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(D) The acceleration due to gravity is $g = \frac{GM}{R^2}$,escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$,and gravitational potential energy is $U =

Vedclass · Accepted Answer

All of the above.

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(D) The acceleration due to gravity is $g = \frac{GM}{R^2}$,escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$,and gravitational potential energy is $U = -\frac{GMm}{R}$.From these relations,we observe that $g \propto \frac{M}{R^2}$,$v_e \propto \sqrt{\frac{M}{R}}$,and $U \propto \frac{M}{R}$.Let the change be $x = 0.5\% = 0.005$. If $M' = M(1+x)$ and $R' = R(1+x)$:$1$. For $g$: $g' = \frac{G M(1+x)}{R^2(1+x)^2} = g(1+x)^{-1} \approx g(1-x)$. Thus,$g$ decreases by $0.5\%$.$2$. For $v_e$: $v_e' = \sqrt{\frac{2G M(1+x)}{R(1+x)}} = \sqrt{\frac{2GM}{R}} = v_e$. Thus,escape velocity remains unchanged.$3$. For $U$: $U' = -\frac{G M(1+x)m}{R(1+x)} = -\frac{GMm}{R} = U$. Thus,potential energy remains unchanged.Therefore,all statements are correct.

The radius and mass of the Earth are both increased by $0.5\%$. Which of the following statements is true at the surface of the Earth?

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