Sun, Earth and a car — comparing gravitational forces

Problem

Sun, Earth and a car. Estimate the Sun-Earth gravitational force ($M_S \approx 2\cdot 10^{30}$ kg, $M_T \approx 6\cdot 10^{24}$ kg, $r \approx 1{,}5\cdot 10^{11}$ m). Compare it with the force due to a $1{,}5$ t car placed $1$ m from a $70$ kg person.

Solution

Sun-Earth force. Law of universal gravitation $F = G\,\dfrac{M_S M_T}{r^2}$: $F_\text{S-T} = 6{,}67\cdot 10^{-11}\cdot \frac{2\cdot 10^{30}\cdot 6\cdot 10^{24}}{(1{,}5\cdot 10^{11})^2} = 6{,}67\cdot 10^{-11}\cdot \frac{1{,}2\cdot 10^{55}}{2{,}25\cdot 10^{22}} \approx 3{,}6\cdot 10^{22}\ \text{N}$

Car-person force. With $m_1 = 1500$ kg, $m_2 = 70$ kg, $d = 1$ m: $F_\text{auto} = 6{,}67\cdot 10^{-11}\cdot \frac{1500 \cdot 70}{1^2} = 6{,}67\cdot 10^{-11}\cdot 1{,}05\cdot 10^5 \approx 7\cdot 10^{-6}\ \text{N}$

Comparison. $\frac{F_\text{S-T}}{F_\text{auto}} = \frac{3{,}6\cdot 10^{22}}{7\cdot 10^{-6}} \approx 5\cdot 10^{27} \sim 10^{28}$

The Sun-Earth force is about $10^{28}$ times larger. Gravity between everyday objects (cars, people) is almost imperceptible ($\sim 10^{-6}$ N, less than the weight of a gnat), while at the astronomical scale it dominates completely.

$\ev{F_\text{S-T} \approx 3{,}6\cdot 10^{22}\ \text{N};\quad F_\text{auto} \approx 7\cdot 10^{-6}\ \text{N};\quad \text{ratio} \sim 10^{28}}$

Links

Topics: Gravitation Concepts: Law of universal gravitation Skills: Fermi estimates