The mass and diameter of a planet have twice | vedclass

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

Question

(B) The acceleration due to gravity $g$ 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

Vedclass · Accepted Answer

$4.9$

Vedclass · Answer

(B) The acceleration due to gravity $g$ 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.Given that the mass of the planet $M' = 2M$ and the diameter $D' = 2D$,which implies the radius $R' = 2R$.The acceleration due to gravity on the planet's surface $g'$ is given by $g' = \frac{GM'}{(R')^2}$.Substituting the given values: $g' = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = \frac{1}{2} \left( \frac{GM}{R^2} \right) = \frac{g}{2}$.Since $g = 9.8 \ m/s^2$ for Earth,we have $g' = \frac{9.8}{2} = 4.9 \ m/s^2$.

The mass and diameter of a planet have twice the value of the corresponding parameters of Earth. Acceleration due to gravity on the surface of the planet is ........ $m/s^2$.

Similar Questions

Assuming the earth to be a sphere of uniform mass density, a body weighs $300 \,N$ on the surface of the earth. How much would it weigh at a depth of $R/4$ below the surface of the earth (in $\,N$)?

The gravitational potential at a point above the surface of the Earth is $-5.12 \times 10^7 \,J/kg$ and the acceleration due to gravity at that point is $6.4 \,m/s^2$. Assume that the mean radius of the Earth is $6400 \,km$. The height of this point above the Earth's surface is: (in $\,km$)

What is the change in mass of a body,when taken $64 \ km$ below the surface of the earth? $[$Take radius of the earth as $6400 \ km]$

As we move above the surface of the Earth,the value of acceleration due to gravity $g$ . . . . . . .

At what height above the earth's surface is the value of $g$ the same as in a $200 \ km$ deep mine?

Vedclass Test Series

Exam Paper Generator

Online Exam Module