Acceleration Due to Gravity (g) — Key Ideas & Friendly Notes

1. Getting started — the big picture

Imagine Earth as a perfectly uniform sphere. Outside that sphere, the pull you feel is exactly the same as if Earth’s whole mass M_E sat at its centre. The gravitational pull on a small mass m at a distance r from the centre is \(F=\dfrac{G\,M_E\,m}{r^{2}}\). :contentReference[oaicite:0]{index=0}

2. g at the surface

At the surface (r = R_E) this force becomes your weight, mg. Equating the two gives \(g=\dfrac{G\,M_E}{R_E^{2}}\). :contentReference[oaicite:1]{index=1}

• The universally accepted gravitational constant is \(G = 6.67\times10^{-11}\,\text{N m}^2\text{/kg}^2\). :contentReference[oaicite:2]{index=2}
• Knowing g, R_E, and G lets us “weigh” Earth: \(M_E=\dfrac{g\,R_E^{2}}{G}\). (That’s why Cavendish is said to have weighed the planet!) :contentReference[oaicite:3]{index=3}

3. g above the surface (height ⩾ 0)

At a height h (so the distance from the centre is R_E+h), the pull drops to \(g(h)=\dfrac{G\,M_E}{(R_E+h)^{2}}\). :contentReference[oaicite:4]{index=4}

For small heights (h ≪ R_E) a handy approximation is \(g(h)\approx g\!\left(1-\dfrac{2h}{R_E}\right)\), meaning g falls by roughly 0.2 % for every kilometre you climb. :contentReference[oaicite:5]{index=5}

4. g inside Earth (depth d below the surface)

Only the mass of the inner sphere (radius R_E−d) pulls you; the outer shell cancels out. For uniform density: \(M_s = M_E\!\left(\dfrac{R_E-d}{R_E}\right)^{\!3}\). :contentReference[oaicite:6]{index=6}

The resulting acceleration is \(g(d)=\dfrac{G\,M_s}{(R_E-d)^{2}}=\dfrac{G\,M_E}{R_E^{3}}\,(R_E-d)=g\!\left(1-\dfrac{d}{R_E}\right)\). So g decreases linearly with depth, becoming zero at the centre. :contentReference[oaicite:7]{index=7}

5. Why hills feel lighter and mines feel lighter too

• On a mountain (positive ), you are farther from Earth’s centre & outer shells do pull you — but the extra distance dominates, so g drops.
• In a mine (depth d), part of Earth’s mass lies above you and cancels out, again reducing g.

6. Inside Earth at any radius r

More generally, at a distance r (< R_E) the pull is \(F=\dfrac{G\,M_E\,m\,r}{R_E^{3}}\) :contentReference[oaicite:8]{index=8}, giving \(g(r)=g\,\dfrac{r}{R_E}\). The field therefore grows linearly from zero at the centre to maximum at the surface.

7. High-yield NEET takeaways

1. Surface formula: \(g=\dfrac{G\,M_E}{R_E^{2}}\) — the go-to relation for weight and planetary mass.
2. Variation with height: \(g(h)\approx g\!\left(1-\dfrac{2h}{R_E}\right)\) for small h.
3. Variation with depth: \(g(d)\approx g\!\left(1-\dfrac{d}{R_E}\right)\) for small d, reaching zero at the centre.
4. Linear field inside Earth: \(g(r)=g\,\dfrac{r}{R_E}\).
5. Weighing Earth: Using \(M_E=\dfrac{g\,R_E^{2}}{G}\) plus Cavendish’s value of G.

Keep these nuggets handy — they pop up again and again in NEET problems. You’ve got this!