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(c) $g = \frac{{GM}}{{{R^2}}}$ $	herefore$ $g \propto \frac{M}{{{R^2}}}$

According to problem ${M_p} = \frac{{{M_e}}}{2}$ and ${R_p} = \frac{{{R_e

Vedclass · Accepted Answer

$19.6$

Vedclass · Answer

(c) $g = \frac{{GM}}{{{R^2}}}$ $	herefore$ $g \propto \frac{M}{{{R^2}}}$

According to problem ${M_p} = \frac{{{M_e}}}{2}$ and ${R_p} = \frac{{{R_e}}}{2}$

$	herefore$ $\frac{{{g_p}}}{{{g_e}}} = \left( {\frac{{{M_p}}}{{{M_e}}}} ight)\;{\left( {\frac{{{R_e}}}{{{R_p}}}} ight)^2} = \left( {\frac{1}{2}} ight)\; 	imes {(2)^2} = 2$

==> ${g_p} = 2{g_e} = 2 	imes 9.8 = 19.6\;m/{s^2}$

If a planet consists of a satellite whose mass and radius were both half that of the earth, the acceleration due to gravity at its surface would be ......... $m/{\sec ^2}$ ($ g$ on earth $= 9.8\, m/sec^2$ )

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