Different planets have masses M_1, M_2, M_3 a | vedclass

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

Question

(A) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$,where $M$ is the mass and $R$ is the radius of the pla

Vedclass · Accepted Answer

$M_3 > M_1 > M_2$

Vedclass · Answer

(A) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$,where $M$ is the mass and $R$ is the radius of the planet.From the graph,we observe that at the surfaces,$g_1 = g_3 > g_2$.For $g_1 = g_3$,we have $\frac{GM_1}{R_1^2} = \frac{GM_3}{R_3^2}$,which implies $\frac{M_1}{R_1^2} = \frac{M_3}{R_3^2}$.Since $R_1 Now,comparing $g_1$ and $g_2$,we have $g_1 > g_2$,so $\frac{GM_1}{R_1^2} > \frac{GM_2}{R_2^2}$.Given the graph,$R_1  g_2$,which leads to $M_1 > M_2$.Combining these results,we get $M_3 > M_1 > M_2$.

Different planets have masses $M_1, M_2, M_3$ and radii $R_1, R_2, R_3$ respectively. The acceleration due to gravity on their surfaces are $g_1, g_2, g_3$ respectively. Based on the given graph,arrange their masses in descending order.

Similar Questions

The weight of a body at the centre of the earth is

The variation of acceleration due to gravity $(g)$ with distance $(r)$ from the center of the earth is correctly represented by ... (Given $R =$ radius of earth)

The height $h$ above the earth's surface at which the value of acceleration due to gravity $g$ becomes $\frac{g}{3}$ is ($R=$ radius of the earth).

The period of oscillation of a simple pendulum of constant length at the Earth's surface is $T$. Its period inside a mine is

What is the time period of a seconds pendulum on a planet whose mass and radius are twice that of the Earth?

Vedclass Test Series

Exam Paper Generator

Online Exam Module