Why is the weight of an object on the moon 1/6th weight on the earth

Let M Ebe the mass of the Earth and mbe an object on the surface of the Earth. Let R Ebe the radius of the Earth. According to the universal law of gravitation, weight W Eof the object on the surface of the Earth is given by, `W_E=(GM_Em)/R_E^2` Let M M and R Mbe the mass and radius of the moon. Then, according to the universal law of gravitation, weightW Mof the object on the surface of the moon is given by:- `W_E=(GM_Mm)/R_M^2` `W_M/W_E=(M_MR_E^2)/(M_ER_M^2)` `"Where, "M_E=5.98xx10^24kg," "M_M=7.36xx10^22kg` `R_E=6.4xx10^6m," "R_M=1.74xx10^6m` `thereforeW_M/W_E=(7.36xx10^22xx(6.37xx10^6)^2)/(5.98xx10^24xx(1.74xx10^6)^2)=0.165approx1/6` Therefore, weight of an object on the moon is `1/6` of its weight on the Earth.

Hello everyone. So today I'm going to solve this question for you. Let me write the solution over here. So basically we know that the weight of a body, it's basically a force, a force that can be expressed as mass times the acceleration. And since the weight of a body is by virtue of the gravitation by virtue of gravity, that is why this acceleration becomes, we take it due to gravity. Okay, so this is acceleration due to gravity. So wait becomes equal to MG where m is the mass of the body? Okay. And in this question we have been asked to tell why the weight of an object…

Solution The mass and radius of moon are less than that of earth due to which the moon exerts a lesser gravitational force on objects (which is 1 6 th of that exerted by the earth). Since the gravitational force of the moon is 1 6 th that of the earth ,therefore , the weight of an object on the moon is 1 6 th of its weight on the earth .

Table of Contents • • • • • • • • • • Why does an object on the moon weigh 1/6th compared to its weight on Earth? The mass of the moon is 1/100 times and its radius 1/4 times that of earth. The moon’s gravitation force is determined by the mass and the size of the moon. Hence, the weight of an object on the moon is 1/6th its weight on the earth. The moon is far less massive than the Earth and has a different radius(R) as well. How is the weight of an object on the moon one sixth of its weight on the Earth? Why is the weight of an object on the moon 1/6th its weight on the earth? The moon’s mass is 1/100 times and 1/4 times the earth’s radius. As a result, when compared to earth, the gravitational attraction on the moon is about one-sixth. Thus, an object’s weight on the moon is 1/6th its earth weight. Does the moon have 1/6 mass of Earth? Because weight is directly dependent upon gravitational acceleration, things on the Moon will weigh only 16.6% (= 1/6) of what they weigh on the Earth. What is its mass on the moon where the force of gravity is 1/6 that of the Earth answer in units of KG? Its mass is still 50 kg but the Moon’s gravitational pull is only about 1/6 of the Earth’s – in other words about 1.6 N/kg. This means that the weight of our crate on the Moon will be 50×1. 6 = 80 N and so it will be much easier to lift up….Mass and weight. Planet Weight of the 50 kg crate Uranus 520 N Neptune 690 N (Pluto) 14.5 N Is the moon’s gravity 1/6 of Earth’s gravity? The Moon’s ...

Step 1: Parameters at Moon Mass of the moon m m = 1 100 m e Radius of the moon R m = 1 4 R e Step 2: Calculate the value of g at moon's surface the value of g at surface of earth, g e = G M e R 2 e the value of gon surface of moon, g m = G M m R 2 m g m = G × 1 100 M e 1 16 × R 2 e g m ≃ 1 6 × G M e R 2 e g m = 1 6 g e Hence, the weight of an object on the moon 1 6 times its weight on the earth.

(a) Newtons Second Law of Motion F = m a ; F = m g Universal Law of Gravitation F = R 2 G M m ​ m g = R 2 G M m ​ , g = R 2 G M ​ (b) g e ​ = g and g m ​ = 6 g ​ Force applied to lift a mass of 15 kg, at the earth F = m g e ​ = 1 5 g N Therefore, the mass lifted by the same force on the moon, m = g m ​ F ​ = 6 g ​ 1 5 g ​ = 9 0 k g Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0 . 0 3 3 7 m / s 2, whereas a point at the poles experiences no centripetal acceleration. If a person at the equator has a mass of 7 5 . 0 k g, calculate (a) the gravitational force (true weight) on the person and (b) the normal force (apparent weight) on the person. (c) Which force is greater? Assume the Earth is a uniform sphere and take g = 9 . 8 0 0 m / s 2.