Question

The ratio of acceleration due to gravity on the surface of mercury with its value on Earth's surface, assuming the radii of mercury and earth is 1:3 and their mean densities in ratio 3:5 would be?​

Answer 1

Given :   radii of mercury and earth is 1:3 and their mean densities in ratio 3:5

To Find :  The ratio of acceleration due to gravity on the surface of mercury with its value on Earth's surface

Explanation:

g = GM/R²

M = Mass of Earth

R = Radius of Earth

D = Density of Earth

radii of mercury and earth is 1:3

Radii of mercury = R/3

Density = Mass / Volume

D = M / (4/3)πR³

=> M = D4πR³/3

densities in ratio 3:5

=> Density of mercury  =  3D/5

Mass of Mercury  =  (3D/5) 4π (R/3)³ /3

= D4πR³/27 * 5

= D4πR³/3 * 45

= M/45

g₁ at Mercury

g₁ = G(M/45)/(R/3)²

= GM/5R²

= g/5

g₁ : g = 1: 5

ratio of acceleration due to gravity on the surface of mercury with its value on Earth's surface  =  1 :  5

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Answer 2

Given:

The radii of mercury and earth is 1:3 and their mean densities in ratio 3:5.

To find:

The ratio of acceleration due to gravity on this two planets ?

Calculation:

The general expression for gravitational acceleration on any planet is given as:

$\rm \therefore \: g = \dfrac{GM}{ {r}^{2} }$

Now, mass can be represented as the product of density and volume:

$\rm \implies \: g = \dfrac{G \bigg( \dfrac{4}{3} \pi {r}^{3} \times \rho \bigg ) }{ {r}^{2} }$

$\rm \implies \: g = G \bigg( \dfrac{4}{3} \pi r \times \rho \bigg )$

Since , G and π are constants:

$\boxed{ \rm \implies \:g \propto (r \times \rho)}$

So, the required ratio:

$\rm \therefore \: \dfrac{g_{m}}{g_{e}} = \dfrac{r_{m} \times \rho_{m}}{r_{e} \times \rho_{e}}$

$\rm \implies \: \dfrac{g_{m}}{g_{e}} = \dfrac{r_{m}}{r_{m}} \times \dfrac{ \rho_{m}}{ \rho_{e}}$

$\rm \implies \: \dfrac{g_{m}}{g_{e}} = \dfrac{1}{3} \times \dfrac{3}{5}$

$\rm \implies \: \dfrac{g_{m}}{g_{e}} = \dfrac{1}{5}$

$\rm \implies \: g_{m}: g_{e} = 1 : 5$

So, final answer is

$\boxed{ \bold{\: g_{m}: g_{e} = 1 : 5}}$