There is a planet which is 8 times more massi | vedclass

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

Question

(C) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.Since density $\rho = \frac{M}{V} = \frac{M}{\frac

Vedclass · Accepted Answer

$g^{\prime} = 18g$

Vedclass · Answer

(C) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.Since density $ho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi R^3}$,we can express the radius $R$ as $R = \left( \frac{3M}{4\pi ho} ight)^{1/3}$.Substituting $R$ into the formula for $g$:$g = \frac{GM}{(\frac{3M}{4\pi ho})^{2/3}} = G M^{1/3} (\frac{4\pi ho}{3})^{2/3}$.This shows that $g \propto M^{1/3} ho^{2/3}$.Given for the planet: $M^{\prime} = 8M$ and $ho^{\prime} = 27ho$.Taking the ratio:$\frac{g^{\prime}}{g} = \left( \frac{M^{\prime}}{M} ight)^{1/3} \left( \frac{ho^{\prime}}{ho} ight)^{2/3}$.$\frac{g^{\prime}}{g} = (8)^{1/3} 	imes (27)^{2/3} = 2 	imes (3^3)^{2/3} = 2 	imes 3^2 = 2 	imes 9 = 18$.Therefore,$g^{\prime} = 18g$.

There is a planet which is $8$ times more massive and $27$ times denser than the Earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surfaces of the planet and the Earth respectively,then:

Similar Questions

The value of $g$ at the surface of the earth is $9.8 \, m/s^2$. Then the value of $g$ at a place $480 \, km$ above the surface of the earth will be nearly .......... $m/s^2$ (radius of the earth is $6400 \, km$).

The acceleration due to gravity at a height of $6400 \,km$ from the surface of the earth is $2.5 \,ms^{-2}$. The acceleration due to gravity at a height of $12800 \,km$ from the surface of the earth is (Radius of the earth $= 6400 \,km$) (in $\,ms^{-2}$)

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

The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)

At what distance from the centre of the earth,the value of acceleration due to gravity $g$ will be half that on the surface? ($R =$ radius of earth)

Vedclass Test Series

Exam Paper Generator

Online Exam Module