Two planets A and B of radii R and 1.5 R have densities ρ and (ρ)/2 respectively. The ratio of acceleration due to gravity at the surface of B to A is:

Correct answer: (C) 3 : 4. The acceleration due to gravity is g=(GMR^2=4πGRρ)/3⇒g∝ρ·R Thus, the ratio is ⇒g_Bg_A=(1.5Rρ)/(2ρR)=1/2×1.5=0.75⇒g_Bg_A=3/4

Q76 / 90 · JEE Main 2023 (13 Apr Shift 2) · Physics

Two planets A and B of radii $R$ and $1.5 R$ have densities $ρ$ and $\frac{ρ}{2}$ respectively. The ratio of acceleration due to gravity at the surface of B to A is:

A
$2 : 3$
B
$2 : 1$
Correct answer
C
$3 : 4$
D
$4 : 3$

Solution

The acceleration due to gravity is
$g = \frac{G M}{R^{2}} = \frac{4 π G R ρ}{3} \Rightarrow g \propto ρ \cdot R$
Thus, the ratio is
$\Rightarrow \frac{g_{B}}{g_{A}} = \frac{1.5 R ρ}{2 ρ R} = \frac{1}{2} \times 1.5 = 0.75 \Rightarrow \frac{g_{B}}{g_{A}} = \frac{3}{4}$

Hence, the correct answer is option: 'C'.

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Two planets A and B of radii R and 1.5 R have densities ρ and ρ2 respectively. The ratio of acceleration due to gravity at the surface of B to A is:

Solution

Relevant Questions

Two planets A and B of radii $R$ and $1.5 R$ have densities $ρ$ and $\frac{ρ}{2}$ respectively. The ratio of acceleration due to gravity at the surface of B to A is: