Imagine a new planet having the same density as that of the earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of the earth is $$g$$ and that on the surface of the new planet is $$g',$$ then
A.
$$g' = 3g$$
B.
$$g' = \frac{g}{9}$$
C.
$$g' = 9g$$
D.
$$g' = 27g$$
Answer :
$$g' = 3g$$
Solution :
The acceleration due to gravity on the new planet can be found using the relation
$$g = \frac{{GM}}{{{R^2}}}\,......\left( {\text{i}} \right)$$
but $$M = \frac{4}{3}\pi {R^3}\rho ,\rho $$ being density.
Thus, Eq. (i) becomes
$$\eqalign{
& \therefore g = \frac{{G \times \frac{4}{3}\pi {R^3}\rho }}{{{R^2}}} = G \times \frac{4}{3}\pi R\rho \cr
& \Rightarrow g \propto R \cr
& \therefore \frac{{g'}}{g} = \frac{{R'}}{R} \cr
& \Rightarrow \frac{{g'}}{g} = \frac{{3R}}{R} = 3 \Rightarrow g' = 3g \cr} $$
Releted MCQ Question on Basic Physics >> Gravitation
Releted Question 1
If the radius of the earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth’s surface would-
If $$g$$ is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth, is-
A geo-stationary satellite orbits around the earth in a circular orbit of radius $$36,000 \,km.$$ Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface $$\left( {{R_{earth}} = 6400\,km} \right)$$ will approximately be-