Question

show that
The weight of an Object on the moon is 1/6
of its total weight
on the earth
give one example also​

Answer 1

Answer:

weight of the moon is 1/6 weight on the earth

Answer 2

$\bold{ \color{darkblue}{QUESTION}}$

show That the weight of an Object on the moon

is 1/6 of its total weight on the earth .

$\bold{ \color{darkblue}{ANSWER}}$

suppose we have an object of mass (m). Let its weight on the moon be $\rm{w_m}$.

Again suppose that the mass of the moon is (M)and it's radius is R .Now according to universal law of gravitation , the weight (force) of the object on moon will be :

$\bold{ \color{red}{weight \: on \: moon \: w_m \: = G \times \frac{M \times m}{ {R}^{2} } }}$

↓____________ Equation (i)

This weight is actually the force with which the moon attracts the object.

Now,

Suppose the weight of the same object on the earth be $\rm{w_e}$.Now, we know that the mass of the earth is 100 times that of the moon and the radius of the earth is 4 times that of the moon.

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Thus, if mass of the moon is M ,then the mass of earth will be 100M and radius of the earth as 4R , and applying universal law of gravitation, the weight of object on the earth will be :

$\bold{ \color{red}{weight \: on \: earth \: w_e \: = G \times \frac{100M \times \: m}{ {4R}^{2} } }}$

$\red \implies\bold{ \color{red}{w_e \: = G \times \frac{100M \times \: m}{16 {R}^{2} } }}$

↓ _____________ Equation (ii)

Now , dividing equation (i) and (ii) , we get :

$\implies\bold{ \color{black}{ \frac{W_m}{W_e} = \frac{G \times M \times m \times {16R}^{2} }{ {R}^{2} \times G \times 100M \times m} }}$

$\implies\bold{ \color{black}{ \frac{W_m}{W_e} = \cancel \frac{16}{100}}}$

$\implies\bold{ \color{black}{ \frac{W_m}{W_e} =\frac{1}{6}}}$

$\implies\bold{ \frac{weight \: on \: moom}{weight \: on \: earth} = \frac{1}{6}}$

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Please note that the mass of the object is same on the moon and the earth but it's weight on the moon is only 1\6 th of the weight on earth.