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(D) The weight of the passenger is the net gravitational force acting on them. Let $M_e$ and $M_m$ be the masses of Earth and Mars,and $R_e$ and $R_m$

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$D$

Vedclass · Answer

(D) The weight of the passenger is the net gravitational force acting on them. Let $M_e$ and $M_m$ be the masses of Earth and Mars,and $R_e$ and $R_m$ be their radii. The gravitational force $F$ at a distance $x$ from Earth along the path is given by $F = |F_e - F_m| = |\frac{G M_e m}{x^2} - \frac{G M_m m}{(d-x)^2}|$,where $d$ is the distance between Earth and Mars.At the surface of Earth $(x = R_e)$,the weight is $W_e = m g_e = 60 	imes 10 = 600 \, N$.At the surface of Mars $(x = d - R_m)$,the weight is $W_m = m g_m = 60 	imes 4 = 240 \, N$.As the spaceship moves from Earth to Mars,the gravitational pull of Earth decreases and the gravitational pull of Mars increases. At some point between the two planets,the gravitational forces from Earth and Mars will be equal in magnitude,making the net gravitational force (weight) zero.Since the spaceship moves with constant velocity,the distance $x$ is proportional to time $t$. Therefore,the graph of weight versus time must decrease from $600 \, N$,reach $0 \, N$ at some intermediate time,and then increase to $240 \, N$ at the arrival time $t_0$. Curve $D$ is the only one that shows the weight becoming zero and then increasing to $240 \, N$.

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