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
527.7k+ 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
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

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

Which among the following are examples of coming together class 11 social science CBSE

Discuss the various forms of bacteria class 11 biology CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw a labelled diagram showing structure of LS of class 11 biology CBSE

