Question

If the mass of a planet is $ 10\% $ less than that of the earth and the radius is $ 20\% $ greater than that of the earth, the acceleration due to gravity on the planet will be:
(A) $ \dfrac{5}{8} $ times of that on the surface of earth
(B) $ \dfrac{3}{4} $ times of that on the surface of earth
(C) $ \dfrac{1}{2} $ times of that on the surface of earth
(D) $ \dfrac{9}{{10}} $ times of that on the surface of earth

Answer 1

Hint
 Initially, calculate the mass and radius of the planet with respect to the mass and radius of the earth. Then using the equation for the acceleration due to gravity, $ g = \dfrac{{Gm}}{{{r^2}}} $ , we get the acceleration due to gravity on the planet.
Formula used: The acceleration due to gravity, $ g = \dfrac{{Gm}}{{{r^2}}} $ , where $ G $ is the gravitational constant, $ m $ is the mass of the planet and $ r $ is the radius of the planet.

Complete step by step answer
It is given that the mass of the planet is $ 10\% $ lesser than the mass of earth. So, the mass of the new planet will be:
 $\Rightarrow {m_p} = {m_e} - \dfrac{{10{m_e}}}{{100}} $ where $ {m_e} $ is the mass of the earth.
On cross-multiplication followed by simplification, we get:
 $\Rightarrow {m_p} = \dfrac{{100{m_e} - 10{m_e}}}{{100}} = \dfrac{{90{m_e}}}{{100}} = 0.9{m_e} $
Similarly, it is also given that the radius of the planet is $ 20\% $ more than the radius of earth. So the radius of the planet will be:
 $\Rightarrow {r_p} = {r_e} + \dfrac{{20{r_e}}}{{100}} $ where $ {r_e} $ is the radius of earth.
This also can be cross-multiplied and simplified as:
 $\Rightarrow {r_p} = \dfrac{{100{r_e} + 20{r_e}}}{{100}} = \dfrac{{120{r_e}}}{{100}} = 1.2{r_e} $
Since we obtained the radius and mass of the planet with respect to those of earth, we can now calculate the acceleration due to gravity.
 $\Rightarrow {g_p} = \dfrac{{G \times 0.9{m_e}}}{{{{\left( {1.2{r_e}} \right)}^2}}} $
This can be further simplified as:
$\Rightarrow {g_p} = \dfrac{{0.9}}{{{{1.2}^2}}} \times \dfrac{{G{m_e}}}{{{r^2}}} = 0.625 \times \dfrac{{G{m_e}}}{{{r^2}}} $
Since $ 0.625 $ can also be written as $ \dfrac{5}{8} $ , the acceleration due to gravity can be estimated as:
$\Rightarrow {g_p} = \dfrac{5}{8}{g_e} $ where $ {g_e} = \dfrac{{G{m_e}}}{{r_e^2}} $ is the acceleration due to gravity on the surface of earth.
Therefore, we found that the acceleration due to gravity on the planet will be $ \dfrac{5}{8} $ times of that on the surface of earth.
Hence, the correct answer is option (A).

Note
Some people may think that if we are taking a body of mass $ m $ to that particular planet's surface, then what will be the mass of the body over there? This is interesting because mass of a body is the measure of how much matter it is contained in, and therefore the mass is always constant, wherever it goes. But, the weight of the body of mass $ m $ is $ W = mg $ and the weight changes as the acceleration due to gravity changes.