Question

if the distance between the earth and moon is 3.84×10⁵ km ,than calculate the exerted by the earth on the moon (take mass of the earth 6.4×10² kg mass of the moon 7.4×10²²kg​

Answer 1

Diagram:

$\setlength{ \unitlength}{1cm} \thicklines \begin{picture}(10,10) \put(5,1){ \circle{4}} \put(10,1){ \circle{0.4}} \put(4.7, - 0.4){ \sf earth} \put(9.7, - 0.3){ \sf moon} \put(5,1){ \line(1,0){5}} \put(7.5,1.2){ \sf d} \put(4.9,1.2){ \sf M} \put(9.7,0.5){ \sf m} \end{picture}$

$\qquad \sf F = \dfrac{GMm}{ {d}^{2} } \\ \\ \underline{ \sf Given} \\ \hookrightarrow \sf \quad G = 6.67 \times {10}^{ - 11} \: N {m}^{2} {kg}^{ - 2} \\ \hookrightarrow \sf \quad M = 6.4 \times {10}^{24} \: kg \\ \hookrightarrow \quad \sf m = 7.4 \times {10}^{22} \: kg \\ \hookrightarrow \quad \sf d = 3.84 \times {10}^{8 } \: m \\ \\ \sf \rightarrow \quad F = \frac{6.67 \times {10}^{ - 11} \times 6.4 \times {10}^{24} \times 7.4 \times {10}^{22} }{ {(3.84 \times {10}^{8} )}^{2} } \\ \\ \sf \rightarrow \quad F = \frac{315.89 \times {10}^{ - 11 + 24 + 22} }{14.74 \times {10}^{16} } \\ \\ \sf \rightarrow \quad F = \frac{315.89}{14.74} \times \frac{ {10}^{35} }{ {10}^{16} } \\ \\ \sf \rightarrow \quad F = 21.43 \times {10}^{35 - 16} \\ \\ \sf \rightarrow \quad F = 21.36 \times {10}^{19} \: N \\ \\ \\ \underline{ \sf Properties \: Used} \\ \star \sf \quad { ({a}^{n}) }^{m} = {a}^{m \times n} \\ \\ \star \quad \sf \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n} \\ \\ \star \quad \sf {a}^{m} \times {a}^{n} = {a}^{m + n}$

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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