The radius of Earth is about 6400 ,km and tha | vedclass

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

Question

(A) Given: Radius of Earth $(R_e)$ $= 6400 \,km$,Radius of Mars $(R_m)$ $= 3200 \,km$.Mass of Earth $(M_e)$ $= 10 M_m$,where $M_m$ is the mass of Mars

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) Given: Radius of Earth $(R_e)$ $= 6400 \,km$,Radius of Mars $(R_m)$ $= 3200 \,km$.Mass of Earth $(M_e)$ $= 10 M_m$,where $M_m$ is the mass of Mars.The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.The ratio of acceleration due to gravity on Mars $(g_m)$ to that on Earth $(g_e)$ is:$\frac{g_m}{g_e} = \frac{G M_m / R_m^2}{G M_e / R_e^2} = \frac{M_m}{M_e} 	imes \left(\frac{R_e}{R_m}ight)^2$.Substituting the given values:$\frac{g_m}{g_e} = \frac{1}{10} 	imes \left(\frac{6400}{3200}ight)^2 = \frac{1}{10} 	imes (2)^2 = \frac{4}{10} = \frac{2}{5}$.The weight of an object is $W = mg$.Given weight on Earth $W_e = m g_e = 200 \,N$.Weight on Mars $W_m = m g_m = m \left(\frac{2}{5} g_eight) = \frac{2}{5} W_e$.$W_m = \frac{2}{5} 	imes 200 \,N = 80 \,N$.

The radius of Earth is about $6400 \,km$ and that of Mars is $3200 \,km$,and the mass of the Earth is about $10$ times the mass of Mars. An object weighs $200 \,N$ on the surface of Earth. Then,its weight on the surface of Mars will be (in $\,N$)

Similar Questions

The length of a seconds pendulum at a height $h=2R$ from the earth's surface will be. (Given: $R =$ radius of earth and acceleration due to gravity at the surface of earth $g = \pi^{2} \ m/s^{2}$)

How does the rotation of the Earth affect the acceleration due to gravity on its surface?

$A$ tunnel is dug along a diameter of the Earth. If $M_e$ and $R_e$ are the mass and radius of the Earth respectively,then the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the center is:

Find the angular speed of rotation of the Earth,so that the apparent $g$ at the equator becomes $(1/6)$th of its original value. $(R = 6.4 \times 10^6 \ m)$

How does the acceleration due to gravity $(g)$ at a location on Earth change with latitude?

Vedclass Test Series

Exam Paper Generator

Online Exam Module