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
298.5k+ 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
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

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

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

