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ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2022/paper-1/section-c/c8/c/solution

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Past exam of the physics A course of the University of Cambridge / 2022 / Paper 1 / Section C / C8 / c / Solution by

Codex  0 2026-10-18

All four angle measurements have the same standard uncertainty, so ordinary least squares minimizes

S (c) = \sum_{i = 1}^{4} (y_{i} - c x_{i})^{2} .

Differentiation gives

\frac{d S}{d c} = - 2 \sum_{i} x_{i} (y_{i} - c x_{i}) = 0, \frac{d ^{2} S}{d c ^{2}} = 2 \sum_{i} x_{i}^{2} > 0.

Thus the unique minimum is

\overset{c}{^} = \frac{\sum _{i} x _{i} y _{i}}{\sum _{i} x _{i}^{2}} = \frac{x y}{xx} .

The overbars denote sample means over the four observations. Numerically,

\sum_{i} x_{i} y_{i} = 542.888740 m^{- 2}, \sum_{i} x_{i}^{2} = 3.00072844 \times 1 0^{7} m^{- 4},

so

\overset{c}{^} = 1.80919 \times 1 0^{- 5} m^{2}, \hat{G} = \frac{κ c ^}{m M L} = 9.59 \times 1 0^{- 11} m^{3} k g^{- 1} s^{- 2} .

The fixed

x_{i}

have negligible uncertainty. Assuming independent errors in the four

y_{i}

, propagation of uncertainty in quadrature gives

σ_{\overset{c}{^}}^{2} = \sum_{i} (\frac{\partial c ^}{\partial y _{i}})^{2} σ_{y}^{2} = \frac{σ _{y}^{2} \sum _{i} x _{i}^{2}}{( \sum _{i} x _{i}^{2} ) ^{2}} = \frac{σ _{y}^{2}}{\sum _{i} x _{i}^{2}} .

Hence

σ_{\overset{c}{^}} = 3.65 \times 1 0^{- 7} m^{2}, \overset{c}{^} = (1.809 \pm 0.037) \times 1 0^{- 5} m^{2} .

If the period measurement is independent of the angle measurements, including its calibration uncertainty gives

\frac{σ _{G}}{G ^} = (\frac{σ _{\overset{c}{^}}}{c ^})^{2} + (\frac{σ _{κ}}{κ})^{2} = 0.0314, \hat{G} = (9.59 \pm 0.30) \times 1 0^{- 11} m^{3} k g^{- 1} s^{- 2} .

This last uncertainty includes the stated random measurement errors only. Whether the model describes the data must be checked before treating it as a reliable measurement of the gravitational constant.

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