Question

What is the magnitude of earth and a 6 x 10^6 kg and radius of the earth is 6.4 x 10 m.) the gravitational force between the 1 kg object on its surface? (Mass of the earth is




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Answer 1

Answer:

Given:- m = 1kg, M = 6×10^24, R = 6.4×10^6

Solution:- $F= G\frac{Mm}{R {}^{2} } \\ \\ F = \frac{6.67 \times 10 {}^{ - 11} \times 6 \times 10 {}^{24} \times 1 }{(6.4 \times 10 {}^{6}) {}^{2} } \\ \\F = \frac{6.67 \times 6 \times 10}{6.4 {}^{2} } \\ \\F = 9.8 \:Newton\: (approx)$

Answer 2

$\begin{gathered}\begin{gathered}⠀⠀⠀⠀\begin{gathered} \\ { { {\boxed {\rm {\blue { F = \dfrac{Gm_1m_2}{{r}^{2} }}}}}}} \\ \\ \\ \bf \: According \: to \: formula, \: \\ \\ \tt { \: F = \dfrac{6.7 \times {10}^{ - 11 } \times 6 \times {10}^{24} \times 1}{( \: 6.4 \times {10}^{6} ) {}^{2} } } \\ \\ \tt { \: F = \dfrac{ 6.7 \times {10}^{ - 11} \times 6 \times {10}^{24} \times 1 }{6.4 \times 6.4 \times {10}^{6} \times {10}^{6} } } \\ \\ \tt { \: F = \dfrac{6.7 \times6 \times {10}^{( - 11 + 24)} }{6.4 \times 6.4 \times {10}^{(6 + 6)} } } \\ \\ \tt { \: F = \dfrac{6.7 \times 6 \times {10}^{13} }{6.4 \times 6.4 \times {10}^{12} } } \\ \\ \tt { \: F = \dfrac{40.2 }{40.96} \times {10}^{(13 - 12)} } \\ \\ \tt { \: F = \dfrac{40.2 }{40.96} \times 10 } \\ \\ \tt { \: F = \dfrac{402 \: \times \: \cancel{ 10 } \: \times 100}{4096 \: \times \: \cancel{10}} } \\ \\ \tt { \: F = \cancel {\dfrac{40200}{4096} } = 9.8 \: N} \\ \\ \\ \end{gathered} \end{gathered} \end{gathered}$

Therefore , magnitude of the gravitational force between the earth and 1 kg object on its surface is 9.8 Newtons.