Gravitation: Common Mistakes and Fixes (7)

Question

A satellite is in a circular orbit of radius $r$ around Earth. By what factor does the orbital speed change if the radius is doubled? Many students answer $\tfrac{1}{2}$. Why is that wrong, and what is the correct factor?

Solution — Step by Step

The correct factor is $\dfrac{1}{\sqrt{2}} \approx 0.707$.

Why This Works

Students mix up two different relationships. Linear momentum scales linearly with $r$ in some contexts, but orbital speed depends on $1/\sqrt{r}$ because gravitational force falls as $1/r^2$, not $1/r$. The square root sneaks in when you solve $v^2 \propto 1/r$.

Compare with time period: from Kepler’s third law, $T^2 \propto r^3$, so $T \to T \times 2^{3/2}$ when $r$ doubles. Different scaling for different quantities — write the formula every time.

Alternative Method

Use Kepler’s third law directly. $T \propto r^{3/2}$, so $T$ becomes $2\sqrt{2}\,T$. Speed $v = 2\pi r / T$, so

$v_{\text{new}} = \frac{2\pi (2r)}{2\sqrt{2}\,T} = \frac{v}{\sqrt{2}}$

Same factor.

Common Mistake

Treating “doubled” linearly without checking the exponent. The fix is a 10-second habit: write $v = \sqrt{GM/r}$ before answering. If $r \to nr$, then $v \to v/\sqrt{n}$. Done.