
Derive relation between $g$ and $G$
Answer
144.6k+ views
Hint: Newton's law of gravitation states that the force between two unknown masses is directly proportional to the force acting between them. The force is inversely proportional to the square of distance between the masses. Also these masses experience acceleration due to gravity as well.
Complete step by step solution:
The acceleration of a body in free fall due to the massive body's gravity is g. The attraction force between two objects with a unit mass divided in some portion of this universe by a unit distance is G.
The gravity of any large body is g. The inertia on an object. A universal gravitational constant denoting G is the attraction force between any two masses divided by unit size. There is no proportional relationship between G and g. That implies that they are distinct bodies.
Let’s consider two bodies of masses $M$ and $m$ kept at distance $r$ from each other, now, according to Newton’s law of gravitation, we know that,
$F = \dfrac{{GMm}}{{{r^2}}}$
$F$ is the force between the two bodies
$G$ is the gravitational constant
$M$ is mass for first body
$m$ is mass of second body
$r$ is distance between two bodies
Let us consider that the first body is earth with mass $M$, $r$ radius. Now the force acting along the body will be
$F = mg$
$g$ is acceleration due to gravity
From the above two equations, we can write that,
$\dfrac {{GMm}} {{{r^2}}} = mg$
$ \Rightarrow g = \dfrac{{GM}}{{{r^2}}}$
Hence we have a relation between $g$ and $G$.
Note: Although the relationship between g and G in physics can be expressed in a shape. Because of the gravity and the universal gravity, there is no relation between the acceleration and the G value. For some point in this world, the value of G is constant. G and g are not mutually based.
Complete step by step solution:
The acceleration of a body in free fall due to the massive body's gravity is g. The attraction force between two objects with a unit mass divided in some portion of this universe by a unit distance is G.
The gravity of any large body is g. The inertia on an object. A universal gravitational constant denoting G is the attraction force between any two masses divided by unit size. There is no proportional relationship between G and g. That implies that they are distinct bodies.
Let’s consider two bodies of masses $M$ and $m$ kept at distance $r$ from each other, now, according to Newton’s law of gravitation, we know that,
$F = \dfrac{{GMm}}{{{r^2}}}$
$F$ is the force between the two bodies
$G$ is the gravitational constant
$M$ is mass for first body
$m$ is mass of second body
$r$ is distance between two bodies
Let us consider that the first body is earth with mass $M$, $r$ radius. Now the force acting along the body will be
$F = mg$
$g$ is acceleration due to gravity
From the above two equations, we can write that,
$\dfrac {{GMm}} {{{r^2}}} = mg$
$ \Rightarrow g = \dfrac{{GM}}{{{r^2}}}$
Hence we have a relation between $g$ and $G$.
Note: Although the relationship between g and G in physics can be expressed in a shape. Because of the gravity and the universal gravity, there is no relation between the acceleration and the G value. For some point in this world, the value of G is constant. G and g are not mutually based.
Recently Updated Pages
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
