Gravitation: Edge Cases and Subtle Traps (1)

For a circular orbit, $KE = \frac{1}{2} \frac{GMm}{r}$ and $PE = -\frac{GMm}{r}$. Total energy $E_1 = -\frac{GMm}{2r} = -\frac{GMm}{4R}$.

KE increases by $50\%$: new $KE' = \frac{3}{2} \times \frac{1}{2}\frac{GMm}{r} = \frac{3GMm}{4r}$. PE is unchanged at the moment of the boost (position has not changed).

New total energy:

$E_2 < 0$, so the satellite remains bound. It now follows an elliptical orbit.

For a bound orbit, $E = -\frac{GMm}{2a}$ where $a$ is the semi-major axis:

The boost happened at $r = 2R$, which is the perigee (closest point) since speed increased there. Perigee distance $r_p = 2R$, apogee $r_a = 2a - r_p = 8R - 2R = 6R$.

Final answer: bound, semi-major axis $a = 4R$, perigee $= 2R$, apogee $= 6R$.

Why This Works

The escape condition is $E \geq 0$. As long as total mechanical energy stays negative, the satellite is bound. The instant after the boost, only KE changes — PE depends on position, which has not changed yet.

For elliptical orbits, the relation $E = -GMm/(2a)$ is the cleanest way to find $a$ without computing speeds at every point.

Alternative Method

You can use angular momentum conservation and vis-viva. New speed at $r = 2R$: $v' = \sqrt{1.5}\, v_{\text{circ}}$. Apply $v^2 = GM(2/r - 1/a)$ to find $a$ — same answer.