The radii of two planets are respectively R_1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{GM}{R^2}$.Since the mass $M$ of a planet can be expressed in

Vedclass · Accepted Answer

$g_1:g_2 = R_1 \rho_1 : R_2 \rho_2$

Vedclass · Answer

(D) The acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{GM}{R^2}$.Since the mass $M$ of a planet can be expressed in terms of its density $ho$ and radius $R$ as $M = ho 	imes V = ho 	imes \frac{4}{3} \pi R^3$,we substitute this into the formula for $g$:$g = \frac{G (ho \cdot \frac{4}{3} \pi R^3)}{R^2} = \frac{4}{3} \pi G ho R$.This shows that $g \propto ho R$.Therefore,the ratio of the accelerations due to gravity for the two planets is $\frac{g_1}{g_2} = \frac{ho_1 R_1}{ho_2 R_2}$.Thus,$g_1 : g_2 = R_1 ho_1 : R_2 ho_2$.

The radii of two planets are respectively $R_1$ and $R_2$ and their densities are respectively $\rho_1$ and $\rho_2$. The ratio of the accelerations due to gravity at their surfaces is

Similar Questions

The amplitude of a wave represented by the displacement equation $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$ will be

In the interval $(-3, 3)$,the function $f(x) = \frac{x}{3} + \frac{3}{x}, x \neq 0$ is :

If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin to a plane,then the equation of that plane is

$A$ prism of refractive index $\mu$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is $A$,then the value of $A$ in terms of $\mu$ is :

The efficiency of a Carnot engine is $50\%$ and the temperature of the sink is $500\, K$. If the temperature of the source is kept constant and its efficiency is to be raised to $60\%$,then the required temperature of the sink will be........ $K$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module