Question bank — Specific orbital energy ε = −GM - 2a

False. $\epsilon$ is energy per unit mass, so $m$ cancels; a feather and a boulder in the identical orbit share one value $-GM/2a$.

False. Speed alone is meaningless — it is largest at perigee and smallest at apogee of the same orbit. Only the sum $\frac{v^2}{2}-\frac{GM}{r}$ (which is $\epsilon$) ranks orbits by energy.

False. The negative sign is a convention: PE is defined zero at infinity, so any finite $r$ sits below that reference. Nothing was lost — it just means work is needed to climb out to infinity.

True. A circle has $r_p=r_a=r$, so $a=\frac{1}{2}(r_p+r_a)=r$. This is the only case where "$a$ is the radius" is correct.

False. From $v=\sqrt{GM/r}$, larger $r$ gives smaller $v$. It has more total energy but stores it as height (PE), leaving less speed (KE).

False. $\epsilon =0$ is a parabola ($a\to \infty$, an open escape trajectory), not a closed circle. The body coasts to infinity arriving with zero speed.

True. For $\epsilon >0$, solving $\epsilon =-GM/2a$ gives $a=-GM/2\epsilon <0$. Negative $a$ is the formal fingerprint of an unbound path (see Conic sections in orbits).

False. Each one changes continuously (they trade back and forth); only their sum $\epsilon$ stays fixed.

True. $|\epsilon |=GM/2a$ is inversely proportional to $a$, so $a\to 2a$ makes $|\epsilon |$ half as large (i.e. $\epsilon$ moves closer to zero, less tightly bound).

The error is confusing $a$ (a fixed geometric property of the whole ellipse) with $r$ (the moving position). $a$ never changes during free flight, so $\epsilon$ is a single constant, independent of position.

Escape speed reaches infinity with zero leftover speed, not any orbit. It comes from setting $\epsilon =0$: $v_{\text{esc}}=\sqrt{2GM/r}$ (see Escape velocity). A stable circular orbit needs less speed, $\sqrt{GM/r}$.

A circle sets $r=a$ (not $0$), giving $v^2=GM(\frac{2}{a}-\frac{1}{a})=GM/a$. Setting $r=0$ would mean sitting at the centre — physically impossible and gives infinity.

Gravity is conservative, so total energy is conserved — it does not decrease. Falling inward converts PE into KE, but the sum $\epsilon$ is unchanged.

Only a prograde burn (along the velocity) at perigee raises apogee. A radial or retrograde burn can raise speed momentarily yet lower or reshape the orbit. Direction, not just speed change, determines the new $a$.

Kinetic energy per mass $\frac{1}{2}{v}^2$ is always $\ge 0$. The whole $\epsilon$ is negative because the negative PE term $-GM/r$ outweighs the positive KE term; neither piece alone is negative in the wrong way.

The derivation shows $\epsilon =-GM/(r_p+r_a)=-GM/2a$: only the sum of the extreme distances enters, and that sum equals $2a$ regardless of how squashed or round the ellipse is. Shape controls how energy is split between KE and PE, not the total.

So the satellite's own mass cancels, letting one number describe the orbit itself — the same reason all objects fall at the same rate. It makes $\epsilon$ a property of the trajectory, not of the payload.

We used $h=r_p{v}_p=r_a{v}_a$ (from Conservation of angular momentum) only as a tool to eliminate the unknown speeds. Once the speeds are gone, geometry remains and the result depends purely on $r_p+r_a=2a$.

As it falls inward, PE ($-GM/r$) drops (becomes more negative), and since $\epsilon$ is fixed, KE must rise, so $v$ increases. The reverse happens climbing to apogee — a continuous energy trade.

Because $\epsilon$ is constant, equating $\frac{v^2}{2}-\frac{GM}{r}=-\frac{GM}{2a}$ links $v$ to just $r$ and $a$. The angular position is already baked into $r$, so no separate angle is required.

$\epsilon =0$ means the total energy just barely reaches the zero reference at infinity — the body arrives at $r\to \infty$ with $v\to 0$. Any less energy ($\epsilon <0$) traps it; any more ($\epsilon >0$) leaves speed to spare (see Escape velocity).

$\epsilon =-GM/2a\to 0$. This is the parabolic escape boundary: the orbit is no longer closed and the body just reaches infinity with zero speed.

It signals a hyperbolic (unbound) flyby with $\epsilon >0$. The "$a$" is now the distance parameter of the hyperbola's geometry, not a loop size, and no apogee exists.

$v^2=GM(\frac{2}{r}-\frac{1}{a})\to \infty$, so speed blows up. Physically the orbit would have to pass through the centre — impossible for a real orbit around a finite body, which is why perigee stays above the surface.

Yes. Any ellipse (however eccentric) with the same semi-major axis $a$ has identical $\epsilon =-GM/2a$ — a nearly circular orbit and a highly elongated one can be energy-twins. This underlies why Kepler's Third Law (period depends only on $a$) works.

Nowhere — $\epsilon$ is the same at every point of that single ellipse. Only KE and PE individually vary; the transfer's total specific energy is one fixed number set by its own $a=(r_p+r_a)/2$.

Not necessarily. Doubling $v$ quadruples KE; if the new $\epsilon =\frac{v^2}{2}-\frac{GM}{r}$ becomes $\ge 0$, the orbit becomes parabolic or hyperbolic and the body escapes.