Question

Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing $W$ on earth will weigh
(A) $W$
(B) $2W$
(C) $W/2$
(D) ${{2}^{{}^{1}/{}_{3}}}W$ at the planet

Answer 1

The weight of any object depends upon the acceleration due to gravity on the surface of the planet; the acceleration of gravity depends on the mass of the planet and the radius of the planet. To find a relation between the mass and the radius of the planets, we will use the information about equal densities provided to us. The detailed solution of the given question is given below.
Formula Used: $density=\dfrac{mass}{volume}$ , ${{V}_{sphere}}={}^{4}/{}_{3}\pi {{R}^{3}}$. , $g=\dfrac{GM}{{{R}^{2}}}$

Complete step by step solution:
Let the mass of the earth and the radius of the earth be ${{M}_{earth}}$ and ${{R}_{earth}}$ respectively
Similarly, let the mass of the planet and the radius of the planet be ${{M}_{planet}}$ and ${{R}_{planet}}$ respectively
We have been given that the density of the earth and the density of the planet are equal, that is ${{\rho }_{planet}}={{\rho }_{earth}}$
We know that density is the ratio of the mass of an object to its volume, that is $density=\dfrac{mass}{volume}$ and the planets are considered to be spherical object; spherical objects have a volume ${{V}_{sphere}}={}^{4}/{}_{3}\pi {{R}^{3}}$ where $R$ is the radius of the sphere
Substituting the values of the mass and volume of the planet and earth in the density equation given above, we get

& {{\rho }_{planet}}={{\rho }_{earth}} \\\ & \Rightarrow \dfrac{{{M}_{planet}}}{{}^{4}/{}_{3}\pi {{\left( {{R}_{planet}} \right)}^{3}}}=\dfrac{{{M}_{earth}}}{{}^{4}/{}_{3}\pi {{\left( {{R}_{earth}} \right)}^{3}}} \\\ & \Rightarrow \dfrac{{{M}_{planet}}}{{{M}_{earth}}}={{\left( \dfrac{{{R}_{planet}}}{{{R}_{earth}}} \right)}^{3}} \\\ & \Rightarrow \dfrac{{{R}_{planet}}}{{{R}_{earth}}}={{\left( \dfrac{{{M}_{planet}}}{{{M}_{earth}}} \right)}^{{}^{1}/{}_{3}}}--equation(1) \\\ \end{aligned}$$ As discussed above, the weight of an object depends on the acceleration due to gravity in a region. So if we find the ratio of the acceleration due to gravity for the planet and the earth, we can find the ratio of the weights of any object on the planet and on earth. We can mathematically express the acceleration due to gravity as $$g=\dfrac{GM}{{{R}^{2}}}$$ where $$G$$ is the universal gravitational constant and the meaning of the other symbols have been discussed above The ratio of acceleration due to gravity on the planet to the acceleration due to gravity on the earth can be given as $$\begin{aligned} & \dfrac{{{g}_{planet}}}{{{g}_{earth}}}=\dfrac{\dfrac{G{{M}_{planet}}}{{{\left( {{R}_{planet}} \right)}^{2}}}}{\dfrac{G{{M}_{planet}}}{{{\left( {{R}_{earth}} \right)}^{2}}}} \\\ & \Rightarrow \dfrac{{{g}_{planet}}}{{{g}_{earth}}}={{\left( \dfrac{{{R}_{earth}}}{{{R}_{planet}}} \right)}^{2}}\left( \dfrac{{{M}_{planet}}}{{{M}_{planet}}} \right) \\\ \end{aligned}$$ Substituting the value from the equation marked one, we get $$\begin{aligned} & \dfrac{{{g}_{planet}}}{{{g}_{earth}}}={{\left( \dfrac{{{M}_{earth}}}{{{M}_{planet}}} \right)}^{{}^{2}/{}_{3}}}\left( \dfrac{{{M}_{planet}}}{{{M}_{planet}}} \right) \\\ & \Rightarrow \dfrac{{{g}_{planet}}}{{{g}_{earth}}}={{\left( \dfrac{{{M}_{planet}}}{{{M}_{planet}}} \right)}^{{}^{1}/{}_{3}}} \\\ \end{aligned}$$ Since the weight is directly proportional to the acceleration due to gravity, we can say that an object weighing $$W$$ on earth will weight $${{2}^{{}^{1}/{}_{3}}}W$$ on the given planet. **Therefore, option (D) is the correct answer.** **Note:** Since the weight of an object is the product of mass and the acceleration of gravity. The point to be careful about is that the mass in the weight equation is of the body whose weight is to be found and not of the planet. Many students are in a hurry and substitute the mass of the earth or the mass of the planet anywhere that mass is mentioned. You should not make that mistake.