 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/section-a/1/solution

g (r) = - GM (< r) r / r^{2}

. There is no enclosed mass at the cavity boundary, whereas the entire shell has mass

M = \frac{4 π ρ}{3} [(2 a)^{3} - a^{3}] = \frac{28 π}{3} ρ a^{3}

. Thus

g (a) = 0, g (2 a) = - \frac{7 π Gρ a}{3} r, g (4 a) = - \frac{7 π Gρ a}{12} r .

Both nonzero fields point towards the centre. The field is continuous at each boundary; there is no surface mass sheet.

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