Radius of a planet

Problem

A planet has mass $M = 5\cdot 10^{23}$ kg and surface gravitational acceleration $g = 3{,}5$ m/s$^2$. Calculate the radius of the planet.

Solution

The surface gravitational acceleration is obtained from the law of gravitation by setting $r = R$: $g = \frac{G M}{R^2}$ Isolating the radius $R$: $R = \sqrt{\frac{G M}{g}}$ Calculating the product $G M = 6{,}67\cdot 10^{-11}\cdot 5\cdot 10^{23} = 3{,}34\cdot 10^{13}$ m$^3$/s$^2$. Substituting: $R = \sqrt{\frac{3{,}34\cdot 10^{13}}{3{,}5}} = \sqrt{9{,}53\cdot 10^{12}} \approx 3{,}09\cdot 10^6~\text{m}$ $\ev{R \approx 3{,}09\cdot 10^6~\text{m} \ (\approx 3090~\text{km})}$

Links

Topics: Gravitation Concepts: Law of universal gravitation · Weight force Skills: Symbolic setup