Explanation
The acceleration due to gravity is: $$g=\dfrac{GM}{R^2}$$
The new value of g would be: $$g'=\dfrac{G(0.99M)}{(0.99R)^2}\simeq1.01\dfrac{GM}{R^2}=1.01g$$
Thus,g would increase by about 1%.The new escape velocity would be:
$${V_e}'=\sqrt{\dfrac{2\times0.99M\times G}{0.99R}}=\sqrt{\dfrac{2MG}{R}}=V_e$$
Thus,the escape velocity will remain unchanged.The potential energy of a body of mass m on earth's surface would be
$$-\dfrac{GM(0.99M)}{(0.99R)}=-\dfrac{GmM}{R}$$
Thus,the potential energy will also remain unchanged.Hence,the correct choices are (b) and (d).
A.The escape velocity for the Moon is $$2.4kms^{-1}$$
D.$$v_0=\sqrt{\dfrac{2GM}{R}}=\sqrt{\dfrac{2G}{R}\times\dfrac{4}{3}\pi {R^3}\rho}$$
or $${v_0}\propto {\sqrt{\rho}}$$
so,the incorrect choices are A and D.
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