
Gravitational on the moon of that on earth?
Answer
512.4k+ 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.
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