Question

The value of acceleration due to gravity at a place is 2% less than its value on Earths surface. Find the height of that place above Earths surface. (Given: Earths radius = 6400 km)

Answer 1

$$\begin{gathered} g\text{'}=g-2\%ofg \\ wheregisthegravityatearthsurface \\ g\text{'}isthegravityatheighthfromtheearthsurface. \\ g\text{'}=g-\frac{2}{100}g \\ g\text{'}=\frac{98}{100}g \\ g=\frac{GM}{R^2} \\ whereRistheradiusoftheearth \\ Misthemassoftheearth. \\ g\text{'}=\frac{GM}{(R+h{)}^2} \\ \frac{g}{g\text{'}}=\frac{\frac{GM}{R^2}}{\frac{GM}{(R+h{)}^2}}=\frac{(R+h{)}^2}{R^2} \\ \frac{g}{\frac{98}{100}g}=\frac{(R+h{)}^2}{R^2} \\ \frac{100}{98}{R}^2=(R+h{)}^2 \\ 1.01R=R+h \\ h=0.01R=0.01\times 6400=64km \end{gathered}$$