
Let us suppose a body weighs $30$kg on earth. Find its weight on a planet whose mass is $\dfrac{1}{9}th$ the mass of earth and the radius is half of the earth.
A. 10kg
B. 15.6kg
C. 13.3kg
D. 8kg
Answer
498.6k+ views
Hint: In this question we are asked to find weight of given body on a planet whose mass is $\dfrac{1}{9}th$ 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 ${W_E}$=$30kg$
$30 \times 9.8 = 294N$
Mass of planet ${M_P} = \dfrac{{{M_e}}}{9}$
$\dfrac{{{M_p}}}{{{M_e}}} = \dfrac{1}{9}$ Equation (1)
Radius of planet ${R_p} = \dfrac{{{R_e}}}{2}$
\[\dfrac{{{R_p}}}{{{R_e}}} = \dfrac{1}{2}\] Equation (2)
To find the weight of body on planet\[{W_p}\], we have
The force of attraction due to earth on body of mass m as
\[F = \dfrac{{G{M_e}m}}{{{R_e}^2}}\] Equation (3)
Also we know that \[F = mg\] Equation (4)
Comparing equation 3 and 4, we get
\[mg = \dfrac{{G{M_e}m}}{{{R_e}^2}}\]
Therefore \[{g_e} = \dfrac{{G{M_e}}}{{{R_e}^2}}\]which is the acceleration due to gravity on surface of earth. Equation(5)
Now we know that \[{W_e} = m{g_e}\] equation (6)
and similarly \[{W_p} = m{g_p}\].
So substituting equation (5) in equation (6), we have
\[{W_e} = m\left( {\dfrac{{G{M_e}}}{{{R_e}^2}}} \right)\] Equation (7)
Similarly \[{W_p} = m\left( {\dfrac{{G{M_p}}}{{{R_p}^2}}} \right)\] Equation (8)
Dividing equation (8) by equation (7) and substituting the values we get
\[
\dfrac{{{W_p}}}{{{W_e}}} = \dfrac{{m\left( {\dfrac{{G{M_p}}}{{{R_p}^2}}} \right)}}{{m\left( {\dfrac{{G{M_e}}}{{{R_e}^2}}} \right)}} \\
\dfrac{{{W_p}}}{{{W_e}}} = \dfrac{{{M_p}}}{{{M_e}}} \times {\left( {\dfrac{{{R_e}}}{{{R_p}}}} \right)^2} \\
\]
\[
\dfrac{{{W_p}}}{{30}} = \dfrac{1}{9} \times {\left( {\dfrac{2}{1}} \right)^2} \\
{W_p} = 13.33kg \\
\]
Or we can say,
\[ {W_p} = 13.33 \times 9.8 \\
= 130.63N \\
\]
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.
Complete step-by-step solution:
Given, weight of the body on earth ${W_E}$=$30kg$
$30 \times 9.8 = 294N$
Mass of planet ${M_P} = \dfrac{{{M_e}}}{9}$
$\dfrac{{{M_p}}}{{{M_e}}} = \dfrac{1}{9}$ Equation (1)
Radius of planet ${R_p} = \dfrac{{{R_e}}}{2}$
\[\dfrac{{{R_p}}}{{{R_e}}} = \dfrac{1}{2}\] Equation (2)
To find the weight of body on planet\[{W_p}\], we have
The force of attraction due to earth on body of mass m as
\[F = \dfrac{{G{M_e}m}}{{{R_e}^2}}\] Equation (3)
Also we know that \[F = mg\] Equation (4)
Comparing equation 3 and 4, we get
\[mg = \dfrac{{G{M_e}m}}{{{R_e}^2}}\]
Therefore \[{g_e} = \dfrac{{G{M_e}}}{{{R_e}^2}}\]which is the acceleration due to gravity on surface of earth. Equation(5)
Now we know that \[{W_e} = m{g_e}\] equation (6)
and similarly \[{W_p} = m{g_p}\].
So substituting equation (5) in equation (6), we have
\[{W_e} = m\left( {\dfrac{{G{M_e}}}{{{R_e}^2}}} \right)\] Equation (7)
Similarly \[{W_p} = m\left( {\dfrac{{G{M_p}}}{{{R_p}^2}}} \right)\] Equation (8)
Dividing equation (8) by equation (7) and substituting the values we get
\[
\dfrac{{{W_p}}}{{{W_e}}} = \dfrac{{m\left( {\dfrac{{G{M_p}}}{{{R_p}^2}}} \right)}}{{m\left( {\dfrac{{G{M_e}}}{{{R_e}^2}}} \right)}} \\
\dfrac{{{W_p}}}{{{W_e}}} = \dfrac{{{M_p}}}{{{M_e}}} \times {\left( {\dfrac{{{R_e}}}{{{R_p}}}} \right)^2} \\
\]
\[
\dfrac{{{W_p}}}{{30}} = \dfrac{1}{9} \times {\left( {\dfrac{2}{1}} \right)^2} \\
{W_p} = 13.33kg \\
\]
Or we can say,
\[ {W_p} = 13.33 \times 9.8 \\
= 130.63N \\
\]
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

