## Tuesday, June 16, 2015

### Easier to do push-ups in a high flying airplane

A few days ago, I was in an airplane flying at an altitude of over 32,000 feet or 10,000 meters. It was a 13-hour flight. To make this journey a little more comfortable, I did some stretching and, with flight attendants' approval, took a tiny corner to do push-ups. It seems to be easier to do push-ups in this high flying airplane. I did 66 without too much effort, almost tying my record set on ground 20+ years ago.

I'm curious to see how much easier to do this exercise at this altitude. So here's my calculation. If I simplify the calculation by assuming all the mass of the earth to be at the center of the earth (as opposed to integrating the mass along the 6371 km radius), Newtonian universal gravitation between me and the earth is reduced by about 3 thousandths at 10000 meters altitude:

```F0 = G (mmemearth)/(6371000)^2

F10k = G (mmemearth)/(6371000+10000)^2
```

where F0 is the force between me and the earth when I'm at sea level, and F10k the force when I'm at 10,000 meters altitude. Calculation of the percent change can omit the constant G and both masses m's:

```1/(6371000)^2 - 1/(6371000+10000)^2
----------------------------------- = .00313189770433034782 i.e. 0.313%
1/(6371000)^2
```

For a person weighing 70 kg or 154 pounds, that translates to 0.22 kg or about half a pound.

What's more, since the plane travels at the speed of 900 km (560 miles) per hour on this inter-continental flight, the centrifugal force [note] additionally reduces a 70 kg person by 686 newtons or 700 grams:

```Fc10k = 70000 * (900000/3600)^2/(6371000+10000) = 685.62921172230057984641 newtons
or 700 grams or 1.54 pounds
```

So in total, the body weight is effectively reduced by about 0.5 + 1.5 = 2 pounds in the plane traveling at 900 km/hour at 10000 meters altitude.

________________
[note] Centrifugal force is imaginary in Newtonian mechanics and is used as a convenient expedient. A physicist may choose to do calculation in Lagrangian mechanics.