Question

The ratio of the weights of a body on the earth’s surface to that on the surface of a planet is 9:4. The mass of the planet is $\dfrac{1}{9}$th of that of the earth. If ‘R’ is the radius of the earth, what is the radius of the planet?
$\text{A}\text{. }\dfrac{R}{3}$
$\text{B}\text{. }\dfrac{R}{2}$
$\text{C}\text{. }\dfrac{R}{4}$
$\text{D}\text{. }\dfrac{R}{9}$

Answer 1

Hint: Weight of a body on a planet is the gravitational force exerted by the planet on the body. Therefore, the ratio of the gravitational forces on both the planets is given. According to the law of gravity, the force on the body of mass m on the surface of a planet is given as $F=\dfrac{GmM}{{{r}^{2}}}$, where, G is the gravitational constant, M is the mass of the planet and r is the radius of the planet. Write the expressions for both the forces and then divide them to find the expression for the radius of the other planet.

Formula used:
$F=\dfrac{GmM}{{{r}^{2}}}$

Complete answer:
When the body is on the surface of earth, the earth will exert a gravitational force F = mg, m is the mass of the body and g is acceleration due to gravity. This gravitational force is equal to the weight of the body.
The mass of the body will not change on the surface of the other planet. However, the acceleration due to gravity will change and hence the gravitational force will change. Thus, the same body will have a different weight on the surface of the other planet. Let the gravitational force exerted by another planet be F’.
It is given that the ratio of F to F’ is 9 : 4.
i.e. $\dfrac{F}{F'}=\dfrac{9}{4}$ …. (i).
According to the law of gravitation, the force on the body of mass m on the surface of a planet is given as $F=\dfrac{GmM}{{{r}^{2}}}$.
where, G is the gravitational constant, M is the mass of the planet and r is the radius of the planet.
Let the masses of earth and the other planet be M and M’ respectively.
The radius of earth is given to be R and let the radius of the other planet be R’.
Then we get that, $F=\dfrac{GmM}{{{R}^{2}}}$ …. (ii).
And $F'=\dfrac{GmM'}{R{{'}^{2}}}$ ….. (iii).
Divide (ii) by (iii).
$\Rightarrow \dfrac{F}{F'}=\dfrac{\dfrac{GmM}{{{R}^{2}}}}{\dfrac{GmM'}{R{{'}^{2}}}}$
$\Rightarrow \dfrac{F}{F'}=\dfrac{MR{{'}^{2}}}{M'{{R}^{2}}}$ … (iv).
It is given that $M'=\dfrac{M}{9}$.
Substitute the values of M’ and $\dfrac{F}{F'}$ in equation (iv).
$\Rightarrow \dfrac{9}{4}=\dfrac{MR{{'}^{2}}}{\dfrac{M}{9}{{R}^{2}}}$
$\Rightarrow \dfrac{9}{4}=\dfrac{9R{{'}^{2}}}{{{R}^{2}}}$
$\Rightarrow R{{'}^{2}}=\dfrac{{{R}^{2}}}{4}$
$\Rightarrow R'=\dfrac{R}{2}$

Hence, the correct option is B.

Note:
Other than using the ratio of the forces, we can use the ratio of the accelerations due to gravities of both the planets.
Let the g and g’ be the gravitational accelerations on earth and on the other planet respectively.
Then $\dfrac{g}{g'}=\dfrac{9}{4}$.
The acceleration due to gravity on the surface of a planet is given as $g=\dfrac{GM}{{{r}^{2}}}$.
Therefore, $g=\dfrac{GM}{{{R}^{2}}}$ and $g'=\dfrac{GM'}{R{{'}^{2}}}$.
Therefore, $\dfrac{g}{g'}=\dfrac{9}{4}=\dfrac{\dfrac{GM}{{{R}^{2}}}}{\dfrac{GM'}{R{{'}^{2}}}}=\dfrac{MR{{'}^{2}}}{M'{{R}^{2}}}$.
Then follow the same procedure as above.