The decrease in the value of $g$ on going to a height $\dfrac{R}{2}$ above the earth’s surface will be:
(A) $\dfrac{g}{2}$
(B) $\dfrac{{5g}}{9}$
(C) $\dfrac{{4g}}{9}$
(D) $\dfrac{g}{3}$
Answer
298.5k+ views
Hint: The acceleration due to gravity is inversely proportional to the square of the radius of the earth. If the gravity is measured at a height then the value of distance will be increased. Thus, the acceleration due to gravity will be decreased.
Complete solution:
Acceleration due to gravity is the rate of change in velocity of a free-falling body under the influence of gravity. The numerical value seems to be a contestant on the surface of earth. And can be called acceleration due to gravity.
The expression for the acceleration due to gravity is given as,
$g = \dfrac{{GM}}{{{R^2}}}...................\left( 1 \right)$
Where, $G$ is the gravitational constant, $M$ is the mass of earth and $R$ is the radius of earth.
When the acceleration due to gravity is calculated at a height of $h$ , then the distance is taken as, $R + h$ .
Then the equation $\left( 1 \right)$ changes to,
$\Rightarrow {g_h} = \dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}$
If we are considering $h = \dfrac{R}{2}$. That is if the height taken is half the radius of earth.
Then, we can write the equation as,
$\Rightarrow {g_h} = \dfrac{{GM}}{{{{\left( {R + \dfrac{R}{2}} \right)}^2}}} \\
\Rightarrow \dfrac{{GM}}{{{R^2} + {R^2} + \dfrac{{{R^2}}}{4}}} \\
\Rightarrow \dfrac{{GM}}{{\dfrac{{9{R^2}}}{4}}} \\
\Rightarrow \dfrac{{4GM}}{{9{R^2}}}$
Substitute the equation (1) in the above equation.
$\Rightarrow {g_h} = \dfrac{{4g}}{9}$
In order to find the decrease in the value of acceleration due to gravity, subtract the above expression from the acceleration due to gravity.
Therefore, $g - {g_h}$
Substitute the value in above expression,
$\Rightarrow g - \dfrac{{4g}}{9} = \dfrac{{5g}}{9}$
The decrease in the value of $g$ on going to a height above the earth’s surface will be $\dfrac{{5g}}{9}$.
The answer is option B.
Note: It is clear that the acceleration due to gravity will decrease when we go up to the surface of earth. At the surface of earth, the acceleration due to gravity is constant. The acceleration due to gravity is directly proportional to the gravitational constant.
Complete solution:
Acceleration due to gravity is the rate of change in velocity of a free-falling body under the influence of gravity. The numerical value seems to be a contestant on the surface of earth. And can be called acceleration due to gravity.
The expression for the acceleration due to gravity is given as,
$g = \dfrac{{GM}}{{{R^2}}}...................\left( 1 \right)$
Where, $G$ is the gravitational constant, $M$ is the mass of earth and $R$ is the radius of earth.
When the acceleration due to gravity is calculated at a height of $h$ , then the distance is taken as, $R + h$ .
Then the equation $\left( 1 \right)$ changes to,
$\Rightarrow {g_h} = \dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}$
If we are considering $h = \dfrac{R}{2}$. That is if the height taken is half the radius of earth.
Then, we can write the equation as,
$\Rightarrow {g_h} = \dfrac{{GM}}{{{{\left( {R + \dfrac{R}{2}} \right)}^2}}} \\
\Rightarrow \dfrac{{GM}}{{{R^2} + {R^2} + \dfrac{{{R^2}}}{4}}} \\
\Rightarrow \dfrac{{GM}}{{\dfrac{{9{R^2}}}{4}}} \\
\Rightarrow \dfrac{{4GM}}{{9{R^2}}}$
Substitute the equation (1) in the above equation.
$\Rightarrow {g_h} = \dfrac{{4g}}{9}$
In order to find the decrease in the value of acceleration due to gravity, subtract the above expression from the acceleration due to gravity.
Therefore, $g - {g_h}$
Substitute the value in above expression,
$\Rightarrow g - \dfrac{{4g}}{9} = \dfrac{{5g}}{9}$
The decrease in the value of $g$ on going to a height above the earth’s surface will be $\dfrac{{5g}}{9}$.
The answer is option B.
Note: It is clear that the acceleration due to gravity will decrease when we go up to the surface of earth. At the surface of earth, the acceleration due to gravity is constant. The acceleration due to gravity is directly proportional to the gravitational constant.
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)

