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Question: Let us suppose a body weighs \(30\)kg on earth. Find its weight on a planet whose mass is \(\dfrac{1...

Let us suppose a body weighs 3030kg on earth. Find its weight on a planet whose mass is 19th\dfrac{1}{9}th the mass of earth and the radius is half of the earth.

A. 10kg
B. 15.6kg
C. 13.3kg
D. 8kg

Explanation

Solution

In this question we are asked to find weight of given body on a planet whose mass is 19th\dfrac{1}{9}th of the mass of the earth and it has radius half of that of earth. Generally mass is defined as the amount of "matter" in an object whereas weight is the force exerted on an object by gravity. Matter is any substance that has mass and occupies space by having volume.

Complete step-by-step solution:
Given, weight of the body on earth WE{W_E}=30kg30kg
30×9.8=294N30 \times 9.8 = 294N
Mass of planet MP=Me9{M_P} = \dfrac{{{M_e}}}{9}
MpMe=19\dfrac{{{M_p}}}{{{M_e}}} = \dfrac{1}{9} Equation (1)
Radius of planet Rp=Re2{R_p} = \dfrac{{{R_e}}}{2}
RpRe=12\dfrac{{{R_p}}}{{{R_e}}} = \dfrac{1}{2} Equation (2)
To find the weight of body on planetWp{W_p}, we have
The force of attraction due to earth on body of mass m as
F=GMemRe2F = \dfrac{{G{M_e}m}}{{{R_e}^2}} Equation (3)
Also we know that F=mgF = mg Equation (4)
Comparing equation 3 and 4, we get
mg=GMemRe2mg = \dfrac{{G{M_e}m}}{{{R_e}^2}}
Therefore ge=GMeRe2{g_e} = \dfrac{{G{M_e}}}{{{R_e}^2}}which is the acceleration due to gravity on surface of earth. Equation(5)
Now we know that We=mge{W_e} = m{g_e} equation (6)
and similarly Wp=mgp{W_p} = m{g_p}.
So substituting equation (5) in equation (6), we have
We=m(GMeRe2){W_e} = m\left( {\dfrac{{G{M_e}}}{{{R_e}^2}}} \right) Equation (7)
Similarly Wp=m(GMpRp2){W_p} = m\left( {\dfrac{{G{M_p}}}{{{R_p}^2}}} \right) Equation (8)
Dividing equation (8) by equation (7) and substituting the values we get

WpWe=m(GMpRp2)m(GMeRe2) WpWe=MpMe×(ReRp)2  \dfrac{{{W_p}}}{{{W_e}}} = \dfrac{{m\left( {\dfrac{{G{M_p}}}{{{R_p}^2}}} \right)}}{{m\left( {\dfrac{{G{M_e}}}{{{R_e}^2}}} \right)}} \\\ \dfrac{{{W_p}}}{{{W_e}}} = \dfrac{{{M_p}}}{{{M_e}}} \times {\left( {\dfrac{{{R_e}}}{{{R_p}}}} \right)^2} \\\ Wp30=19×(21)2 Wp=13.33kg  \dfrac{{{W_p}}}{{30}} = \dfrac{1}{9} \times {\left( {\dfrac{2}{1}} \right)^2} \\\ {W_p} = 13.33kg \\\

Or we can say,

=130.63N  = 130.63N \\\

So, weight of body on the planet is Option(C) i.e. 13.3 kg

Note: The SI unit of mass is Kilogram (Kg) and of weight is Newton (N).
Mass does not depend upon gravity and is constant everywhere whereas Weight is dependent on gravity and so, it varies from place to place.
Weight can be zero where there is no gravity but mass can’t be zero.