
At a height equal to earth’s radius, above the earth’s surface, the acceleration due to gravity is:
A) $g$.
B) $\dfrac{g}{2}$.
C) $\dfrac{g}{4}$
D) $\dfrac{g}{5}$.
Answer
219.9k+ views
Hint: Due to the large mass of the earth there is a force that pulls the objects towards the centre of the earth and that force is gravitational force. The gravitational force attracts the object towards the centre of the earth with an acceleration known as acceleration due to gravity.
Complete step by step answer:
The formula of the acceleration due to gravity is given by,
$g = G\dfrac{{{M_e}}}{{{d^2}}}$
Where acceleration due to gravity is g the universal gravitational constant is G the mass of earth is ${M_e}$ and the distance of the object from the centre of the earth is d.
Step by step solution:
It is given in the problem that at height R from the surface of the earth an object is placed and we need to find the value of acceleration due to gravity on that object where the radius of the earth is equal to R.
The formula of the acceleration due to gravity is given by,
$g = G\dfrac{{{M_e}}}{{{d^2}}}$
Where acceleration due to gravity is g the universal gravitational constant is G the mass of earth is ${M_e}$ and the distance of the object from the centre of the earth is d.
The acceleration due to gravity at the surface of the earth is equal to,
$ \Rightarrow g = G\dfrac{{{M_e}}}{{{d^2}}}$
$ \Rightarrow g = G\dfrac{{{M_e}}}{{{R^2}}}$………eq. (1)
As the object is R distance from the surface of the earth where the radius of the earth is R the total distance of the object from the centre of the earth is 2R, then the acceleration due to gravity is given by,
$ \Rightarrow {g_1} = G\dfrac{{{M_e}}}{{{d^2}}}$
$ \Rightarrow {g_1} = G\dfrac{{{M_e}}}{{{{\left( {2R} \right)}^2}}}$
$ \Rightarrow {g_1} = G\dfrac{{{M_e}}}{{4{R^2}}}$………eq. (2)
On comparing the equation (1) and equation (2) we get.
\[ \Rightarrow \dfrac{{{g_1}}}{g} = \dfrac{{G\left( {\dfrac{{{M_e}}}{{4{R^2}}}} \right)}}{{G\left( {\dfrac{{{M_e}}}{{{R^2}}}} \right)}}\]
\[ \Rightarrow \dfrac{{{g_1}}}{g} = \dfrac{{\left( {\dfrac{1}{4}} \right)}}{1}\]
\[ \Rightarrow \dfrac{{{g_1}}}{g} = \dfrac{1}{4}\]
\[ \Rightarrow {g_1} = \dfrac{g}{4}\].
So the acceleration due to gravity is equal to \[{g_1} = \dfrac{g}{4}\].
The correct answer for this problem is option C.
Note: The acceleration due to gravity varies with the distance from the centre of the earth and the acceleration due to gravity is inversely proportional to the distance of the object from the centre of the earth and therefore more far the object is placed from the centre of the earth the less the acceleration due to gravity acts on the object.
Complete step by step answer:
The formula of the acceleration due to gravity is given by,
$g = G\dfrac{{{M_e}}}{{{d^2}}}$
Where acceleration due to gravity is g the universal gravitational constant is G the mass of earth is ${M_e}$ and the distance of the object from the centre of the earth is d.
Step by step solution:
It is given in the problem that at height R from the surface of the earth an object is placed and we need to find the value of acceleration due to gravity on that object where the radius of the earth is equal to R.
The formula of the acceleration due to gravity is given by,
$g = G\dfrac{{{M_e}}}{{{d^2}}}$
Where acceleration due to gravity is g the universal gravitational constant is G the mass of earth is ${M_e}$ and the distance of the object from the centre of the earth is d.
The acceleration due to gravity at the surface of the earth is equal to,
$ \Rightarrow g = G\dfrac{{{M_e}}}{{{d^2}}}$
$ \Rightarrow g = G\dfrac{{{M_e}}}{{{R^2}}}$………eq. (1)
As the object is R distance from the surface of the earth where the radius of the earth is R the total distance of the object from the centre of the earth is 2R, then the acceleration due to gravity is given by,
$ \Rightarrow {g_1} = G\dfrac{{{M_e}}}{{{d^2}}}$
$ \Rightarrow {g_1} = G\dfrac{{{M_e}}}{{{{\left( {2R} \right)}^2}}}$
$ \Rightarrow {g_1} = G\dfrac{{{M_e}}}{{4{R^2}}}$………eq. (2)
On comparing the equation (1) and equation (2) we get.
\[ \Rightarrow \dfrac{{{g_1}}}{g} = \dfrac{{G\left( {\dfrac{{{M_e}}}{{4{R^2}}}} \right)}}{{G\left( {\dfrac{{{M_e}}}{{{R^2}}}} \right)}}\]
\[ \Rightarrow \dfrac{{{g_1}}}{g} = \dfrac{{\left( {\dfrac{1}{4}} \right)}}{1}\]
\[ \Rightarrow \dfrac{{{g_1}}}{g} = \dfrac{1}{4}\]
\[ \Rightarrow {g_1} = \dfrac{g}{4}\].
So the acceleration due to gravity is equal to \[{g_1} = \dfrac{g}{4}\].
The correct answer for this problem is option C.
Note: The acceleration due to gravity varies with the distance from the centre of the earth and the acceleration due to gravity is inversely proportional to the distance of the object from the centre of the earth and therefore more far the object is placed from the centre of the earth the less the acceleration due to gravity acts on the object.
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

States of Matter Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Atomic Structure for Beginners

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

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

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

