# Energy to "nudge" a planet to a smaller orbit

Suppose we wanted to move a mass-$$m$$ planet to a much smaller new orbit around our mass-$$M$$ Sun. We'll assume the initial and final orbits are circles of radii $$r_i,\,r_f$$ with $$r_f\ll r_i$$, so GPE is lost. I seek a first-order approximation of the energy needed, with the intuition one can lightly "nudge" an object to destabilize its orbit, so it moves to lower GPE.

My idea is to follow a path of length $$\sim r_i$$ between the orbits by mimicking an increase in solar gravity with a force $$\sim GMmr_f^{-2}$$ in a suitable direction, doing work $$\sim GMmr_ir_f^{-2}$$. This is the GPE loss $$\sim GMmr_i^{-1}$$ times $$(r_i/r_f)^2$$. Moving Neptune to an orbit the size of Earth's, the latter factor $$\approx10^{-3}$$.

Am I on the right lines, or out by a power of $$r_i/r_f$$? It's also possible the planetary radius $$A$$ is relevant for reasons I haven't identified, if only because it naturally occurs in e.g. the torque due to tidal acceleration, which introduces a $$(A/r)^5$$ factor with $$r_f\le r\le r_i$$.

Edit: the first answer here has pointed out the planet's orbital angular momentum contributes a kinetic energy term that needs to increase by $$\sim\tfrac12GMmr_i^{-1}$$ because$$L=rmv,\,\tfrac{mv^2}{r}=\tfrac{GMm}{r^2}\implies\tfrac{L^2}{2mr^2}\sim\tfrac12mv^2\sim\tfrac{GMm}{2r}.$$But an object knocked off a high shelf on Earth manages to gain plenty of KE it was never gifted, due to GPE conversion. I'd appreciate an answer that addresses in detail whether an analogous trick with orbits is possible, or whether the velocity-acceleration misalignment makes them so different to falling that there's no way to significantly reduce the required input.

• "an object knocked off a high shelf on Earth manages to gain plenty of KE it was never gifted". It was "gifted" when it was lifted up onto the shelf. Aug 14 at 16:20
• @OrganicMarble By "gifted" I mean "by anyone seeking to relocate the object with their own energy source". You don't need to explain $T+V$ energy conservation to me.
– J.G.
Aug 14 at 16:40
• Isn't this just a standard Hohmann transfer? Aug 16 at 0:12
• What about an object knocked off a shelf on the ISS ? Surely that it a better analogy, since the whole point of a planet orbiting any star is that the planet is in continuous free-fall? Aug 16 at 8:40

The relevant quantity to consider here is the specific orbital energy:

The specific orbital energy $$\epsilon$$ (or vis-viva energy) of two orbiting bodies is the constant sum of their mutual potential energy ($$\epsilon_p$$) and their total kinetic energy ($$\epsilon_k$$), divided by the reduced mass.

\begin{align} \epsilon &= \epsilon_k + \epsilon_p\\ &= \frac{v^2}{2} - \frac{\mu}{r} = -\frac{\mu}{2a} \end{align}

where

• $$v$$ is the relative orbital speed;
• $$r$$ is the orbital distance between the bodies;
• $$\mu = G(m_1+m_2)$$ is the sum of the standard gravitational parameters of the bodies;
• $$a$$ is the semi-major axis.

For an elliptic orbit the specific orbital energy is the negative of the additional energy required to accelerate a mass of one kilogram to escape velocity (parabolic orbit).

A body that's gravitationally bound has $$\epsilon<0$$ and if you increase its specific orbital energy to zero it will become unbound and either escape the system or fall into the primary, depending on the direction it's heading.

For a circular orbit, $$r = a$$, and $$\epsilon_k$$ and $$\epsilon_p$$ are both constant. So $$\epsilon = \epsilon_p/2 = -\epsilon_k$$

Rearranging the earlier equation, we get the celebrated vis-viva equation: $$v^2 = \mu\left(\frac{2}{r} -\frac{1}{a}\right)$$

We can use this equation to calculate the speed changes necessary to move from one circular orbit to another. The simplest way to do that is the Hohmann transfer orbit. It's also the transfer which has the minimum energy requirements. (In the real world of non-circular, non-Keplerian orbits, there can be lower energy solutions). The Hohmann transfer uses (half of) an elliptic trajectory with a major axis equal to the sum of the radii of the two circular orbits: At the start of the manoeuvre, you do a tangential burn which changes the orbital speed and hence the orbital energy. At the other end, you do another tangential burn to circularize the orbit... unless you want to loop back to your starting point.

To drop from the orbit of Neptune (~30 au) to Earth orbit, you need to shed most of your orbital speed. That's not a major ordeal because mean orbital speed is fairly sedate at that distance. However, when you get to 1 au your speed will be huge, and it takes a lot of delta-vee to circularize your orbit.

It might be more intuitive to consider the reverse manoeuvre; the energy requirements are identical because gravitation is a conservative force. To send a body from Earth orbit to Neptune takes a lot of energy. Your almost kicking it out of the system, and then there's the additional energy needed to circularize the orbit when you reach your destination.

Here are the relevant numbers, calculated in Python, mostly using the vis-viva equation.

Hohmann transfer
Earth
Period 365.250 d = 1.000 y
Speed 29784.878 m/s
Specific energy -443569480.467 (m/s)²
Neptune
Period 60160.792 d = 164.711 y
Speed 5433.605 m/s
Specific energy -14762032.097 (m/s)²
Hohmann transfer ellipse
Period 22340.642 d = 61.165 y
(The transfer time is half the period).
Specific energy -28573147.290 (m/s)²
Transfer from Neptune
Speed 1379.070 m/s
delta_v -4054.535 m/s
Transfer to Earth
Speed 41438.287 m/s
delta_v -11653.409 m/s

Here's a live version of the Python script, running on the SageMathCell server.

At the start of the transfer, you have to shed ~75% of your current orbital speed. And when you reach 1 au, your speed is almost 40% higher than Earth's so you have a lot of kinetic energy you need to shed, otherwise you'll loop back to Neptune.

• How does this compare with the expense estimate in the other answer?
– J.G.
Aug 16 at 10:58
• @J.G. Mine's a little smaller. ;) SE's $GMmr_i^{-1}/2$ is just the negative of the specific orbital energy of Earth, times $m$, the payload mass. My table gives that as ~443 million joules/kg. But we can reduce that by Neptune's $-\epsilon$ of ~14.7 MJ/kg to get the actual specific energy required, which is ~428.8 MJ/kg. (You get the same number if you add up the absolute values of the specific kinetic energy changes at the two burns). Sorry, I should've included that subtraction in my program & table. I'll update it tomorrow. Aug 16 at 12:46
• I guess I should also to mention en.wikipedia.org/wiki/Delta-v#Delta-v_budgets "It is not possible to determine delta-v requirements from conservation of energy by considering only the total energy of the vehicle in the initial and final orbits since energy is carried away in the exhaust" Aug 16 at 12:52
• OK. It looks like there's no way to turn lost GPE into angular-momentum KE to make a significant "nudging" saving.
– J.G.
Aug 16 at 12:55
• @J.G. Right. The vis-viva relation means that GPE gets converted into KE, which you then need to get rid of when you reach your destination. OTOH, you can save some energy if you don't mind ending up in an eccentric orbit. Aug 16 at 13:14

A problem with this analysis is that increasing the radial force preserves angular momentum, while two orbits of different radius do not have the same angular momentum.

Your process does thus not end up with a lower circular orbit, but rather a hyperbolic orbit with low perihelion. These are different enough that claiming one is an approximation for the other is not obvious, or indeed, not even correct. (Needing a "fudge factor" of $$(r_i/r_f)^2$$ should give a strong hint that something is off)

The condition $$r_f\ll r_i$$ can be used more efficiently by observing that the orbital energy of the lower orbit is much larger (but negative) than the higher orbit $$|\epsilon_f| \ll |\epsilon_i|$$. So your first order approximation can simply be $$\epsilon_i$$, ignoring $$\epsilon_f$$ completely.

$$\epsilon_f - \epsilon_i \approx - \epsilon_i =GMmr_i^{-1}/2$$

• So is there no way not to need energy comparable to the lost GPE? That's not only unfortunate but surprising. After all, it can take very little energy to initiate a fall on Earth. (Of course, in that case velocity is approximately parallel to acceleration, whereas for orbits it's approximately perpendicular, but you'd still think that since we want an overall energy reduction it should be cheap to push the planet, say, sideways to address the angular momentum concern.)
– J.G.
Aug 14 at 11:10
• (I realize you're estimating the increase in $L^2/(2mr^2)$, but my hope was this would increase because of the GPE reduction, rather than our input.)
– J.G.
Aug 14 at 11:24