Ranking gravitational forces

Problem

Four pairs of bodies (with $G = 6{,}67\cdot 10^{-11}$ N m$^2$/kg$^2$): (A) $M = 10^{24}$ kg, $m = 1$ kg, $r = R_T = 6{,}37\cdot 10^6$ m; (B) $M = 10^{24}$ kg, $m = 1$ kg, $r = 2R_T$; (C) $M = 10^{30}$ kg, $m = 1$ kg, $r = 10^9$ m; (D) $M = 10^{30}$ kg, $m = 1$ kg, $r = 10^{11}$ m. Rank the gravitational forces from smallest to largest.

Solution

Applying the law of universal gravitation to each pair: $F = G\frac{M m}{r^2}$ (A) $r^2 = (6{,}37\cdot 10^6)^2 = 4{,}06\cdot 10^{13}$ m$^2$: $F_A = \frac{6{,}67\cdot 10^{-11}\cdot 10^{24}\cdot 1}{4{,}06\cdot 10^{13}} \approx 1{,}64~\text{N}$ (B) same $M$ but double $r$, so a force $1/4$ of A’s: $F_B = \frac{F_A}{4} \approx 0{,}41~\text{N}$ (C) $r^2 = (10^9)^2 = 10^{18}$ m$^2$: $F_C = \frac{6{,}67\cdot 10^{-11}\cdot 10^{30}\cdot 1}{10^{18}} \approx 66{,}7~\text{N}$ (D) same $M$ but $r$ 100 times greater, so a force $1/10^4$ of C’s: $F_D = \frac{6{,}67\cdot 10^{-11}\cdot 10^{30}\cdot 1}{10^{22}} \approx 6{,}67\cdot 10^{-3}~\text{N}$ Ranking from smallest to largest value: $F_D \approx 0{,}0067$ N, $F_B \approx 0{,}41$ N, $F_A \approx 1{,}64$ N, $F_C \approx 66{,}7$ N. $\ev{D < B < A < C}$

Links

Topics: Gravitation Concepts: Law of universal gravitation Skills: Ranking reasoning