
The moon’s radius is \[1/4\] that of the earth and its mass \[1/80\] times that of the earth. If \[g\] represents the acceleration due to gravity on the surface of the earth, then on the surface of the moon its value is:
(A) \[\dfrac{g}{4}\]
(B) \[\dfrac{g}{5}\]
(C) \[\dfrac{g}{6}\]
(D) \[\dfrac{g}{8}\]
Answer
233.1k+ views
Hint: We have been provided with the information about the ratio of the masses of the moon and the earth and the value of the ratio of the radius of the moon to that of the earth. Since no actual data has been provided to us, we can say that we have to solve this question by comparison of the two scenarios. Let’s see how it will be done.
Formula Used:
\[g=\dfrac{GM}{{{r}^{2}}}\]
Complete step by step answer:
The acceleration due to gravity on the surface of the earth is given as
\[{{g}_{earth}}=\dfrac{G\times mas{{s}_{earth}}}{radiu{{s}_{earth}}^{2}}\] where \[G\] is the universal gravitational constant
Similarly, the acceleration due to gravity on the surface of the moon will be given as
\[{{g}_{moon}}=\dfrac{G\times mas{{s}_{moon}}}{radiu{{s}_{moon}}^{2}}\] where \[G\] is the universal gravitational constant
Taking the ratio of the acceleration due to the gravity of the moon and the earth, we get
$\implies$\[\dfrac{{{g}_{moon}}}{{{g}_{earth}}}=\dfrac{\dfrac{G\times mas{{s}_{moon}}}{radiu{{s}_{moon}}^{2}}}{\dfrac{G\times mas{{s}_{earth}}}{radiu{{s}_{earth}}^{2}}}\]
The universal gravitational constant will get cancelled out, leaving us with
$\implies$\[\dfrac{{{g}_{moon}}}{{{g}_{earth}}}=\dfrac{mas{{s}_{moon}}}{mas{{s}_{earth}}}\times {{\left( \dfrac{radiu{{s}_{earth}}}{radiu{{s}_{moon}}} \right)}^{2}}\]
Now we have been told that the moon’s radius is one-fourth the radius of the earth; this means that
$\implies$\[\dfrac{radiu{{s}_{earth}}}{radiu{{s}_{moon}}}=4\]
Also, the mass of the earth is eighty times that of the moon, so
$\implies$\[\dfrac{mas{{s}_{moon}}}{mas{{s}_{earth}}}=\dfrac{1}{80}\]
Substituting these values in our expression for \[g\], we get
\[\begin{align}
& \dfrac{{{g}_{moon}}}{{{g}_{earth}}}=\dfrac{1}{80}\times {{(4)}^{2}} \\
& \Rightarrow \dfrac{{{g}_{moon}}}{{{g}_{earth}}}=\dfrac{16}{80} \\
& \Rightarrow {{g}_{moon}}=\dfrac{{{g}_{earth}}}{5} \\
\end{align}\]
Since the value \[{{g}_{earth}}\] is given to be \[g\], the values \[{{g}_{moon}}\] will be \[\dfrac{g}{5}\] .
Hence option (B) is the correct answer.
Additional Information: The expression for the acceleration due to gravity is derived from Newton’s law of gravitation formula. Newton’s law of gravitation gives us the force of gravity acting between two particles of a certain mass and kept at a certain distance from each other.
Note: We can simplify the calculation procedure of the above question by assuming the value of the earth’s mass and radius to have certain values. Then we can find the values of the mass of the moon and the radius of the moon with respect to those values. Upon substituting those values in the expression for the acceleration due to gravity, we’ll get very simple terms. The steps of the procedure will remain the same so don’t worry about that.
Formula Used:
\[g=\dfrac{GM}{{{r}^{2}}}\]
Complete step by step answer:
The acceleration due to gravity on the surface of the earth is given as
\[{{g}_{earth}}=\dfrac{G\times mas{{s}_{earth}}}{radiu{{s}_{earth}}^{2}}\] where \[G\] is the universal gravitational constant
Similarly, the acceleration due to gravity on the surface of the moon will be given as
\[{{g}_{moon}}=\dfrac{G\times mas{{s}_{moon}}}{radiu{{s}_{moon}}^{2}}\] where \[G\] is the universal gravitational constant
Taking the ratio of the acceleration due to the gravity of the moon and the earth, we get
$\implies$\[\dfrac{{{g}_{moon}}}{{{g}_{earth}}}=\dfrac{\dfrac{G\times mas{{s}_{moon}}}{radiu{{s}_{moon}}^{2}}}{\dfrac{G\times mas{{s}_{earth}}}{radiu{{s}_{earth}}^{2}}}\]
The universal gravitational constant will get cancelled out, leaving us with
$\implies$\[\dfrac{{{g}_{moon}}}{{{g}_{earth}}}=\dfrac{mas{{s}_{moon}}}{mas{{s}_{earth}}}\times {{\left( \dfrac{radiu{{s}_{earth}}}{radiu{{s}_{moon}}} \right)}^{2}}\]
Now we have been told that the moon’s radius is one-fourth the radius of the earth; this means that
$\implies$\[\dfrac{radiu{{s}_{earth}}}{radiu{{s}_{moon}}}=4\]
Also, the mass of the earth is eighty times that of the moon, so
$\implies$\[\dfrac{mas{{s}_{moon}}}{mas{{s}_{earth}}}=\dfrac{1}{80}\]
Substituting these values in our expression for \[g\], we get
\[\begin{align}
& \dfrac{{{g}_{moon}}}{{{g}_{earth}}}=\dfrac{1}{80}\times {{(4)}^{2}} \\
& \Rightarrow \dfrac{{{g}_{moon}}}{{{g}_{earth}}}=\dfrac{16}{80} \\
& \Rightarrow {{g}_{moon}}=\dfrac{{{g}_{earth}}}{5} \\
\end{align}\]
Since the value \[{{g}_{earth}}\] is given to be \[g\], the values \[{{g}_{moon}}\] will be \[\dfrac{g}{5}\] .
Hence option (B) is the correct answer.
Additional Information: The expression for the acceleration due to gravity is derived from Newton’s law of gravitation formula. Newton’s law of gravitation gives us the force of gravity acting between two particles of a certain mass and kept at a certain distance from each other.
Note: We can simplify the calculation procedure of the above question by assuming the value of the earth’s mass and radius to have certain values. Then we can find the values of the mass of the moon and the radius of the moon with respect to those values. Upon substituting those values in the expression for the acceleration due to gravity, we’ll get very simple terms. The steps of the procedure will remain the same so don’t worry about that.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

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

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

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

