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What will be the weight of a person on earth who weighs $9\,N$ on the moon ?
A. $3\,N$
B. $15\,N$
C. $45\,N$
D. $54\,N$

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Last updated date: 16th Jul 2024
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Answer
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Hint: Before going to the question let us know about gravitation. The force of attraction that exists between all masses in the universe, especially the attraction of the earth's mass to bodies near its surface. “Geometry between two bodies is proportional to the product of their masses and inversely proportional to the square of their separation.”

Complete step by step answer:
The average gravity on the surface of the Earth is \[9.8\] metres per second per second. When an object is thrown from a building's top or a cliff's apex, for example, it travels at \[9.8\] metres per second per second toward the earth. The gravity on the Moon's surface is about \[1/6th\] as strong, or about \[1.6\] metres per second per second.

Given, weight of the person on the moon(${W_M}) = 9\,N$ and we have to find the weight of a person on earth i.e, ${W_E}$. According to the question; as we know that the gravity of the moon is \[1/6th\] of earth’s gravity.So the weight of a person is 1/6th on moon as compared to earth.Therefore,
${W_M} = \dfrac{{W_E}}{6} \\$
$\Rightarrow 9=\dfrac{W_E}{6} \\$
$\Rightarrow {W_E} = 9 \times 6 \\$
$\therefore {W_E}=54\,N$
Hence, the person's weight on the earth will be $54\,N$ .

Hence, the correct option is D.

Note: We don't feel the Moon's gravitational force on us because it is much weaker than Earth's, but we can see its impact on the oceans' liquid water. The seas are slightly pushed into the Moon's gravity, resulting in a bulge or high tide on the Earth's side nearest to the Moon.