The Gravitational Constant \(G\)
1 · Why we care about \(G\)
\(G\) links two masses through Newton’s universal gravitation law:
\(F = \dfrac{G\,M\,m}{d^{2}}\) :contentReference[oaicite:0]{index=0}
Here \(M\) and \(m\) are the masses, \(d\) is the distance between their centres, and \(F\) is the attractive force.
2 · Shell ideas you can trust
- Outside a hollow shell: A point mass outside acts as if the shell’s entire mass sits at its centre. :contentReference[oaicite:1]{index=1}
- Inside a hollow shell: A point mass feels zero net force; pulls from every part cancel perfectly. :contentReference[oaicite:2]{index=2}
3 · Cavendish’s clever torsion-balance experiment (1798)
Henry Cavendish pinned down \(G\) with a light rod (AB) carrying two tiny lead balls. The rod hung from a thin wire.
- He rolled two big lead spheres close to the small ones on opposite sides. Equal pulls created a twist but no net push. :contentReference[oaicite:3]{index=3}
- The twist gave a torque \(F\,L\) that the wire resisted with a restoring torque \(\tau\,\theta\) (τ = torsional constant, \(\theta\) = twist angle). :contentReference[oaicite:4]{index=4}
- At equilibrium the torques balance,
\(\tau\,\theta = \dfrac{G\,M\,m\,L}{d^{2}}\) :contentReference[oaicite:5]{index=5}
Measuring \(\theta\), \(M\), \(m\), \(L\), \(d\), and \(\tau\) lets you solve for \(G\).
4 · Current numerical value
\(G = 6.67 \times 10^{-11}\,\text{N·m}^{2}\text{/kg}^{2}\) :contentReference[oaicite:6]{index=6}
5 · Quick recap for NEET
- \(G = 6.67 \times 10^{-11}\,\text{N·m}^{2}\text{/kg}^{2}\) (know the number and units).
- Universal gravitation: \(F = G\,M\,m/d^{2}\).
- Shell theorem: zero force inside a hollow shell; outside force acts from the centre.
- Torsion-balance idea: wire twist angle \(\theta\) leads to \(\tau\,\theta = G\,M\,m\,L/d^{2}\).
- Cavendish’s 1798 setup is the classic way questions test \(G\).
