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How much will a person will $72\,N$ weight on Earth, weight on the moon?
A. $12\,N$
B. $36\,N$
C. $21\,N$
D. $63\,N$

Answer
VerifiedVerified
510.6k+ views
Hint: The gravitational field is a model used to explain the influence that a massive body extends into the space around itself, producing a force on another massive body. Thus, the gravitational field is used to explain gravitational phenomena and is measured in newton per kilogram.

Complete step by step solution:
Let ${g_m}$ and ${g_e}$ are gravitational field intensities of the moon and earth surface respectively.
The gravity of the moon is $\dfrac{1}{6}$ times the gravity of earth.
Hence, \[{g_m} = \dfrac{{{g_e}}}{6}.......\left( i \right)\]
Multiplying the equation by the mass of the person, we have
$m{g_m} = \dfrac{{m{g_e}}}{6}$
where, $m{g_m} = {w_m}$ is the weight of the person on the surface of moon and
\[m{g_e} = {w_e}\] is the weight of the person on the surface of earth.
Therefore, ${w_m} = \dfrac{{{w_e}}}{6}........\left( {ii} \right)$
Now it is given that the weight of person on the surface of earth is $72\,N$,
Hence, ${w_e} = 72\,N$.

Therefore, from equation (ii),
${w_m} = \dfrac{{{w_e}}}{6} = \dfrac{{72}}{6}$
${w_m} = 12\,N$
Therefore, the weight of the person on the surface of the moon will be $12\,N$.

Hence, the correct option is (A).

Additional Information: A gravitational field is a field of force caused by the attraction of the earth and by the centrifugal force that results from its diurnal rotation. This field also depends on the attraction of the moon, the sun and other heavenly bodies and masses in the terrestrial atmosphere.

Note: Students should be more careful regarding gravitational field intensity of the moon and earth. The gravitational field intensity of the moon is almost equal to $\dfrac{1}{6}$ parts of the gravitational field intensity of the earth.