Question

A planet has twice the mass of earth and of identical size. What will be the height above the surface of the planet where its acceleration due to gravity reduces by 36% of its surface?

Answer 1

$$\begin{gathered} Gravityonthesurfaceg=\frac{2GM}{R^2}=2\times 9.8=19.6m/s^2 \\ whereMisthemassoftheearthandristheradiusoftheearth. \\ Histheheightwheregravityisreducedby36\%ofg. \\ g\text{'}=g(1-\frac{2h}{R}) \\ g\text{'}=g-36\%ofg \\ g\text{'}=g-\frac{36}{100}g \\ g\text{'}=\frac{64}{100}g \\ g(1-\frac{2h}{R})=\frac{64}{100}g \\ 1-\frac{64}{100}=\frac{2h}{R} \\ \frac{36}{100}=\frac{2h}{R} \\ h=\frac{18}{100}R \\ Atheightequaltothe18\%ofRgravityisreduecdby36\%ofg. \end{gathered}$$