More practice questions

4. Q4. A satellite in a circular orbit of radius r has kinetic energy K. How much energy is needed to move it to a circular orbit of radius 2r?

   Show answerHide answer

   E = −K at radius r. At 2r the kinetic energy is K/2, so E = −K/2. The energy needed is −K/2 − (−K) = K/2.

5. Q5. Air drag removes 1.0 × 10^⁸ J from a satellite in a low circular orbit, and it settles into a slightly smaller circular orbit. How do its kinetic and potential energies change?

   Show answerHide answer

   E falls by 1.0 × 10^⁸ J. Since K = −E, the kinetic energy rises by 1.0 × 10^⁸ J; since U = 2E, the potential energy falls by 2.0 × 10^⁸ J. The satellite speeds up.

6. Q6. A 100 kg satellite moves on an ellipse whose nearest and farthest points are 7.0 × 10^⁶ m and 9.0 × 10^⁶ m from the Earth’s centre. Using E = −GM_Em/(2a), find its total energy and its speed at the nearest point. Take GM_E = 4.01 × 10^¹⁴ m^³ s^⁻².

   Show answerHide answer

   The semi-major axis is a = (7.0 + 9.0)/2 × 10^⁶ = 8.0 × 10^⁶ m, so E = −4.01 × 10^¹⁴ × 100/(1.6 × 10^⁷), about −2.51 × 10^⁹ J. At the nearest point, ½v^² − GM_E/r = −GM_E/(2a) gives v^² = GM_E(2/r − 1/a), about 6.44 × 10^⁷ m^² s^⁻², so v ≈ 8.03 km/s.