
Gravitational force on the surface of the moon is only $\dfrac{1}{6}$ as strong as gravitational force on the earth. What is the weight in Newton of a 10Kg object on the moon and on the earth?
Answer
140.4k+ views
Hint: In this question consider the gravitational force on the surface of the moon be ${g_m}$ that is ${g_m} = \dfrac{g}{6} = \dfrac{{9.8}}{6}$. Then use that the force acting upon the body is actually the weight of the body and is the product of the mass and acceleration due to gravity that is $F = W = Mg$. This will help approaching the problem.
Complete step-by-step solution -
According to Newton’s second law of motion the force acting on the body is the product of the mass of the body and the acceleration by which the body is moving.
In case if a body is at a rest position the force acting on the body is equal to the weight of the body which is acting downward on the body and this is equal to the product of mass of the body and the acceleration due to gravity.
$ \Rightarrow F = W = Mg$Newton (N)..................... (1)
Where, W = weight of the body, M = mass of the body and g = acceleration due to gravity.
Now it is given that the mass (M) of an object = 10Kg.
And gravitational force on the surface of the moon is (1/6) times the gravitational force on the earth.
As we know on the earth surface the gravitational force, g = 9.8 m/s2.
So on the surface of the moon the gravitational force will be, ${g_m} = \dfrac{g}{6} = \dfrac{{9.8}}{6}$ m/s2.
Now from equation (1) weight of the object on the surface of the earth is,
$ \Rightarrow {W_e} = 10 \times 9.8 = 98N$
And the weight of the object on the surface of the moon is,
$ \Rightarrow {W_m} = 10 \times \dfrac{{9.8}}{6} = 16.66N$
So this is the required weight of the object on the surface of the earth and on the moon respectively.
Note – The main reason behind the lesser gravity on the surface of the moon as compared to the earth is that moon’s size is comparatively very less as compared to the earth moreover moon is also less dense as compared to that of earth. Less dense allows weak force of gravitational attraction on the moon.
Complete step-by-step solution -
According to Newton’s second law of motion the force acting on the body is the product of the mass of the body and the acceleration by which the body is moving.
In case if a body is at a rest position the force acting on the body is equal to the weight of the body which is acting downward on the body and this is equal to the product of mass of the body and the acceleration due to gravity.
$ \Rightarrow F = W = Mg$Newton (N)..................... (1)
Where, W = weight of the body, M = mass of the body and g = acceleration due to gravity.
Now it is given that the mass (M) of an object = 10Kg.
And gravitational force on the surface of the moon is (1/6) times the gravitational force on the earth.
As we know on the earth surface the gravitational force, g = 9.8 m/s2.
So on the surface of the moon the gravitational force will be, ${g_m} = \dfrac{g}{6} = \dfrac{{9.8}}{6}$ m/s2.
Now from equation (1) weight of the object on the surface of the earth is,
$ \Rightarrow {W_e} = 10 \times 9.8 = 98N$
And the weight of the object on the surface of the moon is,
$ \Rightarrow {W_m} = 10 \times \dfrac{{9.8}}{6} = 16.66N$
So this is the required weight of the object on the surface of the earth and on the moon respectively.
Note – The main reason behind the lesser gravity on the surface of the moon as compared to the earth is that moon’s size is comparatively very less as compared to the earth moreover moon is also less dense as compared to that of earth. Less dense allows weak force of gravitational attraction on the moon.
Recently Updated Pages
Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

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

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
