Is it true that the ratio of acceleration due to gravity of earth and moon $\frac{g_{earth}}{g_{moon}} = 6$ ? (Given gravitation constant G $= 6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}$ and use standard values)

Question

Is it true that the ratio of acceleration due to gravity of earth and moon gearthgmoon=6 ? (Given gravitation constant G =6.67×10-11Nm2kg-2 and  use standard values)

Accepted Answer

It is true that the ratio of acceleration due to gravity of earth and moon

\frac{g_{earth}}{g_{moon}} = 6

. Acceleration due to gravity on the surface of the moon can be written as

g_{moon} = \frac{4}{3} π R_{m} G ρ_{m} - - - (i)

R_{m}

is the radius of the moon

ρ_{m}

is the density
G is the gravitational constant

\frac{4}{3} π

is constant
Acceleration due to gravity on the earth surface is

g_{earth} = \frac{Gm}{R^{2}} - - - (a)

m

is the mass of earth
We know that mass of earth is

m = \frac{4}{3} π R^{2} ρ_{e} - - - (b)

ρ_{e}

is the density

R

is the radius of earth
Substituting the value of m from (b) in (a)
We get

g_{earth} = \frac{G (\frac{4}{3} π R^{2} ρ_{e})}{R^{2}}

g_{earth} = \frac{4}{3} πR {Gρ}_{e} - - - (ii)

We know that
Density of earth

ρ_{e} = 5515 kg m^{3}

Density of moon

ρ_{m} = 3340 kg m^{3}

Radius of earth

R = 6.4 \times 10^{6} m

Radius of moon

R_{m} = 1.75 \times 10^{6} m

Now we can divide (ii) with (i)
We get

\frac{g_{earth}}{g_{moon}} = \frac{\frac{4}{3} πRG ρ_{e}}{\frac{4}{3} π R_{m} G ρ_{m}}

\Rightarrow \frac{g_{earth}}{g_{moon}} = \frac{R \times ρ_{e}}{R_{m} ρ_{m}}

\Rightarrow \frac{g_{earth}}{g_{moon}} = \frac{6.4 \times 10^{6} \times 5515}{1.75 \times 10^{6} \times 3340}

\Rightarrow \frac{g_{earth}}{g_{moon}} = 6.03

Hence the ratio of acceleration due to gravity of earth and moon

\frac{g_{earth}}{g_{moon}}

is

6

.

Is it true that the ratio of acceleration due to gravity of earth and moon $\frac{g_{earth}}{g_{moon}} = 6$ ? (Given gravitation constant G $= 6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}$ and use standard values)

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