Variation Of G With Depth

Inside a uniform Earth only the mass enclosed within the radius of the measurement point contributes to gravity, because the outer spherical shell exerts no net force on an interior point. This leads to $g_d=g\left(1-\frac{d}{R}\right)$, where $d$ is the depth below the surface. With $d=R/2$, the factor becomes $1-\frac{R/2}{R}=1-0.5=0.5$, so $g_d=9.8\times 0.5=4.90$ m/s^2. The option $2.45$ m/s^2 wrongly applies the squared altitude formula instead of the linear depth formula. The option $7.35$ m/s^2 uses an incorrect depth fraction of one-quarter. The option $9.80$ m/s^2 ignores the depth correction. The linear behaviour arises because the enclosed mass grows as the cube of the inner radius while the gravitational pull from that mass falls as the inverse square of the same radius, and the net combination of $r^3/r^2$ produces a result directly proportional to $r$. Consequently gravity is maximum at the surface, decreases steadily inward, and becomes exactly zero at the Earth's centre, where a body would feel no net gravitational force from the surrounding symmetric shells. This is the standard NCERT shell-theorem result, where gravity falls linearly to zero at the centre rather than following the inverse-square pattern of altitude. A plausibility check confirms that at half the radius the enclosed mass and shrinking distance combine to give exactly half the surface gravity.