
A man weighs $\text{200Pounds}$ on earth. How much would he weigh on a planet four times as massive as the earth with a radius five times as great?
A. $\text{4000Pounds}$
B. $\text{250Pounds}$
C. $\text{160Pounds}$
D. $\text{10Pounds}$
E. $\text{32Pounds}$
Answer
510.3k+ views
Hint: Weight of a body is the product of the mass of the body and acceleration due to gravity of the planet. Mass of the body is constant irrespective of the place and height from the surface of the planet.
Complete step by step solution:
Let the mass of man is $m$.
The acceleration due to gravity of earth is ${{g}_{e}}$
The acceleration due to gravity of the other planet is ${{g}_{p}}$
Let mass of earth is ${{M}_{e}}$
And the mass of the other planet is ${{M}_{p}}$
Radius of the earth is ${{R}_{e}}$
Radius of the other planet is ${{R}_{p}}$
It is given that the other planet is 4 times massive as the earth,
$\Rightarrow {{M}_{p}}=4{{M}_{e}}$
The radius of the other planet is 5 times the radius of the earth,
$\Rightarrow {{R}_{p}}=5{{R}_{e}}$
As we know that the acceleration due to gravity is given as,
$g=\dfrac{GM}{{{R}^{2}}}$
Where,
$G=$Universal gravitational constant
$M=$Mass of the planet
$R=$Radius of the planet
Then,
Acceleration due the gravity of earth ${{g}_{e}}=\dfrac{G{{M}_{e}}}{R_{e}^{2}}$
Acceleration due the gravity of other planet ${{g}_{p}}=\dfrac{G{{M}_{p}}}{R_{p}^{2}}$
Weight of the body $W=mg$
If ${{W}_{e}}$ is the weight of man on earth,
${{W}_{e}}=m{{g}_{e}}=\dfrac{G{{M}_{e}}m}{R_{e}^{2}}\ldots \ldots \left( i \right)$
If ${{W}_{p}}$ is the weight of man on the other planet,
${{W}_{p}}=m{{g}_{p}}=\dfrac{G{{M}_{p}}m}{R_{p}^{2}}\ldots \ldots \left( ii \right)$
If it is given that the weight of man on earth is 200 pounds
Dividing equation $\left( i \right)$ and $\left( ii \right)$
$\begin{align}
& \dfrac{{{W}_{e}}}{{{W}_{p}}}=\dfrac{\left( \dfrac{G{{M}_{e}}m}{R_{e}^{2}} \right)}{\left( \dfrac{G{{M}_{p}}m}{R_{p}^{2}} \right)} \\
& =\left( \dfrac{{{M}_{e}}}{{{M}_{p}}} \right){{\left( \dfrac{{{R}_{p}}}{{{R}_{e}}} \right)}^{2}} \\
& =\left( \dfrac{{{M}_{e}}}{4{{M}_{e}}} \right){{\left( \dfrac{5{{R}_{e}}}{{{R}_{e}}} \right)}^{2}} \\
& =\dfrac{25}{4} \\
& \dfrac{200\text{Pounds}}{{{W}_{p}}}=\dfrac{25}{4} \\
& {{W}_{p}}=\left( \dfrac{200\times 4}{25} \right)\text{Pounds} \\
& =32\text{Pounds}
\end{align}$
Therefore, the weight of man on the planet is 32 pounds.
Note: The mass is a scalar quantity and the weight is a vector quantity.
The S.I. unit of the mass is kg and the S.I. the unit of the weight is N.
Complete step by step solution:
Let the mass of man is $m$.
The acceleration due to gravity of earth is ${{g}_{e}}$
The acceleration due to gravity of the other planet is ${{g}_{p}}$
Let mass of earth is ${{M}_{e}}$
And the mass of the other planet is ${{M}_{p}}$
Radius of the earth is ${{R}_{e}}$
Radius of the other planet is ${{R}_{p}}$
It is given that the other planet is 4 times massive as the earth,
$\Rightarrow {{M}_{p}}=4{{M}_{e}}$
The radius of the other planet is 5 times the radius of the earth,
$\Rightarrow {{R}_{p}}=5{{R}_{e}}$
As we know that the acceleration due to gravity is given as,
$g=\dfrac{GM}{{{R}^{2}}}$
Where,
$G=$Universal gravitational constant
$M=$Mass of the planet
$R=$Radius of the planet
Then,
Acceleration due the gravity of earth ${{g}_{e}}=\dfrac{G{{M}_{e}}}{R_{e}^{2}}$
Acceleration due the gravity of other planet ${{g}_{p}}=\dfrac{G{{M}_{p}}}{R_{p}^{2}}$
Weight of the body $W=mg$
If ${{W}_{e}}$ is the weight of man on earth,
${{W}_{e}}=m{{g}_{e}}=\dfrac{G{{M}_{e}}m}{R_{e}^{2}}\ldots \ldots \left( i \right)$
If ${{W}_{p}}$ is the weight of man on the other planet,
${{W}_{p}}=m{{g}_{p}}=\dfrac{G{{M}_{p}}m}{R_{p}^{2}}\ldots \ldots \left( ii \right)$
If it is given that the weight of man on earth is 200 pounds
Dividing equation $\left( i \right)$ and $\left( ii \right)$
$\begin{align}
& \dfrac{{{W}_{e}}}{{{W}_{p}}}=\dfrac{\left( \dfrac{G{{M}_{e}}m}{R_{e}^{2}} \right)}{\left( \dfrac{G{{M}_{p}}m}{R_{p}^{2}} \right)} \\
& =\left( \dfrac{{{M}_{e}}}{{{M}_{p}}} \right){{\left( \dfrac{{{R}_{p}}}{{{R}_{e}}} \right)}^{2}} \\
& =\left( \dfrac{{{M}_{e}}}{4{{M}_{e}}} \right){{\left( \dfrac{5{{R}_{e}}}{{{R}_{e}}} \right)}^{2}} \\
& =\dfrac{25}{4} \\
& \dfrac{200\text{Pounds}}{{{W}_{p}}}=\dfrac{25}{4} \\
& {{W}_{p}}=\left( \dfrac{200\times 4}{25} \right)\text{Pounds} \\
& =32\text{Pounds}
\end{align}$
Therefore, the weight of man on the planet is 32 pounds.
Note: The mass is a scalar quantity and the weight is a vector quantity.
The S.I. unit of the mass is kg and the S.I. the unit of the weight is N.
Recently Updated Pages
Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

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

Trending doubts
Explain why it is said like that Mock drill is use class 11 social science CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

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

Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

Net gain of ATP in glycolysis a 6 b 2 c 4 d 8 class 11 biology CBSE
