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Question: The mass of the earth is \(6 \times {10^{24\,}}\,kg\) and that of the moon is \(7.4 \times {10^{22\,...

The mass of the earth is 6×1024kg6 \times {10^{24\,}}\,kg and that of the moon is 7.4×1022kg7.4 \times {10^{22\,}}\,kg distance between the earth and the moon be 3.84×105km3.84 \times {10^5}\,km ,calculate the force exerted by the earth on the moon. (G=6.7×1011Nm2kg2)(G = 6.7 \times {10^{ - 11}}\,N{m^2}k{g^{ - 2}})

Explanation

Solution

According to Newton's universal law of gravitation, every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between the two masses.
Mathematically it is written as
F=Gm1m2r2F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}
Where m1{m_1}and m2{m_2}are the mass of objects, GGis the gravitational constant having valueG=6.7×1011Nm2kg2G = 6.7 \times {10^{ - 11}}\,N{m^2}k{g^{ - 2}}., rr is the distance between the two objects.

Complete step by step answer:
Given,
Mass of earth, Me=6×1024kg{M_e} = 6 \times {10^{24\,}}\,kg
Mass of moon,Mm=7.4×1022kg{M_m} = 7.4 \times {10^{22\,}}\,kg
Distance between earth and moon,
d=3.84×105kmd = 3.84 \times {10^5}\,km
d=3.84×105×103m\Rightarrow d = 3.84 \times {10^5} \times {10^3}\,m
d=3.84×108m\Rightarrow d = 3.84 \times {10^8}\,m
Gravitational constant,G=6.7×1011Nm2kg2G = 6.7 \times {10^{ - 11}}\,N{m^2}k{g^{ - 2}}
According to Newton's universal law of gravitation, every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between the two masses.
Mathematically it is written as
F=Gm1m2r2F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}
Where m1{m_1} and m2{m_2} are the mass of objects, GG is the gravitational constant having valueG=6.7×1011Nm2kg2G = 6.7 \times {10^{ - 11}}\,N{m^2}k{g^{ - 2}}., rr is the distance between the two objects.
Therefore, force between earth and moon can be given as
F=GMeMmd2F = G\dfrac{{{M_e}{M_m}}}{{{d^2}}}
Substitute the given values,
F=6.7×1011Nm2kg2×6×1024kg×7.4×1022kg(3.84×108m)2F = 6.7 \times {10^{ - 11}}\,N{m^2}k{g^{ - 2}} \times \dfrac{{6 \times {{10}^{24\,}}\,kg \times 7.4 \times {{10}^{22\,}}\,kg}}{{{{\left( {3.84 \times {{10}^8}\,m} \right)}^2}}}
F=2.01×1020N\therefore F = 2.01 \times {10^{20}}\,N
This is the force exerted by the earth on the moon.

Note: It is due to the earth's gravitational pull that the moon revolves around the earth. From the equation for gravitational force we can see that it is inversely related to the square of distance between the masses. So if the moon was much farther then it would not have felt the strong gravitational attraction of the earth and it would just float away into space. But since the distance is considerably less so that it experiences a strong pull from the earth, it keeps revolving around the earth.