Variation Of G With Altitude

The acceleration due to gravity at height $h$ above the surface decreases because the body is farther from Earth's centre, following $g_h=g{\left(\frac{R}{R+h}\right)}^2$, derived directly from the inverse-square law applied at distance $R+h$. Here the altitude equals the Earth's radius, so $h=R$ and $R+h=2R$. Substituting gives $g_h=g{\left(\frac{R}{2R}\right)}^2=g\times \frac{1}{4}=\frac{9.8}{4}=2.45$ m/s^2. The option $4.90$ m/s^2 corresponds to halving $g$, which would wrongly use $g(R/(R+h))$ without squaring. The option $3.27$ m/s^2 corresponds to a factor of one-third and has no physical basis here. The option $9.80$ m/s^2 ignores the altitude entirely. It is important to note that this altitude formula differs fundamentally from the depth formula: above the surface the full mass of the Earth remains below the body, so the inverse-square law applies with the increased distance, whereas below the surface the enclosed mass itself shrinks. For small heights $h\ll R$, this expression can be approximated as $g_h\approx g(1-2h/R)$, but here the height is comparable to $R$, so the exact squared form must be retained. This matches the NCERT result that gravity falls to one-quarter of its surface value at one Earth radius up. A magnitude check confirms the reduction is large yet still positive, consistent with weakening but non-zero gravity.