Question 120: 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: