
Gravitational on the moon of that on earth?
Answer
519k+ views
Hint : We solve this question by using the formula of gravitation. Substituting all the values with respect to the moon we get the gravitational force on the surface of the moon then comparing this with the gravity on earth we get the relation between gravity on the surface of the earth and moon.
Formula used: Surface gravity equation is $ g = \dfrac{{GM}}{{{r^2}}} $
Here, gravity on the surface of moon is represented by $ g $
Gravitation constant $ 6.673 \times {10^{ - 11}} $ is represented by $ G $
Mass of moon is represented by $ M $
Radius of the moon is represented by $ r $ .
Complete step by step answer
The formula for gravity at the surface of a body is $ g = \dfrac{{GM}}{{{r^2}}} $
Since we are finding gravity at the surface of the moon. We take the values with respect to moon
$ M = 7.324 \times {10^{22}}kg $
$ r = 1.74 \times {10^6}m $
Substituting the values in surface gravity equation
$ g = \dfrac{{6.673 \times {{10}^{ - 11}} \times 7.324 \times {{10}^{22}}}}{{{{(1.74 \times {{10}^6})}^2}}} $
$ \Rightarrow g = 1.625m/{s^2} $
Hence gravity on surface of moon is equal to $ 1.625m/{s^2} $
We know that the gravity of earth is equal to $ 9.8m/{s^2} $
Dividing gravity of moon with gravity of earth we get
$ \dfrac{{1.625m/{s^2}}}{{9.8m/{s^2}}} = 0.16 $
By this we can say that gravity on the surface of the moon is $ \dfrac{1}{6} $ of that of earth. Or $ 16.7\% $ of that of earth.
Additional Information
Gravitation is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward (or gravitate toward) one another.
Note
We can also solve this question by multiplying the value of gravity on earth with $ \dfrac{1}{6} $ since it is known that the gravity on the moon is $ \dfrac{1}{6} $ of that on earth. From the solution, we can say that gravity on the surface of a body is directly proportional to the ratio of the mass of the body to the square of its radius. Greater this value of ratio greater the gravitation on the surface.
Formula used: Surface gravity equation is $ g = \dfrac{{GM}}{{{r^2}}} $
Here, gravity on the surface of moon is represented by $ g $
Gravitation constant $ 6.673 \times {10^{ - 11}} $ is represented by $ G $
Mass of moon is represented by $ M $
Radius of the moon is represented by $ r $ .
Complete step by step answer
The formula for gravity at the surface of a body is $ g = \dfrac{{GM}}{{{r^2}}} $
Since we are finding gravity at the surface of the moon. We take the values with respect to moon
$ M = 7.324 \times {10^{22}}kg $
$ r = 1.74 \times {10^6}m $
Substituting the values in surface gravity equation
$ g = \dfrac{{6.673 \times {{10}^{ - 11}} \times 7.324 \times {{10}^{22}}}}{{{{(1.74 \times {{10}^6})}^2}}} $
$ \Rightarrow g = 1.625m/{s^2} $
Hence gravity on surface of moon is equal to $ 1.625m/{s^2} $
We know that the gravity of earth is equal to $ 9.8m/{s^2} $
Dividing gravity of moon with gravity of earth we get
$ \dfrac{{1.625m/{s^2}}}{{9.8m/{s^2}}} = 0.16 $
By this we can say that gravity on the surface of the moon is $ \dfrac{1}{6} $ of that of earth. Or $ 16.7\% $ of that of earth.
Additional Information
Gravitation is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward (or gravitate toward) one another.
Note
We can also solve this question by multiplying the value of gravity on earth with $ \dfrac{1}{6} $ since it is known that the gravity on the moon is $ \dfrac{1}{6} $ of that on earth. From the solution, we can say that gravity on the surface of a body is directly proportional to the ratio of the mass of the body to the square of its radius. Greater this value of ratio greater the gravitation on the surface.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

While covering a distance of 30km Ajeet takes 2 ho-class-11-maths-CBSE

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is the opposite of entropy class 11 chemistry CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE
