Suppose the acceleration due to gravity at the earth's surface is \(g \text{ m/s}^2\) and at the surface of moon it is \(g' \text{ m/s}^2\). An M kg passenger goes from the earth to the moon in a spaceship moving with a constant velocity (Neglect all other objects in the sky). Which curve best represents the weight (net gravitational force) as a function of time?

Let the mass of the Earth be \(M_e\), the mass of the Moon be \(M_m\), and the distance between their centers be \(d\).

When the spaceship is at a distance \(r\) from the center of the Earth, the net gravitational force on the passenger of mass \(M\) is given by:
\(F_{net} = \dfrac{G M_e M}{r^2} - \dfrac{G M_m M}{(d-r)^2}\)

Since the spaceship moves with a constant velocity, the distance \(r\) increases linearly with time \(t\).

At the surface of the Earth (\(t = 0\)), the gravitational pull of the Moon is negligible, so the net force is approximately the weight on Earth:
\(F_{net} \approx \dfrac{G M_e M}{R_e^2} = Mg\)

As the spaceship travels towards the Moon, the Earth's gravitational pull decreases while the Moon's pull increases. At a specific point called the neutral point, the two gravitational forces become equal in magnitude and opposite in direction, making the net gravitational force zero.

Since the Earth is much more massive than the Moon (\(M_e \approx 81 M_m\)), this neutral point lies much closer to the Moon. Thus, the net force becomes zero at a time closer to \(t_0\).

Beyond the neutral point, the Moon's gravitational pull dominates. At the surface of the Moon (\(t = t_0\)), the net force is approximately the weight on the Moon:
\(|F_{net}| \approx \dfrac{G M_m M}{R_m^2} = Mg'\)

The weight of the passenger is the magnitude of the net gravitational force, which cannot be negative. It starts at \(Mg\), decreases to zero at the neutral point (closer to \(t_0\)), and then increases to \(Mg'\).

Curve C is the only graph that correctly represents this behavior, as it touches zero and then rises to \(Mg'\).

Answer: C