Question

calculate the gravitational force of attraction between the moon (mass=7.36×10²² kg)and the earth (mass=5.96×10²⁴kg) taking the distance of the moon from the earth as 3.8×100000000m.​

Answer 1

Answer:

Let vector be R

Then R^2=A^2+B^2 for magnitude

By the Triangle Law of Vectors

I think

Hope this helped

Explanation:

Answer 2

$\huge \sf {\fbox{\fbox{\red{\fbox{Answer:}}}}}$

$\huge \bf{Given:-}$

$\sf {→Mass \: of \: the \: earth,m_1 = 6 \times{{10}^{24} }}$

$\sf {→Mass \: of \: the \: moon,m_2 = 7.4\times{{10}^{22} }}$

$\sf {→Distance \: between \: the \: earth \: and \: moon,r = 3.84\times{{10}^{5} }}$

$\sf {→Distance \: between \: the \: earth \: and \: moon,r = (3.84\times{{10}^{5} \times 1000)m }}$

$\sf {→Distance \: between \: the \: earth \: and \: moon,r = 3.84\times{{10}^{8}m}}$

$\huge \bf{To \: find:- }$

$\sf{⇒Force \: exerted \: to \: one \: body \: to \: another \: body.}$

$\huge \bf{Formula \: used: - }$

$\leadsto\sf{F =G \times \frac{m_1 \times m_2}{r^2} }$

$\huge \bf{Concept \: used: - }$

$\sf{Gravitational \: constant,G=6.7 \times {10}^{- 11N} N{m}^{2}k {g}^{ - 2} }$

$\huge\bf{According \: to \: Question:-}$

$\bf{F = \frac{6.7 \times {10}^{ - 11} \times 6 \times {10}^{24} \times 7.4 \times {10}^{22}} {(3.84 \times {10}^{8} )^{2} } = 2.01 \times {10}^{20} newtons}$

$\sf\longmapsto{2.01 \times {10}^{20} newtons}$

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