Gravitation Part 1

At neutral point g = 0 so graph (C) is correct Given,Acceleration due to gravity on Earth, $g_{E}=10 \text{ m/s}^2$and acceleration due to gravity on Mars, $g_{M}=4 \text{ m/s}^2$We know that,$g_{E}$ (at height $h)=g_{\;\text{Earth }\;}\left(1-\frac{2h}{R}\right)$$\therefore$ Weight at Earth,$m g_{E}=100 \times 10 = 1000 \text{ N}$As the spaceship moves far away from Earth, the value of $g_{E}$ decreases to zero at a point where $g_{E}+g_{M}=0$ and hence weight will also be zero.This point is called neutral point and is shown by graph (c) in the given figure.Then, $g_{E}$ increases $4 \text{ ms}^{-2}$ at Mars surface and weight becomes $400 \text{ N}$ which is also exhibited by graph (c).