8
$\begingroup$

Would the Oberth effect (as explained as an answer to a gravity assist maneuver) apply to an earthbound partial orbit simulator in a strong gravity field, i.e. a playground swing?

Imagine a swing with a rocket engine mounted on the seat or strap or tire with the exhaust pointed tangentially (perpendicular to the support chains or rope) and in the direction of travel (in the case of two-chain swing, in a plane).

Would an impulse burn of the rocket impart more energy (and a higher arc) if fired at the point of maximum velocity in the pendulum arc, i.e. the point closest to ground, instead of the usual push at or shortly after the stationary peak at one end?

I would try this experimentally, but my wife would object to a rocket engine on the swing. (Daughter might be game, but the park managers would not). I could test this by positioning myself next to swing at the center and pushing when she hits the bottom of the arc, but it would be difficult (especially without instrumentation) to guarantee a push on her moving tush would be imparting the same impulse as the normal push at the peak of the arc.

The swing is one of my go-tos for explaining Newtonian physics to my daughter. (Does she like that? Not sure.)

$\endgroup$
4
  • 1
  • 1
    $\begingroup$ I can confirm this anecdotally. We have a very tall (10m) tire swing, and although I can definitely exert more force towards the stationary peak by using my legs, back, etc., when I squat on the ground underneath the height minimum and push only with my arms, even though I'm unquestionably using less force, the swing goes higher than otherwise. $\endgroup$ Commented May 5, 2021 at 16:01
  • $\begingroup$ WE'RE USING AN OBERTH POWERED SWING TO EELOO IN KERBAL SPACE PROGRAM, WHOOOOOOOOO $\endgroup$ Commented May 5, 2021 at 18:09
  • $\begingroup$ @SpaceLawyer that is fascinating. This seems to illustrate that kinetic energy really is related to the square of velocity, so adding additional speed at peak v really pays off. $\endgroup$
    – wistlo
    Commented May 5, 2021 at 21:53

1 Answer 1

7
$\begingroup$

Yes.

Firing a reaction engine imparts more kinetic energy to a vehicle when it's already at high speed. That's true of any reaction engine, any vehicle, any environment. All that matters is that the reaction mass is carried onboard, so its own kinetic energy is greater at high vehicle speed.

How much more kinetic energy depends on on the engine's impulse, the vehicle's mass, the difference between high speed and low speed, and enough other things that we won't find a published graph of rocket behavior that we can meaningfully extrapolate to a swingset. (Maybe a low-thrust ion drive.)

One could measure this experimentally with a pop bottle rocket (mounted radially with a 90 degree bend in the exhaust, to avoid ullage problems), a hand-operated radio-controlled solenoid, and a tripod-mounted videocamera to measure the speed. For a measurable difference, one might have to discharge the propellant all at once rather than over successive swings, in which case a fuse and a hobby rocket motor would work too.

I'd try this in my back yard now, but the snowstorm means that someone in gentler climes may beat me to it.


You might be able to impart a repeatable tush-push if you hold something that compresses a spring (a slice of pool noodle?) against an end stop, which in turn pushes the tush. Ease off when the end stop is impacted, to get a constant force. Don't vary the duration that the end stop is in contact, to get a constant duration.

But that still isn't quite the same thing as onboard reaction mass.

$\endgroup$
1
  • 2
    $\begingroup$ Might be worth tossing in the delta-momentum and delta-KE equations to make it clear why a fixed exhaust mass & speed (in the rocket's rest frame) leads to different delta-KE at different vehicle speeds. $\endgroup$ Commented Feb 11, 2021 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.