Question

If acceleration due to gravity on the surface pf moon is 1/5 th that at the surface of the earth and radius of moon is 1/4th that of the earth then the ratio of the mass of the earth to mass of moon is

Answer 1

Answer & Explanation:

The gravitational force on a body of mass $m$ at the surface of the Earth is

$F = \dfrac{ G M_E m }{ R_E^2 }$

where $M_E$ is Earth's mass and $R_E$ is Earth's radius. Applying Newton's second law to the body as it is subjected to the acceleration of gravity on Earth $g_E$ we can write

$\dfrac{ G M_E m }{ R_E^2 } = m g_E$

so that the mass $m$ of the body cancels out and the acceleration of gravity is given by

$g_E = \dfrac{ G M_E }{ R_E^2 } \, \, .$

A similar reasoning gives the acceleration of gravity on the surface of the Moon, $g_M$, as

$g_M = \dfrac{ G M_M }{ R_M^2 } \, \, .$

where $M_M$ is Moon's mass and $R_M$ is Moon's radius.

Now we can use the ratios,

$\dfrac{ g_E }{ g_M } = \dfrac{ G M_E }{ G M_M } \dfrac{ R_M^2 }{ R_E^2 } = \dfrac{ M_E }{ M_M } \dfrac{ R_M^2 }{ R_E^2 } = 5$

and since

$R_M = R_E / 4$

we have

$\dfrac{ R_M^2 }{ R_E^2 } = \dfrac{ 1 }{ 16 }$

and we find

$\dfrac{ M_E }{ M_M } \dfrac{ R_M^2 }{ R_E^2 } = \dfrac{ M_E }{ M_M } \dfrac{ 1 }{ 16 }$

and

$\dfrac{ M_E }{ M_M } = 16 \times 5 = 80$