If a body has a greater than one thrust to weight ratio, then is the centrifugal force acting perpendicular to the Earth's gravitational pull that allows it to fly ?

Or are there other forces that I am not aware of being at play here ?

Further Clarification

So, when a body is moving on the surface of the Earth, we can assume that it adheres to some semblance of circular motion.

Any body in circular motion will experience an outward force perpendicular to the direction of motion.


So, given enough thrust, can any body fly because of the centrifugal force created against Earth's gravitational force ?

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    $\begingroup$ I don't think I get what you are trying to say. Can you please clarify it? $\endgroup$ Jun 5 '16 at 17:50
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    $\begingroup$ It's basically a question clarifying the force acting on, say, a rocket or VTOL aircraft - they're confusing centrifugal force/centripetal force, and the idea of any other force counter-acting gravity $\endgroup$
    – Jon Story
    Jun 5 '16 at 19:21
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    $\begingroup$ I'm voting to close this question as off-topic because it belongs either to physics or space exploration. $\endgroup$
    – mins
    Jun 5 '16 at 20:45
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    $\begingroup$ The question title is an oxymoron. If the object is non-aerodynamic then it can't produce lift at all because lift is an aerodynamic force. I agree this is off-topic $\endgroup$
    – TomMcW
    Jun 5 '16 at 21:42
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    $\begingroup$ All upward acceleration is not necessarily lift. Lift is specifically an aerodynamic force. Rockets create no lift whatsoever. They accelerate upward purely due to thrust. $\endgroup$
    – TomMcW
    Jun 5 '16 at 21:55

Basically, you're asking 'Can you achieve orbital velocity in atmospheric conditions?'.

Orbital velocity is approximately equal to

$$v_o \approx \sqrt{\dfrac{GM}{r}}$$

which, if we fill it in at Wolfram.alpha, yields a velocity of almost 8km/s (yes, per second), or almost 18000mph, or a Mach number of 23.23.

The current speed record for sustained atmospheric flight is just shy of 2200mph (which, interestingly, stems from 1976 and was set by the SR-71 Blackbird). This means we're still a factor 8 short. Since drag is roughly proportional to the square of your velocity, this means we're a factor 64 short of thrust on a similarly designed airplane. But I think by that time you'll be running into all sorts of interesting exotic effects, not the least of which is that your airplane will desintegrate by the massive amount of compressive heating (basically, you're doing a 'sustained' orbital re-entry maneuver in a much thicker part of the atmosphere).

The answer is then, hell no. Not as long as Earth has an atmosphere.

Note: if you're asking about vacuum conditions, then the figure you're looking for is delta-V, not thrust-to-weight ratio. The delta-v of your craft must exceed aforementioned orbital velocity. Your T/W ratio doesn't come in to play at all.


It is not centrifugal force that allows a non-aerodynamic body not designed to be aerodynamic to achieve lift by harnessing the energy from thrust unless it is going so fast that it is in orbit.

If an object is moving fast enough, it would eventually get to the point where it would be in a low orbit; in that case I suppose you could say it is the centrifugal force keeping it at that height (orbiting), and of course, the lower it is the faster it must move, and the more thrust is required to counter the air resistance.

Typically when we talk about something moving through the air, the thing providing a constant height would be the aerodynamic force acting on it. Even a brick can fly aerodynamically by creating lift from striking air molecules if it has enough thrust to keep its speed high enough to do so.

It is still due to the air's Newtonian action-reaction on the object. Unless of course, we are talking about vertical flight, in which case the lift comes from the Newtonian action-reaction from the thrust, not the air.

I don't mean to exclude thrust vectors here. Of course, the angle of thrust will play a part even when there is Newtonian aerodynamic lift.

If you're talking about orbital properties, check this out in the space section of SE.

So, given enough thrust, can any body fly because of the centrifugal force created against Earth's gravitational force?

Yes, but it would be called orbiting, not flying.

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    $\begingroup$ I've updated my query. I think I'm using centrifugal force in the incorrect context. But could you tell me why my reasoning is wrong ? Thanks. $\endgroup$ Jun 5 '16 at 20:15
  • $\begingroup$ @SanjeetSuhag The more I gather on your question, it sounds like you're asking if the reason something that isn't designed to be aerodynamic can fly is because it is essentially in a low orbit. If you set the conditions so that it had zero lift by changing its angular orientation to the relative wind as it passes through the air, eventually, yes it would be in a high-air-resistance, low orbit if it were going at the correct speed for that orbital altitude, which would be very very fast if it were still in the lower atmosphere. $\endgroup$ Jun 5 '16 at 20:22
  • $\begingroup$ So, I'm incorrect in my usage of the circular motion in this context, right ? $\endgroup$ Jun 5 '16 at 20:35
  • $\begingroup$ @SanjeetSuhag No, not necessarily. Centrifugal force certainly keeps any object at a constant altitude at the right speed even with 0 aerodynamic lift. $\endgroup$ Jun 5 '16 at 20:38
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    $\begingroup$ "Yes, but it would be called orbiting, not flying": There is a big difference. Flying requires constant energy (thrust) to be converted into lift to fight gravity. Orbiting needs no energy, and no lift (Newton's 1st law). $\endgroup$
    – mins
    Jun 5 '16 at 20:39

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