Option 2 : is reduced to quarter
10 Questions 10 Marks 10 Mins
CONCEPT:
\(F = G\frac{m_1m_2}{r^2}\) G is the Universal Gravitational Constant. which is the same throughout the universe. m1 and m2 are two masses, r is the distance between two masses, F is the gravitational force between m1 and m2 CALCULATION: So, the Gravitational force between two masses m1 and m2 at distance r is \(F = G\frac{m_1m_2}{r^2}\) -- (1) If the distance is doubled, the new distance is r' = 2r and new force s F' \(F' = G\frac{m_1m_2}{r'^2}\) \(F'= G\frac{m_1m_2}{(2r)^2}\) \(F' = G\frac{m_1m_2}{4r^2}\) \(F'=\frac{1}{4} G\frac{m_1m_2}{r^2}\) \(F'=\frac{F}{4}\) (Using Equation 1) So, F' = F / 4 So, the gravitational force becomes one-fourth or quarter of the original value. Additional Information The value of universal Gravitational constant (G) is G = 6.674×10−11 m3 kg−1 s−2 India’s #1 Learning Platform Start Complete Exam Preparation
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