Why is the weight of an object on the moon $( \frac{1}{6})^{th}$ it's weight on the earth?

As we know, the weight of an object $W=mg$

Here, $m\rightarrow$mass of the object

$g\rightarrow$ gravitational acceleration

And we know that gravitational acceleration $g'$ on the moon is $\frac{1}{6}$ of the gravitational acceleration on the earth.

So, the weight of the object on the earth $W_{earth}=mg$

Weight of the object on the moon $W_{moon}=mg'=m\times\frac{1}{6}\times g$

$=\frac{mg}{6}$

$=\frac{W_{earth}}{6}$   [Because $W_{earth}=mg$]

Therefore, we come to know that weight of an object on the moon $( \frac{1}{6})^{th}$ its weight on the earth.

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Updated on: 10-Oct-2022

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