6 \times {10^{24}}\,Kg$ and the radius of the earth is $6 \times {10^6}\,m$ . Gravitational constant is $6.67 \times {10^{ - 11}}\,N{m^2}K{g^{ - 2}}$ .
Answer
249.3k+ views
Hint Use the formula of the acceleration due to gravity and substitute the gravitational constant, mass of the earth and the radius of the earth in it. The simplification of it provides the answer for the acceleration due to gravity of the objects in the earth.
Useful formula:
The formula of the acceleration due to gravity is given by
$g = \dfrac{{GM}}{{{r^2}}}$
Where $g$ is the acceleration due to gravity, $G$ is the gravitational constant, $M$ is the mass of the earth and $r$ is the radius of the earth.
Complete step by step answer
It is given that the
The mass of the earth , $M = 6 \times {10^{24}}\,Kg$
radius of the earth, $r = 6 \times {10^6}\,m$
Gravitational constant, $G = 6.67 \times {10^{ - 11}}\,N{m^2}K{g^{ - 2}}$
Let us consider the formula of the acceleration due to gravity,
$g = \dfrac{{GM}}{{{r^2}}}$
Substitute the gravitational constant, mass of the earth and the radius of the earth in the above formula, we get
$g = \dfrac{{6.67 \times {{10}^{ - 11}}\, \times 6 \times {{10}^{24}}\,}}{{{{\left( {6 \times {{10}^6}\,} \right)}^2}}}$
By performing the various basic arithmetic operations, we get
$g = \dfrac{{6.67 \times {{10}^8}}}{{6 \times {{10}^6}}}$
By further simplification of the above equation, we get
$g = 9.8\,m{s^{ - 2}}$
Hence the acceleration due to gravity of the earth is obtained as the $9.8\,m{s^{ - 2}}$ .
Note The gravitational force of the earth is mainly due to the presence of the molten iron and the nickel in the inner core of the earth. It is constant at all the surface of the earth. As there is the gravitational force of the pull between the object on the earth and the earth, there is also an attraction between the two objects on the surface of the earth. The gravitational force of the moon is six times less than the gravitational force of the moon.
Useful formula:
The formula of the acceleration due to gravity is given by
$g = \dfrac{{GM}}{{{r^2}}}$
Where $g$ is the acceleration due to gravity, $G$ is the gravitational constant, $M$ is the mass of the earth and $r$ is the radius of the earth.
Complete step by step answer
It is given that the
The mass of the earth , $M = 6 \times {10^{24}}\,Kg$
radius of the earth, $r = 6 \times {10^6}\,m$
Gravitational constant, $G = 6.67 \times {10^{ - 11}}\,N{m^2}K{g^{ - 2}}$
Let us consider the formula of the acceleration due to gravity,
$g = \dfrac{{GM}}{{{r^2}}}$
Substitute the gravitational constant, mass of the earth and the radius of the earth in the above formula, we get
$g = \dfrac{{6.67 \times {{10}^{ - 11}}\, \times 6 \times {{10}^{24}}\,}}{{{{\left( {6 \times {{10}^6}\,} \right)}^2}}}$
By performing the various basic arithmetic operations, we get
$g = \dfrac{{6.67 \times {{10}^8}}}{{6 \times {{10}^6}}}$
By further simplification of the above equation, we get
$g = 9.8\,m{s^{ - 2}}$
Hence the acceleration due to gravity of the earth is obtained as the $9.8\,m{s^{ - 2}}$ .
Note The gravitational force of the earth is mainly due to the presence of the molten iron and the nickel in the inner core of the earth. It is constant at all the surface of the earth. As there is the gravitational force of the pull between the object on the earth and the earth, there is also an attraction between the two objects on the surface of the earth. The gravitational force of the moon is six times less than the gravitational force of the moon.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

