Ranking orbital speeds

Problem

A probe orbits a planet of mass $M = 5\cdot 10^{24}$ kg at distances: (A) $r = R = 6\cdot 10^6$ m; (B) $r = 2R$; (C) $r = 5R$; (D) $r = 10R$. Recalling that $v_\text{orb} = \sqrt{GM/r}$, rank the orbital speeds from highest to lowest.

Solution

The orbital speed depends only on the distance $r$: $v_\text{orb} = \sqrt{\frac{GM}{r}}$ It is a decreasing function of $r$: the wider the orbit, the slower the probe. So it suffices to order by increasing radius. With $GM = 6{,}67\cdot 10^{-11}\cdot 5\cdot 10^{24} = 3{,}34\cdot 10^{14}$ m$^3$/s$^2$: $v_A = \sqrt{\frac{3{,}34\cdot 10^{14}}{6\cdot 10^6}} \approx 7{,}46~\text{km/s}$ $v_B = \sqrt{\frac{3{,}34\cdot 10^{14}}{1{,}2\cdot 10^7}} \approx 5{,}27~\text{km/s}$ $v_C = \sqrt{\frac{3{,}34\cdot 10^{14}}{3\cdot 10^7}} \approx 3{,}33~\text{km/s}$ $v_D = \sqrt{\frac{3{,}34\cdot 10^{14}}{6\cdot 10^7}} \approx 2{,}36~\text{km/s}$ The smallest radius (A) gives the highest speed, the largest (D) the lowest. $\ev{A > B > C > D}$

Links

Topics: Gravitation Concepts: Orbits · Uniform circular motion Skills: Ranking reasoning