Question

Show that the weight of the body on the moon is $\dfrac{1}{6}$ times its weight on earth.

Answer 1

Hint: A mass or object's weight is a measure of how much gravity pushes on it. Because there is less gravitational pressure on objects on the moon, they weigh less. If a rock that weighs one pound on Earth is transported to the moon, it will weigh significantly less.

Formula used:
$F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$
$\Rightarrow F = \,force$
$\Rightarrow G\, = \,gravitational\,constant$
$\Rightarrow {m_1}\, = \,mass\,of\,object\,1$
$\Rightarrow {m_2}\, = \,mass\,of\,object\,2$
$\Rightarrow r = dis\tan ce\,between\,centers\,of\,the\,masses$

Complete step by step answer:
The moon's mass is \[\dfrac{1}{100}\] that of the planet, and its radius is $\dfrac{1}{4}$ that of the earth. As a result, the moon's gravitational attraction is around one-sixth that of Earth.The gravitational force of the moon is determined by its mass and scale. As a result, the weight of an object on the moon is $\dfrac{1}{6}$ that of an object on Earth. The moon is much less massive and has a different radius(R) than the Earth.

Let us get a clear idea about this. Weight of the object on the earth is:
${W_e} = $ Is the force with which the earth attracts the object
${W_e} = G \times \dfrac{{{M_e} \times m}}{{R_e^2}}$
$\Rightarrow m = \,mass\,of\,the\,object$
$\Rightarrow {M_e} = \,mass\,of\,the\,earth$
$\Rightarrow {R_m} = radius\,of\,the\,earth$

Weight of the object on the moon is:
${W_m} = $ Is the force with which the moon attracts the object
${W_m} = G \times \dfrac{{{M_m} \times m}}{{R_m^2}}$
$\Rightarrow {M_m} = $Mass of the moon
$\Rightarrow {R_m} = $Radius of the moon.
Now, by dividing both the equations:
$\dfrac{{{W_m}}}{{{W_e}}} = \dfrac{{G \times {M_m} \times m}}{{R_m^2}} \times \dfrac{{R_e^2}}{{G \times {M_e} \times m}}$
$\Rightarrow \dfrac{{{W_m}}}{{{W_e}}} = \dfrac{{{M_m}}}{{{M_e}}} \times {\left( {\dfrac{{{R_e}}}{{{R_m}}}} \right)^2} \\ $
As we discuss above that:
${M_e} = 100\,{M_m}\,\,\,\,and\,\,\,{R_e} = 4{R_m}$
Now,
$\dfrac{{{W_m}}}{{{W_e}}} = \dfrac{{{M_m}}}{{100\,{M_m}}} \times {\left( {\dfrac{{4{R_m}}}{{{R_m}}}} \right)^2}$
$\Rightarrow \dfrac{{{W_m}}}{{{W_e}}} = \dfrac{{16}}{{100}} \\
\therefore \dfrac{{{W_m}}}{{{W_e}}} \simeq \dfrac{1}{6}$

Note: Gravity is the attraction force that attracts two mass objects together. The force of gravity is proportional to the mass of each object. An object with twice the mass will have twice the gravitational attraction on other objects. The gravitational pull increases as an object's size grows.