
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
164.4k+ 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
Uniform Acceleration - Definition, Equation, Examples, and FAQs

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

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

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

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

Charging and Discharging of Capacitor

A body of mass 3Kg moving with a velocity of 4ms towards class 11 physics JEE_Main

Class 11 JEE Main Physics Mock Test 2025

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Other Pages
JEE Advanced 2025 Notes

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks

NEET Marks vs Rank 2024|How to Calculate?
