A) g1 = 3g
B) g1=9g
C) \[\operatorname{g}1\frac{g}{9}\]
D) g1=27g
Correct Answer: A
Solution :
\[\operatorname{g}=\frac{GM}{{{R}^{2}}}\] \[\operatorname{g}=\frac{G\ell V}{{{R}^{2}}}\] \[{{\ell }_{\operatorname{e}}}={{\ell }_{\operatorname{p}}}\,\,\,\operatorname{Rp}=3\operatorname{Re}\] Volume\[=\frac{4}{3}\operatorname{G}\ell \pi \times R\]Use this in equation (1) we get \[\operatorname{g}'=\frac{4}{3}G\ell \pi R\] ?.(2) For planet \[\operatorname{g}'=\frac{4}{3}G\ell \pi \left( 3R \right)\] ?.(3) by (2) & (3) g? = 3gYou need to login to perform this action.
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