Question

find the gravitational force of attraction between the moon and the earth if the man of moon is 1/81 times the mass of earth . ( given : G = 6.67*10^-11 Nm^2 kg ^-2 , radius of the moon's orbit = 3.58*10^8 m , and the mass of the earth is 6.10^24 kg anwer fast i will surely mark u as a brainliest

Answer 1

$\huge\bold{\underbrace{\red{SOLUTION:-}}}$

GIVEN :-

✍️ The mass of the moon is 1/81 times the mass of the earth .

• Gravitational Constant [G] = $\rm{6.67\times{10^{-11}}\:N.m^2.kg^{-2}}$

• radius of moon's orbit [d] = $\rm{3.58\times{10^8}\:m}$

• mass of earth (m) = $\rm{6\times{10^{24}}\:kg}$

TO FIND :-

• The gravitational force of attraction between the moon and the earth .

CALCULATION :-

Let,

✍️ The mass of the earth be “m” .

According to the question,

• $\rm\bold{mass\:of\:moon\:=\:\dfrac{1}{81}\:\times{mass\:of\:earth}\:}$

• $\rm\blue{\implies\:mass\:of\:moon\:=\:\dfrac{m}{81}\:}$

✍️ Using Newton's law of Gravitation;

$\orange\bigstar\:\rm{\purple{\boxed{\red{Force\:=\:G\:\dfrac{m_1\:m_2}{d^2}\:}}}}$

Here,

• $\rm{m_1\:=\:mass\:of\:earth\:=\:m\:}$

• $\rm{m_2\:=\:mass\:of\:moon\:=\:\dfrac{m}{81}\:}$

• d = radius of moon's orbit .

$\rm{\implies\:F\:=\:G\:\times{\dfrac{m\:\times{\dfrac{m}{81}}}{d^2}}\:}$

$\rm{\implies\:F\:=\:6.67\times{10^{-11}}\:\times\:\dfrac{6\times{10^{24}}\times{\dfrac{6\times{10^{24}}}{81}}}{(3.58\times{10^8})^2}\:}$

$\rm{\implies\:F\:=\:\dfrac{240.12\times{10^{37}}}{1038.1284\times{10^{16}}}\:}$

$\rm{\pink{\implies\:F\:=\:2.313\times{10^{20}}\:N\:}}$

$\green\therefore$ The gravitational force of attraction between moon and earth is “$\underline\bold\red{2.313\times{10^{20}}\:N\:}$” .