Question

Acceleration due to gravity on the moon is
1/6th of the acceleration due to gravity on the
earth. If the ratio of densities of the earth and
the moon is 5/3, then radius of the moon in
terms of radius of earth will be​

Answer 1

Answer:

A = k M / R^2

a = k m / r^2      where capital refer to earth and small letters the moon

A / a = (M / m) * (r^2 / R^2)

M = 4/3 pi D R^3      where D refers to density

M / m = D R^3 / d r^3

A / a = (D / d) * (R^3 / r^3) * (r^2 / R^2) = (D / d) * (R / r)

6 = 5/3 * R / r

r = 5 / 18 * R

Answer 2

GIVEN :-

• Acceleration due to gravity on moon is 1/6th the acceleration due to gravityon the earth

• Ratio of densities of earth and moon = 5:3

TO FIND :-

• Radius of moon interms of Radius of earth

SOLUTION :-

We have ,

• If acceleration on earth is 'g' Then acceleration of moon is 'g/6'

• Ratio of densities of earth and moon = 5:3

Let ,

• Density of earth = 5x
• Densityvof moon = 3x

Using the relation ,

$\large {\underline {\boxed {\red {\bf{g = \dfrac{4}{3} \pi G\rho R}}}}}$

Where ,

• G is gravitational constant
• R is radius of the planet
• ρ is density of the planet
• π is constant

Let gravity of earth be 'g'

$\implies \bf g_e = \frac{4}{3} \pi \:G \rho R \\ \\ \implies {\underline {\boxed{ \pink{\bf {\: g_e = \frac{4}{3} \pi G( 5x )R_e}}}}} \longrightarrow \: eq(1)$

Where ,

• G is gravitational constant
• Rₑ is radius of earth
• ρ is density of earth
• π is constant

Then acceleration on moon is ,

$\implies \sf \: g_m = \frac{4}{3} \pi G \rho R_m \\ \\ \implies {\underline {\boxed {\pink {\bf{ \frac{g_e}{6} = \frac{4}{3}\pi G (3x)R }}}}} \longrightarrow \: eq(2)$

Dividing eq(1) by eq(2) ,

$\implies \bf \: \dfrac{g_e}{ ( \frac{g_e}{6} ) } = \bigg[ \dfrac{ \frac{4}{3}\pi G(5x)R_e}{ \frac{4}{3}\pi G(3x)R_m} \bigg] \\ \\ \implies \bf \: \frac{6g_e}{g_e} = \bigg[ \dfrac{ \cancel{ \frac{4}{3}\pi G} \: (5 \cancel{x}) R_e}{ \cancel{\frac{4}{3} \pi \: G}(3 \cancel{x})R_m} \bigg] \\ \\ \implies \bf \: \frac{6 \cancel{g_e}}{ \cancel{g_e}} = \frac{5R_e}{3R_m} \\ \\ \implies \bf \: \frac{6}{1} = \frac{5R_e}{3R_m}$

$\implies \bf \: 6 \times 3R_m = 5R_e \\ \\ \implies \bf \: 18R_m = 5R_e \\ \\ \implies {\underline {\boxed {\blue{\bf{R_m = \frac{5}{18}R_e}}}}}$

∴ The radius of moon interms of earth is $\large{\sf{R_m=\dfrac{5}{18}R_e}}$