
The radius and acceleration due to gravity of the moon are$\dfrac{1}{4}$ and$\dfrac{1}{5}$ that of the earth, the ratio of the mass of earth to the mass of the moon is:
(A). 1:80
(B). 80:1
(C). 1:20
(D). 20:1
Answer
597.9k+ views
Hint: To solve this problem firstly we should understand the term Gravitation, its concept and formula then by using the information and concept behind it, we can easily approach our answer. Newton’s law of gravitation is an attractive force acting between two masses${m_{1\,}}\,and\,{m_2}$ separated by a distance $r$ . This force is $ \propto $ to the ${m_1} \times {m_{2\,}}$ and inversely proportional to ${r^2}$ .
Complete step-by-step solution -
In this problem it has been given that the moon’s surface the gravitational acceleration due to moon are $\dfrac{1}{4}$and $\dfrac{1}{5}$that of earth, we have to calculate the ratio of mass of earth to the mass of the moon. We know that the surface gravitational acceleration at any planet is g.
$g = \dfrac{{GM}}{{{R^2}}}$ l
Where the M is Mass of the planet and R is the radius of panet.
For earth,
${g_e} = \dfrac{{G{M_e}}}{{R_e^2}}$
${M_e} = $ Mass of earth
${R_e} = $ Radius of earth
For moon,
${g_m} = \dfrac{{G{M_m}}}{{R_m^2}}$
${M_m} = $ Maas on moon
${R_m} = $ Radius of moon
${g_m} = \dfrac{1}{5}{g_e}$
${R_m} = \dfrac{1}{4}{R_e}$
Putting these value in
${g_m} = \dfrac{{G{M_m}}}{{R_m^2}}$
\[{g_m} = \dfrac{{G{M_m}}}{{{{\left( {\dfrac{1}{4}{R_e}} \right)}^2}}} = \dfrac{{{g_e}}}{5}\]
$ \Rightarrow \dfrac{{16G{M_m}}}{{R_e^2}} = \dfrac{{{g_e}}}{5}$
$\dfrac{{16G{M_m}}}{{R_e^2}} = \dfrac{1}{5}\dfrac{{G{M_e}}}{{R_e^2}}$
$ \Rightarrow 80{M_m} = {M_e}$
$\dfrac{{{M_e}}}{{{M_m}}} = \dfrac{{80}}{1}$
Hence, the correct answer for this problem is option B. $80:1$
Note: The force of attraction between any two masses compares the law of gravitation. The importance of Universal law of gravitation lies in the fact that it successfully explains many phenomena such as how different objects in this universe affect others, how it is responsible for weight of the body and keeps us on the ground. Some examples of the force of gravity include: the force that holds the gases in the sun. The force that throws the ball in the air to come down again.
Complete step-by-step solution -
In this problem it has been given that the moon’s surface the gravitational acceleration due to moon are $\dfrac{1}{4}$and $\dfrac{1}{5}$that of earth, we have to calculate the ratio of mass of earth to the mass of the moon. We know that the surface gravitational acceleration at any planet is g.
$g = \dfrac{{GM}}{{{R^2}}}$ l
Where the M is Mass of the planet and R is the radius of panet.
For earth,
${g_e} = \dfrac{{G{M_e}}}{{R_e^2}}$
${M_e} = $ Mass of earth
${R_e} = $ Radius of earth
For moon,
${g_m} = \dfrac{{G{M_m}}}{{R_m^2}}$
${M_m} = $ Maas on moon
${R_m} = $ Radius of moon
${g_m} = \dfrac{1}{5}{g_e}$
${R_m} = \dfrac{1}{4}{R_e}$
Putting these value in
${g_m} = \dfrac{{G{M_m}}}{{R_m^2}}$
\[{g_m} = \dfrac{{G{M_m}}}{{{{\left( {\dfrac{1}{4}{R_e}} \right)}^2}}} = \dfrac{{{g_e}}}{5}\]
$ \Rightarrow \dfrac{{16G{M_m}}}{{R_e^2}} = \dfrac{{{g_e}}}{5}$
$\dfrac{{16G{M_m}}}{{R_e^2}} = \dfrac{1}{5}\dfrac{{G{M_e}}}{{R_e^2}}$
$ \Rightarrow 80{M_m} = {M_e}$
$\dfrac{{{M_e}}}{{{M_m}}} = \dfrac{{80}}{1}$
Hence, the correct answer for this problem is option B. $80:1$
Note: The force of attraction between any two masses compares the law of gravitation. The importance of Universal law of gravitation lies in the fact that it successfully explains many phenomena such as how different objects in this universe affect others, how it is responsible for weight of the body and keeps us on the ground. Some examples of the force of gravity include: the force that holds the gases in the sun. The force that throws the ball in the air to come down again.
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