Pair | Masses | Separation |
A | 20kg, 30kg | 1m |
B | 40kg, 60kg | 0.5m |
C | 30kg, 50kg | 2m |
D | 10kg, 40kg | 2.5m |
Answer:
\[{{F}_{A}}=\frac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\] \[=\frac{6.67\times {{10}^{-11}}\times 20\times 30}{{{1}^{2}}}=40\times {{10}^{-9}}=4\times {{10}^{-8}}N\] \[{{F}_{B}}=\frac{6.67\times {{10}^{-11}}\times 40\times 60}{{{(0.5)}^{2}}}\] \[=\frac{160}{0.25}\times {{10}^{-9}}=64\times {{10}^{-8}}N=64\times {{10}^{-8}}N\] \[{{F}_{C}}=\frac{6.67\times {{10}^{-11}}\times 30\times 50}{{{2}^{2}}}\] \[=25\times {{10}^{-9}}=2.5\times {{10}^{-8}}N=2.5\times {{10}^{-8}}N\] \[{{F}_{D}}=\frac{6.67\times {{10}^{-11}}\times 10\times 40}{{{(2.5)}^{2}}}\] \[=4.27\times {{10}^{-9}}N=0.427\times {{10}^{-8}}N\] \[{{F}_{A}}=40\times {{10}^{-9}}N;\,\,{{F}_{B}}=64\times {{10}^{-8}};\] \[{{F}_{C}}=2.5\times {{10}^{-8}}N;\,\,{{F}_{D}}=0.427\times {{10}^{-8}}N\] \[{{T}_{1}}=T\]Pairs in the decreasing order of magnitude of gravitational force is B, A, C, D.
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