80 Questions 80 Marks 50 Mins
Concept:
\(Gravitational\;Force = F \propto \frac{{{m_1}{m_2}}}{{{r^2}}} \Rightarrow F = G\frac{{{m_1}{m_2}}}{{{r^2}}}\) Where G is gravitational constant, r is the distance between two masses, m1 and m2 are the masses. Explanation: Let the masses of two balls are m1 and m2 and they are separated by distance r \(Gravitational\;Force = F = G\frac{{{m_1}{m_2}}}{{{r^2}}}\) Now given that: Mass of each balls and distance between them is doubled: Mass of first ball (m1’) = 2 m1 Mass of second ball (m2’) = 2 m2 Distance between two balls (r’) = 2 r \(New\;Gravitational\;Force\;\left( {F'} \right) = G\frac{{m_1'm_2'}}{{{{\left( {r'} \right)}^2}}} = G\frac{{\left( {2{m_1}} \right)\left( {2{m_2}} \right)}}{{{{\left( {2r} \right)}^2}}}\) \( \implies F' = G\frac{{4{m_1}{m_2}}}{{4{r^2}}} = G\frac{{{m_1}{m_2}}}{{{r^2}}} = F\) Hence, it will remain same. India’s #1 Learning Platform Start Complete Exam Preparation
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