If density of a planet is double that of the earth and the radius 1.5 times that of the earth , the acceleration due to gravity on the surface of the planet is:
A
$$\dfrac{3}{4}$$ times that on the surface of the earthB
3 times that on the surface of the earthC
$$\dfrac{4}{3}$$ times that on the surface of the earthD
6 times that on the surface of the earth
The correct option is B
We have,
The density of the planet is double of the earth
So, $$D_p=2D_e$$
The radius of the planet is equal to 1.5 times the radius of the earth
So, $$R_p=1.5R_e$$
Since$$ D_p=2D_e$$
$$\dfrac{M_p}{\dfrac{4}{3}\pi R_p^3}=2\times\dfrac{M_e}{\dfrac{4}{3}4R_e^3}$$
$$M_p=2\times(1.5)^3M_e$$
So,
$$\dfrac{g_p}{g_e}=\dfrac{\dfrac{GM_p}{R_P^2}}{\dfrac{GM_e}{R_e}^2}$$
$$g_P=3g_e$$