 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/14/a/ii/solution

d F = 0

in the relative total differential and use the two equal

1%

mass increases:

0 = 0.01 + 0.01 - 2 \frac{d r}{r} ⟹ \frac{d r}{r} = 0.01.

Thus the separation must increase by $1%$ . In this special case the result is also exact: the mass product increases by

1.0 1^{2}

, and preserving the inverse-square force requires

r_{new}^{2} = 1.0 1^{2} r^{2}

, hence

r_{new} = 1.01 r

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