A) \[{{g}_{p}}=\frac{{{g}_{e}}}{2}\]
B) \[{{g}_{p}}={{g}_{e}}\]
C) \[{{g}_{p}}=2{{g}_{e}}\]
D) \[{{g}_{p}}=\frac{{{g}_{e}}}{\sqrt{2}}\]
Correct Answer: A
Solution :
Acceleration due to gravity is given by \[g=\frac{GM}{{{R}^{2}}},\] where G is gravitational constant. For earth: \[{{g}_{e}}=\frac{G{{M}_{e}}}{{{R}_{e}}}\] For planet: \[{{g}_{p}}=\frac{G{{M}_{p}}}{R_{p}^{2}}\] Therefor, \[\frac{{{g}_{e}}}{{{g}_{p}}}=\frac{G{{M}_{e}}/R_{e}^{2}}{G{{M}_{p}}/R_{p}^{2}}\] or \[\frac{{{g}_{e}}}{{{g}_{p}}}\frac{{{M}_{e}}}{{{M}_{p}}}\times \frac{R_{p}^{2}}{R_{e}^{2}}\] ?(i) Given, \[{{M}_{p}}=2{{M}_{e}},{{R}_{p}}=2{{R}_{e}}\] Putting the values in the Eq. (i), we obtain \[\frac{{{g}_{e}}}{{{g}_{p}}}=\frac{{{M}_{e}}}{2{{M}_{e}}}\times \frac{\left( 2R_{e}^{2} \right)}{R_{e}^{2}}\] \[=\frac{1}{2}\times \frac{4}{1}\] \[=2\] \[\therefore \] \[{{g}_{p}}=\frac{{{g}_{e}}}{2}\]
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