A) \[\text{1}\text{.3}0\text{4 }\times \text{ 1}{{0}^{\text{4}}}\text{m}/\text{sec}\]
B) \[\text{13}.0\text{4 }\times \text{1}{{\text{0}}^{4}}\text{m}/\text{sec}\]
C) \[\text{1}\text{.3}0\text{4 }\times \text{1}{{\text{0}}^{\text{6}}}\text{m}/\text{sec}\]
D) \[1.304\,\times {{10}^{2}}m/sec\]
Correct Answer: A
Solution :
\[F=\frac{GMM}{{{r}^{2}}}\] \[=\frac{6.67\times {{10}^{-11}}\times (1.99\times {{10}^{30}})\times (1.9\times {{10}^{27}})}{{{(7.8\times {{10}^{11}})}^{2}}}\] \[=4.14\times {{10}^{23}}N\] Now let its speed of Jupiter be\[v\], then \[F=\frac{m{{v}^{2}}}{R}\] \[\Rightarrow \] \[v=\sqrt{\frac{FR}{m}}\] \[=\sqrt{\frac{(7.8\times {{10}^{11}})\times (4.10\times {{10}^{23}})}{(1.9\times {{10}^{27}})}}\] \[=\mathbf{1}\mathbf{.34\times 1}{{\mathbf{0}}^{\mathbf{4}}}\mathbf{m/sec}\]
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