Question 104: Gravitation

Question

The escape velocities of two planets $\mathrm{A}$ and $\mathrm{B}$ are in the ratio $1: 2$. If the ratio of their radii respectively is $1: 3$, then the ratio of acceleration due to gravity of planet A to the acceleration of gravity of planet B will be :

Accepted Answer

The escape velocity of a planet is given by the formula:
$\mathrm{v}_{	ext{escape}} = \sqrt{\frac{2GM}{R}}$
where G is the gravitational constant, M is the mass of the planet, and R is its radius.
If the escape velocity of planet A is v
A
and the escape velocity of planet B is v
B
, then we can write the following relationship:
${{{v_B}} \over {{v_A}}} = {{\sqrt {{{2G{M_B}} \over {{R_B}}}} } \over {\sqrt {{{2G{M_A}} \over {{R_A}}}} }} = \sqrt {{{{R_A}{M_B}} \over {{M_A}{R_B}}}} $
$ \Rightarrow $ $\sqrt {{{{R_A}{M_B}} \over {{M_A}{R_B}}}}  = 2$
$ \Rightarrow $ ${{{M_B}} \over {{M_A}}} 	imes {1 \over 3} = 4$
$ \Rightarrow $ ${{{M_B}} \over {{M_B}}} = 12$
We know, The acceleration due to gravity on a planet can be calculated using the formula:
$\mathrm{g} = \frac{\mathrm{G} \mathrm{M}}{\mathrm{R}^2}$
where G is the gravitational constant, M is the mass of the planet, and R is its radius.
${{{g_A}} \over {{g_B}}} = {{{M_A}R_B^2} \over {{M_B}R_A^2}} = {1 \over {12}} 	imes {\left( {{3 \over 1}} ight)^2} = {9 \over {12}} = {3 \over 4}$

Answer

$\frac{4}{3}$

Answer

$\frac{2}{3}$

Answer

$\frac{3}{4}$

Answer

$\frac{3}{2}$

Gravitation

The escape velocities of two planets \mathrm{A} and \mathrm{B} are in the ratio 1: 2. If the ratio of their radii respectively is 1: 3, then the ratio of acceleration due to gravity of planet A to the acceleration of gravity of planet B will be :

Choose the correct answer: