Weight of the object on Moon — lesson. Science CBSE, Class 9.

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We have learnt that the weight of an object depends on location in which it is placed. The force acting on the apple on the earth surface varies on the moon surface. This is because the mass of the moon is less than that of the earth. Due to this, the moon exerts a lesser force of attraction on objects. So the weight of the apple will be less on the moon surface.

Let us compare the weight of an astronaut near the earth surface and the moon surface,

technology-star-atmosphere-space-darkness-float-1404943-pxhere.com.jpg

Astronaut near the surface of moon

Assume,

The mass of the astronaut(object) be

m

The weight of the astronaut on the moon be

W_{m}

The mass of the moon be

M_{m}

The radius of the moon be

R_{m}

The weight of the object on the moon will be calculated using the universal law of gravitation.

W_{m} = \frac{G M_{m} m}{{R_{m}}^{2}} \to 1

Let us consider the weight of the same object on the earth be

W_{e}

. The mass of the earth as \(M\), and its radius as \(R\).

W_{e} = \frac{GMm}{R^{2}} \to 2

Divide the equation 1 by 2,

\begin{array}{l} \frac{W_{m}}{W_{e}} = \frac{M_{m} R^{2}}{M {R_{m}}^{2}} \\ Apply the known values, \\ Mass of the moon = 7.36 \times 10^{22} \\ Mass of the earth = 5.98 \times 10^{24} \\ Radius of the earth = 6.37 \times 10^{6} \\ Radius of the moon = 1.74 \times 10^{6} \\ \frac{W_{m}}{W_{e}} = \frac{7.36 \times 10^{22} \times {(6.37 \times 10^{6})}^{2}}{5.98 \times 10^{24} \times {(1.74 \times 10^{6})}^{2}} \\ \frac{W_{m}}{W_{e}} = \frac{2.431 \times 10^{10}}{1.474 \times 10^{11}} \\ \frac{W_{m}}{W_{e}} = 0.165 \approx \frac{1}{6} \\ or \\ \frac{Weight of the object on the moon}{Weight of the object on the earth} = \frac{1}{6} \\ or \\ Weight of the object on the moon = \frac{Weight of the object on the earth}{6} \end{array}

Reference:

https://pxhere.com/en/photo/1404943

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