Question

A body weighs $3.5$ kgwt, on the surface of the earth, what will be its weight on the surface of a planet whose mass is $\left( {\dfrac{1}{7}} \right)th$ of the mass of the earth and radius half of that of the earth?
A. $3kgwt$
B. $2kgwt$
C. $7kgwt$
D. $4kgwt$

Answer 1

Mass of any object is the actual amount of matter present in the body whereas the weight of the force exerted by the gravity of an object. Every object in the universe is under the force of the gravitational. Any two objects attract each other and are directly proportional to the product of the mass and inversely proportional to the square of distance between the two.

Formula used:
$F = G\dfrac{{Mm}}{{{R^2}}}$
where g is the gravity of acceleration and G is the gravitational force.

Complete step by step answer:
Let us assume that the Weight of the object on Earth is ${W_E}$
The Mass of the object on Earth is ${M_E}$
The Radius of the object on Earth is ${R_E}$
Similarly for the planet –
Weight of the object on planet is ${W_P}$
The Mass of the object on Planet is ${M_P}$
The Radius of the object on Planet is ${R_P}$
Given that: Weight of the body on the surface of the Earth $= mg = 3.5{\text{ }}kgwt$
Now, when Mass is $\dfrac{1}{7}$ and radius is $\dfrac{1}{2}$ of that of the earth.
Therefore,
${M_P}{\text{ = }}\dfrac{{{M_E}}}{7}{\text{ and }}{{\text{R}}_P}{\text{ = }}\dfrac{{{R_E}}}{2}$ ...... (i)
The force of the attraction due to gravity on the surface of the earth can be expressed as –
$F = \dfrac{{G{M_E}m}}{{{R_E}^2}}$
Also, Force is the product of mass and acceleration due to gravity.
$m{g_E} = \dfrac{{G{M_E}m}}{{{R_E}^2}}$
Like terms from both the sides of the equation cancel each other. Therefore remove “m” from both the sides.
${g_E} = \dfrac{{G{M_E}}}{{{R_E}^2}}$
Now, weight of the object can be expressed as –
${W_E} = m{g_E} \\\ \Rightarrow {W_E} = \dfrac{{G{M_E}m\,{\text{ }}}}{{{R_E}^2}}\;{\text{ }}.....{\text{ (a)}} \\\$
Similarly for the weight of an object on any planet-
${W_P} = m{g_P} \\\ \Rightarrow {W_p} = \dfrac{{G{M_P}m\,{\text{ }}}}{{{R_p}^2}}\;{\text{ }}.....{\text{ (b)}} \\\$
Take ratio of the equation (a) and (b)
$\dfrac{{{W_E}}}{{{W_p}}} = \dfrac{{\dfrac{{G{M_E}m\,{\text{ }}}}{{{R_E}^2}}}}{{\dfrac{{G{M_p}m\,{\text{ }}}}{{{R_p}^2}}}}\;{\text{ }}$
Denominator’s denominator goes to the numerator and vice-versa. Remove the same terms from the denominator and the numerator since they cancel each other.
$\dfrac{{{W_E}}}{{{W_p}}} = \dfrac{{{M_E}{R_p}^2}}{{{M_p}{R_E}^2}}$
Place the values using the equation (i) and given data
$\dfrac{{3.5}}{{{W_p}}} = \dfrac{{{M_E}{{\left( {\dfrac{{{R_E}}}{2}} \right)}^2}}}{{\dfrac{{{M_E}}}{7}{R_E}^2}}$
Simplify the above equation and remove the same terms from the numerator and the denominator.
$\dfrac{{3.5}}{{{W_p}}} = \dfrac{{\left( {\dfrac{1}{4}} \right)}}{{\dfrac{1}{7}}}$
Do-cross- multiplication and simplify the above equation and make the required term the subject.

$$\Rightarrow {W_p} = \dfrac{{3.5 \times 4}}{7} \\\ \Rightarrow {W_p} = 2{\text{ kgwt}} \\\$$

So, the correct answer is “Option B”.

Note:
Remember the difference between the g (gravitational acceleration) and the G (Gravitational constant). Since in G only magnitude is important it is the scalar quantity whereas, in g both the magnitude and the direction are important, and therefore it is vector quantity. The value of “G” remains constant throughout the globe whereas, the value of “g” changes from every place on the planet. Go through certain basic parameters and the physical quantities to solve these types of word problems.