 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/section-a/a2/solution

v^{2} / r = GM / r^{2}

GM = g R^{2}

. The orbital radius is

r = R + h = 7.4 \times 1 0^{6}

m, so

v = \frac{g R ^{2}}{R + h} = \frac{10 ( 6.4 \times 1 0 ^{6} ) ^{2}}{7.4 \times 1 0 ^{6}} ≃ 7.44 \times 1 0^{3} m s^{- 1} .

This is about 7.4 km/s, under the circular-orbit and spherical-Earth approximation. Using surface

g

unchanged at orbital height would overestimate the centripetal acceleration; its role here is to determine

GM

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