Question

physics gravitation numerical

Answer 1

Given that :-

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• Mass of the Earth = m₁ = 6 × 10²⁴ kg
• Mass of the Sun = m₂ = 2 × 10³⁰ kg
• Average distance = r = 1.5 × 10¹¹ m

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To find :-

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• Force of gravitation = F = ??

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Answer :-

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• Force of gravitation between Earth and Sun is 3.557 × 10²² N

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Solution :-

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We know that ;

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$\sf{\large{\dag{\underline{\boxed{ F = G \dfrac{m_1 . m_2}{r^2}}}}}}$

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Where ;

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G = 6.67 × 10⁻¹¹ Nm² kg m⁻²

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Now ;

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$\sf : \implies \: { F = \dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{30} }{ ( 1.5 \times 10^{11} )} }$

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$\sf : \implies \: { F = \dfrac{6.67 \times 6 \times 2 \times 10^{-11 } \times 10^{24} \times 10^{30} }{ ( 1.5 \times 10^{11} )^2 } }$

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• From law of exponent ; aᵐ × aⁿ = aᵐ⁺ⁿ

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$\sf : \implies \: { F = \dfrac{6.67 \times 6 \times 2 \times 10^{24 + 30 +( -11) } }{ ( 1.5 \times 10^{11} )^2} }$

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$\sf : \implies \: { F = \dfrac{6.67 \times 6 \times 2 \times 10^{54 - 11} }{ ( 1.5 \times 10^{11} )^2} }$

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$\sf : \implies \: { F = \dfrac{6.67 \times 6 \times 2 \times 10^{43} }{ ( 1.5 \times 10^{11} )^2} }$

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$\sf : \implies \: { F = \dfrac{6.67 \times 6 \times 2 \times 10^{43} }{ 1.5 \times 1.5 \times 10^{11} \times 10^{11} } }$

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$\sf : \implies \: { F = \dfrac{6.67 \times 6 \times 2 \times 10^{43} }{ 1.5 \times 1.5 \times 10^{11 + 11}} }$

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$\sf : \implies \: { F = \dfrac{6.67 \times 6 \times 2 \times 10^{43} }{ 1.5 \times 1.5 \times 10^{22} } }$

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• From law of exponent ; aᵐ / aⁿ = aᵐ⁻ⁿ

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$\sf : \implies \: { F = \dfrac{6.67 \times 6 \times 2 \times 10^{43 - 22 } }{ 1.5 \times 1.5 } }$

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$\sf : \implies \: { F = \dfrac{6.67 \times 6 \times 2 \times 10^{21 } }{ 1.5 \times 1.5 } }$

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$\sf : \implies \: { F = \dfrac{6.67 \times 6 \times 2 \times 10^{21 } }{ 2.25} }$

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$\sf : \implies \: { F = \dfrac{6.67 \times 2 \times 2 \times 10^{21 } }{ 0.75} }$

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$\sf : \implies \: { F = \dfrac{6.67 \times 4 \times 10^{21 } }{ 0.75} }$

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$\sf : \implies \: { F = \dfrac{26.68 \times 10^{21 } }{ 0.75} }$

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$\sf : \implies \: { F = \dfrac{2668 \times 10^{21 } }{ 75} }$

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$\sf : \implies \: { F = \dfrac{533.6 \times 10^{21 } }{ 15} }$

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$\sf : \implies \: { F = \dfrac{106.72 \times 10^{21 } }{ 3} }$

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$\sf : \implies \: { F = \dfrac{35.573\times 10^{21 } }{ 1 } }$

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$\sf : \implies \: { F = 3.557 \times 10^{22 } N }$

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$\sf\: \: \:\: \: \: \: \: \: \: \bigstar{\underline{\boxed{ F = 3.557 \times 10^{22} N}}}$