Question

A body weighs 700 gram on the surface of the Earth. How much will it weigh on the surface of a planet whose mass is 1/7th and radius is half that of the Earth?
A) $400\,{\text{gram wt}}$
C) $300\,{\text{gram wt}}$

Answer 1

Use the relation between the acceleration due gravity, universal gravitational constant, mass of the planet and radius of the planet. Calculate the value of acceleration due to gravity on the surface of the planet and then the weight of the body on the planet.

Formulae used:
The weight of the body is given by
$W = mg$ …… (1)
Here, $W$ is the weight of the body, $m$ is the mass of the body and $g$ is the acceleration due to gravity.
The relation between the acceleration due to gravity $g$, universal gravitational constant $G$, mass of the planet $M$ and radius $R$ of the planet is
$g = \dfrac{{GM}}{{{R^2}}}$ …… (2)
Here, $F$ is the gravitational force of attraction between the body and the planet, $G$ is the universal gravitational constant, $M$ is the mass and $R$ is the radius of the planet respectively.

Complete step by step answer:
The mass of the body on the surface of the Earth is $700\,{\text{g}}$.
The mass $M'$ of the other planet is 1/7 th of the mass $M$ of the Earth and the radius $R'$ of the planet is half of the radius $R$ of the Earth.
$\Rightarrow M' = \dfrac{1}{7}M$
$\Rightarrow R' = \dfrac{R}{2}$
Rewrite equation (2) for the acceleration due to gravity $g$ on the surface of the Earth.
$\Rightarrow g = \dfrac{{GM}}{{{R^2}}}$ …… (3)
Rewrite equation (2) for the acceleration due to gravity $g'$ on the surface of the planet.
$\Rightarrow g' = \dfrac{{GM'}}{{R{'^2}}}$ …… (4)
Divide equation (4) by equation (3).
$\Rightarrow \dfrac{{g'}}{g} = \dfrac{{\dfrac{{GM'}}{{R{'^2}}}}}{{\dfrac{{GM}}{{{R^2}}}}}$
$\Rightarrow \dfrac{{g'}}{g} = \dfrac{{M'{R^2}}}{{MR{'^2}}}$
Substitute $\dfrac{1}{7}M$ for $M'$ and $\dfrac{R}{2}$ for $R'$ in the above equation.
$\Rightarrow \dfrac{{g'}}{g} = \dfrac{{\left( {\dfrac{1}{7}M} \right){R^2}}}{{M{{\left( {\dfrac{R}{2}} \right)}^2}}}$
$\Rightarrow \dfrac{{g'}}{g} = \dfrac{4}{7}$
Rearrange the above equation for $g'$.
$\Rightarrow g' = \dfrac{4}{7}g$
Calculate the weight of the body on the planet.
Rewrite equation (1) for the weight $W$ of the body on the planet.
$\Rightarrow W = mg'$
Substitute $700\,{\text{g}}$ for $m$ and $\dfrac{4}{7}g$ for $g'$ in the above equation.
$\Rightarrow W = \left( {700\,{\text{g}}} \right)\left( {\dfrac{4}{7}g} \right)$
$\Rightarrow W = 400g$
$\Rightarrow W = 400\,{\text{gram wt}}$
Therefore, the weight of the body on the surface of the planet is $400\,{\text{gram wt}}$.
Hence, the correct option is A.

Note: Here in this question,the difference in weight of the body occurs due to change of acceleration due to gravity and also remember that mass and weight are the two different quantity.Weight of a body can be zero but mass of a body can not be zero.