Gravitation: Speed-Solving Techniques (2)

Question

A satellite orbits Earth at height $h = R$ above the surface, where $R$ is Earth’s radius. Find its orbital speed and time period in terms of $g$ (surface gravity) and $R$. Use the trick that lets you skip $G$ and $M$ entirely.

Solution — Step by Step

Why This Works

The substitution $GM = gR^2$ converts every orbital-mechanics expression into one involving only $g$ and $R$ — both of which you know from class 9. JEE problems almost always give you $g$ or expect you to use $g = 10$ m/s$^2$, so this saves time and reduces the chance of arithmetic errors with $G$.

Alternative Method

Use Kepler’s third law: $T^2 \propto r^3$. Compare with a satellite at $r = R$ (skimming the surface, period $T_0 = 2\pi\sqrt{R/g}$). Then $T = T_0 \cdot (2)^{3/2} = 2\pi\sqrt{R/g} \cdot 2\sqrt{2} = 4\pi\sqrt{2R/g}$. Same answer in two lines.

Final answer: $v_o = \sqrt{gR/2}$ and $T = 4\pi\sqrt{2R/g}$.