Question

A body weighs 3.5kgwt on the surface of the earth. What will be its weight on the surface of a planet whose mass is $\dfrac{1}{7}th$ of the mass of the earth and radius half of that of the earth?

Answer 1

Mass is the actual amount of matter present in a body and weight is the force exerted by the gravity on that object. Thus the weight of an object varies from one planet to another, but mass does not change.

Complete step by step answer:
Given, weight on the surface of the earth,
${W_E} = 3.5kgwt$
$= 3.5 \times 9.8$
$= 34.3N$
Mass of the planet,${M_p} = \dfrac{{{M_E}}}{7}$
$\dfrac{{{M_P}}}{{{M_E}}} = \dfrac{1}{7}$ …………………. (1)
Radius of the planet,${R_P} = \dfrac{{{R_E}}}{2}$
$\dfrac{{{R_P}}}{{{R_E}}} = \dfrac{1}{2}$ …………………(2)
Here we need to find weight of the given body on the surface of the planet,${W_P} = ?$
Consider a mass of earth be ${M_E}$ and radius ${R_E}$. The force of attraction due to earth on a body of mass m on the surface of earth is given by,
$F = \dfrac{{G{M_E}m}}{{{R_E}^2}}$ ………………… (a)
This force produces an acceleration g during motion of a body under gravity. Then,
$F = mg$ ………………… (b)
Compare equation (a) and (b), we get
$mg = \dfrac{{G{M_E}m}}{{{R_E}^2}}$
$\therefore {g_E} = \dfrac{{G{M_E}}}{{{R^2}_E}}$ ………………… (3) this is an equation for acceleration due to gravity on the surface of earth.
We know that,${W_E} = m{g_E}$
${W_P} = m{g_P}$
Substitute equation (3) in the above equations,
${W_E} = m\left( {\dfrac{{G{M_E}}}{{{R^2}_E}}} \right)$ ………………………. (4)
Similarly, ${W_P} = m\left( {\dfrac{{G{M_P}}}{{{R^2}_P}}} \right)$ ……………………. (5)
Now divide equation ( 5) by (4) we get,
$\dfrac{{{W_P}}}{{{W_E}}} = \dfrac{{m\left( {\dfrac{{G{M_P}}}{{{R^2}_P}}} \right)}}{{m\left( {\dfrac{{G{M_E}}}{{{R^2}_E}}} \right)}}$
Simplifying the above equation we get,
$\dfrac{{{W_P}}}{{{W_E}}} = \dfrac{{{M_P}}}{{{M_E}}} \times {\left( {\dfrac{{{R_E}}}{{{R_P}}}} \right)^2}$
Now substitute equation (1) and (2),
We get, $\dfrac{{{W_P}}}{{3.5}} = \dfrac{1}{7} \times {\left( {\dfrac{2}{1}} \right)^2}$
$= \dfrac{4}{7}$
Then, ${W_p} = \dfrac{4}{7} \times 3.5$
$= 2kgwt$
$= 2 \times 9.8$
$= 19.6N$

So, Therefore, the weight of the body on the surface of the planet is 2kg.

Note:
1. Mass of an object is the same as the measure of its inertia. Greater the mass, the greater is the inertia.
2. Mass remains the same whether an object is on the earth, the moon or even in outer space. Mass of the object will not change from place to place.
3. Weight depends on its location. It has both magnitude and direction. The weight is a force acting vertically downwards.