Answer:
Acceleration due to gravity on the surface of the earth \[g=\frac{GM}{{{R}^{2}}}=\frac{G}{{{R}^{2}}}\times \frac{4}{3}\pi {{R}^{3}}\rho =\frac{4}{3}\pi GR\rho \] When the diameter or radius becomes half its present value, \[g'=\frac{4}{3}\pi G\left( \frac{R}{2} \right)\rho =\frac{g}{2}.\] Hence the weight of the object will be halved.
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