Physics, asked by flashcycling56, 6 months ago

calculate the force exerted by the Earth on the Moon if the mass of the earth is 6 into 10 to the power 24 kg and the mass of the moon is 7.4 into 10 to the power 22 kg distance between them is 3.84 into 10 to the power 5 kilometer and gravitational constant equal to 6.67 into 10 to the power minus 11 Newton meter square by kg square ​

Answers

Answered by Anonymous
1

Answer:

Given that,

Mass of the Earth m

1

=6×10

24

Kg

Mass of the Moon m

2

=7.4×10

22

kg

Distance between the Earth and the Moon d=3.84×10

5

km=3.84×10

8

m

Gravitational Constant G=6.7×10

−11

Nm

2

/kg

2

Now, by using Newton’s law of gravitation

F=

r

2

Gm

1

m

2

F=

(3.84×10

8

)

2

6.7×10

−11

×6×10

24

×7.4×10

22

F=

14.8225×10

16

297.48×10

35

F=20.069×10

19

F=20.1×10

19

N

Hence, the gravitational force of attraction is 20.1×10

19

N

Explanation:

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Answered by shaktisrivastava1234
5

 \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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