The rotational velocity at the equator is about 1000mph. (That’s how Concorde could arrive “ahead of the time” going west).
The escape velocity is about 16000mph and that gives you zero G. (Orbital radius is an insignificant increment over the earth’s radius).
So, why aren’t we 1/16 lighter if weighed on the equator?
If this is true, at what latitude is the MTOW specified?
Although I forgot how it is calculated I remember thatyou are only 0.5 percent lighter on the equator than in our latitudes. And since it only changes the weight but not the mass it cannot replace my diet …
Today it’s only MTOM, right? Haven’t seen MTO*W* for some time.
So, why aren’t we 1/16 lighter if weighed on the equator?
The theory is correct, but the numbers are wrong. Due to the centrifugal force you are about 0,3 percent lighter on the equator than on the pole.
Your horizontal velocity is at right angle (orthogonal) to the escape velocity, therefore it does not “count”.
But the effect you describe is actually exploited when launching satellites into (equatorial) orbits: For an object to stay in low earth orbit, it requires a miniumum horizontal velocity of around 8km/s. At that speed, the centrifugal force at an altitude of 200km cancels the gravitational force. Launching eastwards from the equator gives the spacecraft a free 0,4km/s (your 1600MPH) towards the orbital speed. Therefore many rocket launching sites are close to the equator (Florida, Kourou). And on east coasts, so that the launches can be performed towards uninhabited terrain/ocean.
AFAI remember that was one reason to select Florida for Cape Canaveral, becasue it is not far form the Equator, and all the SATURN Vs to the Moon were launched eastwards.
It is precisely the equatorial launches (which I do understand) which made me wonder about the apparent weight loss. Surely if the earth rotated 16x faster then we would all weigh nothing, because we would be moving at the escape velocity?
Mass is different of course and doesn’t vary.
Actually there will be 2nd order effects in the G force anyway, because the earth is not a sphere.
Mass is different of course and doesn’t vary.
The equation to fulfill is Centrifugal force = gravitational force. Since mass is a linear factor on both sides of this equation, it can be removed and the equation become centrifugal acceleration = gravitational acceleration.
Surely if the earth rotated 16x faster then we would all weigh nothing, because we would be moving at the escape velocity?
No. We (at least those of us standing on the equator) would only move at the speed required for a stable low earth orbit (approx 8km/s) and hover weightlessly above the surface, together with a lot of dirt and rocks and all the water from the oceans… Escape velocity (11,2 km/s) would require about 50 percent more speed.
We (at least those of us standing on the equator) would only move at the speed required for a stable low earth orbit (approx 8km/s) and hover weightlessly above the surface
OK, we agree on that.
I misunderstood “escape velocity”. I meant a stable low earth orbit.
The weather would be “interesting” but that’s not the point 
But, if the 16x increase would make us weightless, why doesn’t the present speed make us 1/16 lighter?
But, if the 16x increase would make us weightless, why doesn’t the present speed make us 1/16 lighter?
Because the dependency between rotational speed and centrifugal acceleration is not linear but quadratic.
OK… a square law relationship would make us 1/256 lighter then, i.e. ~0.4%, which is roughly what you posted earlier
One learns something every day. That also blows up my next business venture… damn.
This explains why when I come back from holiday somewhere sunny I always feel heavier…
