Physics, asked by yugbaliyan81, 5 months ago

the mass of earth is 6×10^24kg and the mass of Moon is 7.4×10^22kg if the distance between the earth and the moon is 3.8×10^5km calculate the force exerted by the earth on the moon.

Answers

Answered by arghyasingh112
0

Answer:

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Answered by shaktisrivastava1234
21

 \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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