A) 1 m
B) 0.8 m
C) 0.5m
D) 1.25m
E) 2m
Correct Answer: C
Solution :
Given: \[\frac{{{R}_{e}}}{{{R}_{p}}}=\frac{2}{3}\] \[\frac{{{d}_{e}}}{{{d}_{p}}}=\frac{4}{5}\] As \[MG={{g}_{e}}R_{e}^{2}\] and \[M={{d}_{e}}\times \frac{4}{3}\pi R_{e}^{2}\] \[\therefore \] \[{{d}_{e}}\times \frac{4}{3}\pi R_{e}^{3}\times G={{g}_{e}}R_{e}^{2}\] Or \[{{d}_{e}}\times \frac{4}{3}\pi {{R}_{e}}\times G={{g}_{e}}\] ...(i) Similarly for planet \[{{d}_{p}}\times \frac{4}{3}\pi {{R}_{p}}G={{g}_{p}}\] ...(ii) Dividing Eq. (i) by Eq. (ii) we get \[\frac{{{g}_{e}}}{{{g}_{p}}}=\frac{{{R}_{e}}}{{{R}_{p}}}\times \frac{{{d}_{e}}}{{{d}_{p}}}\] \[\frac{{{g}_{e}}}{{{g}_{p}}}=\frac{2}{3}\times \frac{4}{5}=\frac{8}{15}\] \[=0.5m\]
You need to login to perform this action.
You will be redirected in
3 sec
